Complex Numbers

lGonmilltuunm

A number in the form of a + ib, where a, b are real numbers and i = J-t is called a complex number. A complex _{number can also be defined as an ordered pair of real numbers a and b and may be} written as (a, b), where the first number denotes _{the real part and the second number denotes the} imaginary _{part. lf z = a + ib, then the real part of z is denoted by Re (z) and the imaginary part by} lm(z). A complex number is said to be purely real if lm(z) = 0, and is said to be purely imaginary if Re(z) = 0. The complex number 0 = 0 + i0 is both purely real and purely imaginary.

Two complex numbers _{are said to be equal if and only if their real parts and imaginary parts are} s e p a r a t e l y e q u a l i . e . a + i b = c + i d i m p l i e s a = c a n d b = d . H o w e v e r , t h e r e i s n o o t h e r r e l a t i o n between complex numbers that is of the type a + ib < (or >) c + id.

Remark:

r C l e a r l y i 2 = - 1 , i 3 = i 2 . i = - i , i a = 1 . I n g e n e r a l , i 4 n = 1 , 14n+1 _{- i , i 4 n * 2 = - 1 , i4n+3} - - i f o r a n

integer n.

Algebra of Gomplex Numbers, modulus and argument, triangular inequality, cube roots of unity.

A complex number z = x + iy, written as an ordered pair (x, y), can be represented by a point P whose Cartesian coordinates are (x, y) referred to axes OX and OY, usually called the real and the imaginary axes. The plane of OX and Oy is called the Argand diagram or the complex plane. Since the origin O lies on both OX and OY, the corresponding complex number z = 0 is both purely realand purely imaginary.

Modulus and Argument of a Complex Number:

we define modulus of the complex number _{z = x + iy to be the real number and denote it} by lzl. lt may be noted that lzl > 0 and lzl = 0 would imply that z = 0.

lf z = x + iy, then angle 0 given by tan 0 = I is said to be the argument or amplitude of the complex X

n u m b e r z a n d i s d e n o t e d b y a r g ( z ) o r a m p ( z ) . l n c a s e o f x = 0 ( w h e r e y + o ) , a r g ( z ) = + n l 2 o r - n l 2 ^ ' * y '

r

RSM-9 1 1 -P3-MA-ComPlex Number

d e p e n d i n g u p o n y > 0 o r y < 0 a n d t h e c o m p l e x n u m b e r i s c a l l e d p u r e l y i m a g i n a r y ' l f y = g ( w h e r e x+ 0), then arg(z) = 0 orn depending upon x > 0 orx < 0 and the complex number is called purely real. The argument of the complex number 0 is not defined'

We can define the argument of a complex number also as any value of the 0 which satisfies the

system

of

equations

_"o"

e =

l;fr;,

sin

0 =

ffi

x 2 + Y '

The argument of a complex number is not unique. lf 0 is a argument of a complex number, then 2nn + 0 (n integer) is also argument of z for various values of n. The value of 0 satisfying the inequality - n < 0 < n is called the principal vatue of the argument.

From figure 1, we can see that oP = Jx2 + y2 = lzl and lf 0 = ZPoM, then tanO = y/x' In other words fzl is the length of Op i.e. the distance of point zfrom the origin and arg(z) is the angle which OP makes with the Positive x-axis.

Trigonometric (or Potar)

form of a Complex Number:

Let OP = r, then x = r cos 0, and Y = r sin e

+ z=x + iy = r cos 0 + ir sin e = r ( cos 0 + i sin 0 ). This is known as Trigonometric (or polar) form of a Complex Number. Here we should take the principalvalue of 0' For general values of the argument

z=rlcos ( 2nn + 0) + isin ( 2nn + 0)l (where n is an integer)

Students should note that sometimes cos 0 + i sin 0 is, in short, written as cis(0)' Eulerrs formula: cos e + isin 0 = ei e'

Remark:

r Method of finding the principal value of the argument of a complex number z = x + iy

Step l: Find tan0 = quadrant.

and this gives the value of 0 in the first

Step ||: Find the quadrant in which z lies, with the he|p of sign of X x and y co-ordinates.

Step lll: Then argument of z according as z lies in quadrant

lllustration 1.

For z = tb - i, rina the principal value arg(z).

tvl

FI

wilf be 0, n - e, e - n, and -0 the first second, third or fourth

Solution: Here x = {3,

+ tano

=

l-I

- 1 I ? 0 = ; b y =

1 l

-=1 ./3 |

+ principal value of atg z= -L . (Since z lies in the fourth quadrant)

RSM-91 1-P3-MA-Complex Number

Unimodular Complex Nu mher:

A complex number z for which lzl = unimodular complex number. _{Since lzl} circle of radius 1 unit and centre (0, 0). lI lzl= 1 + z= cos 0 + i sin 0,

+ 1lz= (cos 0 + i sinefl = cos 0 - i sinO.

is'said to be 1, z lies on a

(a + ib) + (c+id) = (a + c) + i ( b+d) (a + ib) - (c+id) = (a - c) + i ( b - d) (a + ib) (c+id; = (ac - bd) + i ( ad + bc)

a + i b c . i d

Algebraic Operations with Complex Numbers:

I I I I Addition Subtraction Multiplication

Division (when at least one of c and d is non-zero)

_ (ac + bd) _ (bc - ad) c 2 + d 2 c 2 + d 2

Geometrical

meaning of Addition and Subtraction:

lel z1 =Xr + iYr and z2 = *, * iy2 be two complex numbers represented by the points P1(x1,y1) and Pz(xz, yz) respectively. By definition 21+22 should be represented by the point P(x1+xr, yt+ yz).This point P is the vertex which completes the parallelogram OP1PP2 with the line segments joining the origin with Pr and Pz as the adjacent sides

= lz1+ zzl= OP.

P(x1+x, _.yt+yz)

P'z('xz, -Yz\

Also by definition z,r - zz should be represented by the point (xr - xz, yr - yz).This point is the vertex Q which completes the parallelogram OPjaPl with the line segments joining _{the origin with Pr and P'2} (where the point P'2 repr€sents -22; the point -22 c?n be obtained by producing the directed line PzO by fength _{lz2) as the adjacent sides = lzt - z2l = OQ = PzPr.}

Remarks:

r In any triangle, sum of any two sides is greater than the third side and difference of any two sldes is l'ess than the third side; we have

l4l+lz2l>-14+zzl; here equality holds when arg(4122) = 0. llzrl)zrll3lzt- z2l; here equality holds when arg(4lz) = 0.

r In the parallelogram OP1PP2, the sum of the squares of its sides is equal to the sum of the squares of its diagonals ; i.e. OP2 + P2P? = OPI +P,P2 +PPl +P2O2

+l z, + zz 12 + | zt - zz f = 2( z'r 12 + | =, lt ).

( i ) (ii ₎

Geometrical

Meaning of Product and Division:

Letzr= 11(cos01+ isin01), z2-- r2(cos 0z + isinOz) be complex numbers represented by Q1 and Qz'

(i) Construction for the Point Representing the Productzlz2: Let L be the point on OX which represents unity, so that OL

= 1. Draw the triangle OQzP directly similar to the triangle OLQr making ZLOQr = 7Q2OP and ZOLQr = ZOQzP-Then point P represents the producl z1z2

Explanation:

Due to similar triangles

oP - oQz . that is oP =t? + op = rrrz

OQ, OL t,r 1

Also ZQzOP = /,LOQI = 0r 3 ZLOP = 0t + 0z

Since zp2 =r1r2 {cos(01 + 0z) + isin (01 + 0z) } , P represents z1z2 '

(ai) Construction for the Point Representing the Quotient4lz2: Draw the triangle OQrP directly similar to the triangle

oQ2L.

Then P represents the quotient zrlzz. Explanation:

From the rast constru " c t l o n ' oQn oP rt oP o e 2 = o L = _{k = 1} l - l t - |

Op = ]-11 = li1-l also IQ,oL = 0r ond IQaOP = 02 l z z l l z z l

:. IPOL = 0r - 0z

OP(z)

= 1(cos(01

- o, ) * isin(0.,

- or)\ = ?

l 2 t . 2 2 + number represented by P =1t . z2

Remark:

t l f z 1 = 1 1 ( C o s 0 r + i s i n 0 1 ) , a n d Z z = r 2 ( c o s 0 2 + i s i n 0 z ) , t h e n Z j z z = r . , r , e i ( e ' * e ' ) S n d

l t - t t ."(et-',).

H e n c e

_{l = . = r l = \ r z : l z t}

_{l l z r l}

a n d

_{H = ! = H , / r l + 0 .}

2 2 1 2 | ' 't _{e 2 l f 2 l z r l} t arg(z'12) = 0r + 0z= arg(zi + ary(7,2)

( z,\

I _{a r g l a | = 0r -02= arg(21)} - a r g ( 2 2 ) \ z z )

Square Root of a ComPIex Number:

Letzl =xr + iyr be the given complex number and we have to obtain its square root. Let x + iy = (xr + iyr)% = x' -f + 2ixy= Xr + iYr

s X 1 = x ' - y ' a n d y 1 = 2 x y = x'- 4=^, 4x' > x 2 -l z t l + \. v = , 1 Y l

2

_{| v r l}

l f y 1 > 0 = x = + _{Y = +} ( R e ( 2 , ) * |., l) 2 l f y l < 0 = x = + ' Y = T (ne(2.' _{)* | t, l)} 2 lllustration 2.

Find the square root of I - 1Si. Solution:

F;t;,

\ l 2,

F;t;;

\ 2 H e r e y = - 1 5 < 0 (

+ v 8 _ r s i = t

I

1 = + _

_ J '

_{(5 - 3i).} (i) ( i i )

An Alternate Method to find the square root of a complex number is as follows:

lf the imaginary part is not even then multiply and divide the given complex number by 2. e.g z= 8-15i, here imagin ary partis not even so write _{.=1ga- 30 i) and let a+ib-16- 30i.}

z

Now divide the numerical value of imaginary _{part of s + ib by 2 and let quotient be p and find} all possible two factors of the number P thus obtained _{and take that pair in which difference} of squares of the numbers _{is equal to the real part of a + ib. e.g. here numerical value of} lm(16 - 30 i) is 30 . Now 30 = 2 x 15. All possible way to express 15 as a product of two are 1 x 1 5 , 3 x 5 , e t c .

here 52 _{- 32 = 16 = Re (16 - 30i) so we will take 5, 3.}

Take i with the smaller or the greater _{factor according as the real part of a + ib is positive or} negative _{and if real part is zero then take equal factors of P and associate i with any one of} them. e.g. Re(16 _{- 30i) > 0, we will take i with 3. Now complete the square and write down} the square root of z.e.g.

== l.ya-30i1

=

Ilu,

.(gi)'

- 2x5x3i]

=

;ts-3i12

+ l; = *S(5-3i).

Solving Complex Equations:

Simple equations in z may be solved by putting _{z = a + ib in the equation and equating the real part} on the L'H.S _{with the real part on the R.H.S and the imaginary part on the L.H.S with the imaginary} part _{on the R.H.S.} (i ii)

I zrl-x,

2

trR"GJ]

, )

RSM-9 1 1 -P3-MA-Complex Number

Illustration 3.

Solve for z i.e., find all complex numbers z which satisfy 1zf - Ziz + 2c(1+ i) =0 where c is real. Solution:

Put z = a + ib then a? + b2 - 2ai + 26 + 2c + 2ci = 0 = ( a 2 + b2 + 26+2c)+ (2c- 2a) i= 0

= a' * b2 + 2b + 2C =0 and 2c - 2a = 0 + a = c. N o w b 2 + 2 b + ( c 2 + 2 c ) = 0 = b = - 1 4 1 - 2 c - c ' z S i n c e b i s r e a l , _{, 1 - 2 c - c 2 > 0 + c e} I - 1 - 1 2 , - 1 + ' l 2 l ' > z = c + i ( - 1 t ' h - z c - c 2 1

solution of the given equation does not exits _{for c e l-1 - \12, -1 +"'121.}

Conjugate of a ComPIex Number:

The conjugate of the complex number z = a + ib is defined as a - ib and is denoted byZ. In other words Z is the mirror image of z in the real axis'

lf z= a + ib, z + 2 = 2 a ( real), z - 2 = 2 i b ( i m a g i n a r y )

and z2 = ( a+ib)(a-ib) = a' + b2 (real '1= 1212 =lZl2 '

A l s o

R e

( 4 =

T , l m ( z )

=

T

Real axis i) (a) ii) ii0 (a)

Find real 0 such that real 3 + 2 i s i n Q . - r s 1 - 2 i s i n 0 (b) purely imaginary

Simptifv

-?9-

* -J-=----

- -J:

v""P"''

Ja -Ji

sJj -2Ji 2J1-Jj'

Find the sgure root of

-8 - 6i (b) x t . y 2 1 ( , y ) I

7 - 7

- i [ t - ,

) - z

Conjugate:

. c4:.

t lzl=l2l

t z+ 2=2Re(z) . z - 2 = 2 i l m ( z ) t lf z is purelY realz= 2 t ll z is purely imaginary z= -2

t

_{z Z : l z l 2 = lzl2}

a

RSI|-91 1 -P3-MA-€ompbx Number

J Z , t * 2 2 = Z t + Z z fn general, 4 + zz +.... + zn = 4 + 22 + .-.+ zn z t - z z = 4 - z z 2122 = \.22 f n general zlz2z3 ..-..2n :4-zz.zz -....1., I I

Explanation:

Let f(z) = oo * d1z + Q2z2 + asz3 +

complex number. Then

t dnzn, where Qo, a1, d2,, . ., an are real numbers and z is a

. zn = (z)"

/ \ / - \ I z t l 1 4 1

r I t - t I l z z ) \ z z )

r lf cr = f(z), theno =t(r)= f (Z), where a = f(z) is a function in complex variable with real coefficients. In otherwords if f(x + iy) = a + ib then f(x- iy) = a - ib.

f ( z ) = a o + a 1 z + a r ( 2 ) 2 + a r ( z ) 3 + ' . . + a n ( Z ) n = a o + a 1 z + a r z 2 + a 3 z 3 + . " + a n z n

Modulus:

r _{l z l = Q +} z = 0 + i 0

) _{lzt - z2 | denotes the distance} between ! and 22 .

t _{-lzl < Re(z) < lzl ; equality holds on right or on left side depending} upon z being purely positive real or negative real.

t _{-lzl < lmz < lzl; equality} holds on right side or on left side depending upon z being purely imaginary and above the real axes or below the real axes.

. _{lzl < lRe(z)l + llm(z)l} < .12121; equality holds on left side when z is purely imaginary or purely real and equality holds on right side _{when lRe(z)l} = llm(z)1.

t _lzl2 = 72 s lzp2l= lzillz2l ln general _{lz. 22 . . .. z"l = lzi lzrl . . .. lz^l} t _{l z " l = l z l n , n e l}

l = , 1

l = , 1

I _{l - l = r i}

l z z l l z z l

. lzt+zzlslzl+ lz2l=lz1+22*

. . . * z n l 3 l z l + l z r l + . . . +

l z n l ;

e q u a l i t y

h o l d s

i f o r i g i n ,

a n d

t h e

points represented bf 21,22,zs, ...,2n ate collinear and are on the same side of the origin. r _{l z t - z " l 2 l l z l - l z " l l ;} e q u a l i t y h o l d s w h e n a r g ( 4 1 2 2 ) = 0 i.e.originandpointsrepresented

bf 21, z2 are collinear and are on the same side of the origin.

I _{l z r + z 2 l 2 = (zr + zz) (2t + 2z)=lzi2 + lzrl'+ ztZz+ zz2t= lzl2 + lzzl2} + 2Re(zt7z) I _{lzt - zzl2} = (zt - zr) (2 t - 2z) = lzi' + lzrl' - ztZz - zz2 t= lziz + lzzl' - 2Re(a7z)

Argument:

t arg(zp2) = 0r + 02= arg(7,1)'+ arg(zz) r arg (z1lz2) = 0r - 0z = arg(z) - arg(22) r arg (zn) = n arg(z), n el

IVofe.'

r In the above result 0r + 0z or 0r - 0z are not necessarily the principal values of the argument of corresponding complex numbers. e.g arg(zn) = n arg(z) only shows that one of the argument of zn is equal to n arg(z) (if we consider arg(z) in the principle range)!!

(i) arg(z) = 0, n =z isa purely real number >z=2'

(ii) arg(z) = nl2, -nl2 + z is a purely imaginary number = z= -7 ' Note that the property of argument is the same as the property of logarithm

Comptex Numbers Represented By Vectors :

It can be easily seen that multiplication by real numbers of a complex humber is subjected to the same rule as the vectors. The addition or the subtraction of two complex numbers is also the same as the addition or the subtraction of two vectors. This fact is fundamental in theory and very useful in practice.

It should be noticed that if a number z is represented by points P and OP by a vectorOP, then lzl is the length OP and arg(z) is the angle which the directed line OP makes with directed OX. '

pfease _{note that if z = x+ iy and P is the point (x, y), a one-to-one} correspondence exists between the number z and any of the following : (i) the point p; (ii) the displacement OP; (iii) the vector ( or directed length _)OP.

Any one of these three things may therefore be said to represent z, or to be represented by z.

lllustration 4. /

For any three complex numbers 21, 22 ottd ft, prove that 4lm(Z2zs) + z2lm(Zrz) + 4lm(4 z) = 0. Solution: t As lm(z) = *e-z) = zlm(zrzs) + z2lm(4zi + 4lm(Zz2) 1 = ! @, (zrzs - z,r2r) + zz (zszr - zs4) + zt (42, - z.a)) 2 i " " " 1 ' l ) = o ' = )r(zFzzs - z4z4 + z2z3z1 - zzzs4 + zs4z2 - 23212 ttlustration 5.

Consider a quadratic equation a/ +bz +c =O where a, b, c, are complex numbers- Find the condition that the equation has

0 one purely imaginary root (iit) two purelY imaginary roots

(it) one purely realroot (iv) two purely realroots

Solution:

(i) Let z1( purely imaginary) be a root of the given equation > zt = --4

and az'tz + [71 +s = Q ... (1)

= a z i a 6 r U " = 0 =

^ 4 ' * 6 4 + c = o

> 62.2 -62, +6 = 0 (as 4 = -21) "'(2) Now (1)and (2)musthave one common root'

RSM-91 1 -P3-MA-Complex Number

. , z , ' 2 1 1

- 2 _ ( o e + c o ) (ca-ac)2

_---=---=_---=---..-( b c + c b ) _---=---=_---=---..-( c a - a c ) - _---=---=_---=---..-( a b + e b ) _ ( a o + a o ) _{( a 6 + a b ) 2} + (oe+ c6)(a6 + eb) + (ca -ac)2 = o

(ii) Let z1( purely real ) be a root of the given equation . = 4 = Z

a n d a z 1 2 + [2., +g = Q _{. . . ( 1 )}

---;-= a z r ' + b z , , + c ---;-= 0 + az,,t + b 4 + c = 0

+ a 2 . , 2 + b z 1 + c = 0 _{( a s 4=z.1)} . . . ( 2 ) Now (1) and (2) must have one common root.

zt2 z1 1

(oe

- 5c) -(ac - ac) (a6

- ao)

Required condition is (nc - c5)(a6 - aO) + (ca - ae)2 = O (iii) Let 21 and z2 be two purely imaginary roots then

Zt = -Zt, ZZ = -ZZ

N o w a z 2 + [ 2 + s = g . . . ( 1 )

= a z 2 + b z + c = 0 +fr2 +bZ+c=0

+ d z 2 - b z + c = o _{. . . ( 2 )}

Equation (1) and (2) must be identical as their root are same a b c

- - = - : = =a b c

(iv) Let 21 and z2 be two purely real roots then 4 = 2,, zz = zz. l n t h i s c a s e a z 2 + b z + c = 0 . . . ( 1 )

= f r 2 + 6 2 + c = 0

* a z 2 + b z + c = 0 . . . ( 2 )

Equation (1) and (2) must be identical, _{as their root are same = 3 = 9 = 9.} a b c lllustration 6.

Let 21, 22, Zs b€ three distinct complex numbers satisfying _{la - 1l = lz2-11=14 -11. Let A, B,} and C be the points represented in the Argand plane corresponding to 21, 22 and zs respectively. _{Prove that z;zr+7, = 3 if and only if AABC is an equilateral triangte.}

Solution:

Ia -11= lz2-11= lzr-11

= The point corresponding to 1(say P) is equidistant from the points A, B and C. =+ P is the circumcentre of the AABC

Now if z1+22t7, = 3 then the point corresponding to centroid of the AABC i" zt + zz + zs - 1 3

= circumcentre and centroid coincide = A ABC is equilateral

Conversely if A ABC is equilateral, then centroid is the same as the circumcentre i.e. p. H e n c e c e n t r o i d z 1 + 2 2 + 2 3 _{= 1 = zt+zz*zg=3.}

3

I a

I

I

RSM-91 1-P3-MA-ComPlex Number l l 1 0 l f n i s a n y i n t e g e r , t h e n ( c o s e + i s i n 0 ) n = c o s n e + i s i n n 0 ' T h i s i s k n o w n a s D e M o i v r e ' s T h e o r e m .

Remarks:

r Writing the binomial expansion of ( cos 0 + isin 0 )n and equating the real part to cosno and the imaginary Part to sin n 0 , we get

cos n 0 = cosi 0- n crcosn-2 0sin2 0+ n c* cosn-4 0sin4 e+""""'

Sinno = n Cl COSn-1 esine- n Ca COSn-3 0Sin3 0+ n C5 CoSn-S 0Sin5 0""""

I lf n is a rational number, then one of the values of (cos0+isin0)n is cosno+ i sinno'

Let n = p/q, where p and q are integers (q > 0) and p' q have no common factor' Then (cos 0 + i sin0)n has q distinct values, one of which is cosnO + i sinn0'

t l f z = r ( c o s 0 + i s i n O ) , a n d n i s a p o s i t i v e i n t e g e r ' t h e n r e k n + 0 * i r i n 2 k n * 0

l , k = 0 , 1 , 2 , . . . n _ 1 . z l t n = r 1 l n l c o s a _{n f l l}

Here it

""n n" noted that any'n' consecutive

values of k will serve the purpose'

Applications of De Moivre's Theorem:

Here we will discuss few of these which This is a fundamental theorem and has various applications'

are important from the examination point of view'

The nh Root of UnitY:

Let x be nth root of unitY. Then

x n = 1 = 1 + i . 0 = c o s O o + i s i n 0 o = c o s ( 2 k l t + 0 ) + i s i n ( 2 k n + 0 ) = cos 2kn + isin 2kn 3 x = "or 2k* _{+ isin} 2kn n n Let o =

"o"?1+ n n isin?1.tnen the nth roots of unity are c't

(l=0,1,2,....n-1), i.e. the nth roots of unityare 1'o'ct2"""ctn-1 ' (where k is an integer)

k = 0 t , 1 , 2 , ' . ' . . ' n - 1 .

show that lez+ s)(Jt - i)l= J-slzz

+ 5l' where

z is a comptex

number'

lf zr and 22 ?rQ two complex numbers such thatlzi = lz'l + lzt - z2l' show that

a r ! 2 1 - 2 r $ Z 2 = 2 n n , n e l '

RSM-g1 1-P3-MA

Sum of the Roots:

1 + a + . - ' * . . . . + o n - r

_{= 1-on =0 +l"or?!n=0 and |"in?E=o}

t-u

_a

n

_ffi

n

Thus the sum of the roots of unity is zero.

Product of the Roots:

n( n-1 l ff n is even, uT = -1

n(n*1) l f n i s odd, aT = 1.

Remarks

:

r The points represented by the 'n' nth roots of unity are located at the vertices of a regular polygon _{of n sides} inscribed in a unit circle having centre at the origin, one vertex being on the positive real axis (Geometrically represented as shown).

r Gube roots of unity :

For n = 3, we get the cube roots of unity and they are l, cos2.! + i sir 21 4n , ^, 4n :

-3 1 - a n d c o s T * I sinf i.e. 1,

-1+ iJ3 _ , -1- iJ3 x'

2 and 2 . They are generally denoted by 1, ar and ro2 and geometrically represented by the vertices of an equilateral triangle whose circumcentre is the origin and circumradius is unity.

( i ) o 3 = 1 a n d 1 + o + r , r 2 = 0 n(n-1) ( _{zn 2r)"[";ll} . = [ c o s - + r s r n - J = cos{n(n_ 1 ) } + i s i n { r u ( n _ 1)} ( i i ) 1 + ( l ) n + ( l ) 2 n = 3 ( i i i ) 1 + c , l n + c o 2 n = 0 (n is a multiple of 3)

(n is an integer, not a multiple of 3)

lllustration 7.

Sotve the equation z7 + I = 0 and deduce that

(i)

"o"!ro"!"or?

= -!,

ti)

cosLc 3n

5r J7

' ' / - - - 7 _{7 7 8} " _{1 4 o " * " o t u = a ' ,}

. r 3 n 5 n 1

Iilt) s/n-srn-srn- = -.

1 4 1 4 1 4 I '

(iv) tan'z

f4+tan2

_,Y*r"n'

_ff

= S.

Solution: z 7 + 1 = O

The roots are -1 , cf,, cf,3, o5, d , a3, crs . Where cr = c o s - +

rstn-7 rstn-7

l'

c r 3 =

c o s ? + i s i n f , c r u =

c o s T + i s i n f

N o w z 7 + l = ( z + 1 X z - c r ) ( z - d X z - o " ' S 1 z - 4 3 ) ( z - o u ; 1 2 - 4 5 ; = t : ! : = _{{ r '} - ( ' + c r ) z + a c r } _{{ r '} - ( o u + 4 5 ) + o u u u } = ( = u '25 +24 -23 +22 -z+1)

=(r'-z=

"orf

* ,)(* -r= .o'f .t)(='-"

[ s i n c e a + c r = 2 c o s ] , ad=11 z 6 + 1 z 5 + z 2 4 + 2 2 z 3

- ="

--V-*

f

-7

/ - \ / ? n \ f .

I r ' - z r " o " ] . , ll

z 2

- z z c o s l

+ t

l l z '

: ; t , - , t

l ' l [

_{\ / \}

' J t

=

_["*+)-

_{(o.]).('.:)-'}

=

_{[=.

_])-,*"]]('.

_])-'*"7]('.

₁₎

-'*'?]

o , , [ , . : ) ' - ' [ ' . : ) - [ ' . ) ' . ' . ( ' . ) - '

=

_[=

_{* )- r*"1)(,.}

_:-'*'+)('.

_{:-' *'?)}

L e t z + ! = 2 x z

we set,

8x3

_{- 6x}

+* * 2+

_{2x -1=' (" - ""t+X- - t"t +)[' - ""tT)}

or, 8x3 4x2 4x+ 1

= r(*

-"*+)["-*"+)('-*'?)

.. .or|, eosf anO

cosf arethe

rootsof

the

equation

8x3-4x2-4x

+ 1 = 0

.'. Product

ot roots

= -*

.'.

cosf

_{. .o.f ' "o"f = -;}

(ii) Putting x = - 1 in (2), weget.

3n)l

-a-

4 +

4 + 1=-a

_[r

+ *'+)[''.

*.+](,..".?)

or,

7 =8'cos'

_{f4 cos'ff}

.or'

_{# * to'ff"o'ffto'ff}

= t{

stnce

0.#,

#, #.;,

we

rejectthe

-ve

sisn

*'?., (1)

'tl

)

-zzcos!*

. . . ( 2 )

I

RSM-91 1-P3-MA-Complex Number

--- T. 3n 5n j7

. ' . c o s * c o s

U " o " U = g . (iii) Putting x = 1 in (2), we get,

8 -4 -4 + 1=

8

["

-*";)[,.

-*"+)(. -*"+) = o

or, 1=82sin2

_'

asin2 !1sin2

A or. sina ^,^ 3n -'- 5r , 1

1 4

1 4

1 4

1 4

s ' n

1 4

s r n

1 4

= ta

since o < a.

3r

.

5n Tc

14 14. 14. r,

we rejecr

me _ve

stgn

s i n a s i n 9 1 s i n 5 n

= 1 .

1 4 1 4 1 4 8

(iv) tan

_{] tan}

_{1 4 1 4 - 1 4}

_{$ t"n $ = 1t 8/ r- - = +. we have}

(1),

/ , 1 7 t 8 J7

#

= (., - zzcosl

.',)(o - zzcos!

. r)(o

- zzcosl + t)

...(3)

L e l z = 1 + Y 1 - y .

z, -2zcos!*.,

= (!!,)' -r(!+J['-t""

#].,

t

_{\ 1 _ v}

₎

_{t r _ v / [ r + tan2}

# )

_

z(t. v')(t.

t',

h)-

z( - v,)(t

- t^,

#)

(t-v)'

_(,."",

_#)

.'. Putting

==

H

in (3),

we get

( 1 + v)7

ffi.t

26

cos2

ftcos"ff"o"'

#(r, +tan2

fi

#)(r

+tan2*)(n

*)(n *t^n'!)

+tan2ff

1 + Y , t

- T I

1 - y

.'. Equating coefficient of ya from both sides, we get

)('"^*' co)

= z[t"n'

fr+tan2#."u

#)

.'.

tan2

_{fr+tan2ft+tan2ff}

=s.

(r - v)u

J,lltCC LH.' ErrrJEE House)9=t+ t61!a:?n|!r

Logarithm of ComPlex Number:

ln order to find log(x + iy), we write log(x + iy) = a + ib

. . . x + i y = g a + i b = e " l c o s b + i s i n b ] = e " ( c o s ( 2 k n + b ) + i s i n ( 2 k n + b ) ) .'.e" cos(2kn+b) = x and e" sin(2kn+b) = y

Sofve for a and b, ;. eu^ = f * f

1 ^ 4

ol' €r =

)tntx' + y21, tan(21n+51= (y/x)'

when k = 0, corresponding values of a and b are refered to as principle values

Method to Find (x + iY)"

*ib :

For evaluating (x + iy)" *'o we write, c + i d = ( x + i y ) " * i b

.'. log(c + id) = (a + ib). log(x + iY)

Now evaluate log (x + iy) and then solve c + id = e(a + ib) ros(x + iv)' Illustration 8.

11 7 = (1+i tan a)1*i, then find the magnitude of z' Solution: ('t +i tanct)1*i = "(1+i)'ln(1+itano) = "t,*,).'nt"- 'cos c) - "(1+iXio-ln coscr) = "-(o+ln

cosc)+ i(ct-ln cosct)

"' lzl = "-(cr+ln cosct)' of in ll z and z' are two complex numbers then

! it tn" angle through which Oz' must z

order that it maY lie along Oz'

z - lzleis - ltl

"'tr-r')

='E1,

",o

z' lr'|"'t' l='l

lz'l

argument be turned Find the integral solution of the equations:

@ ( t * [ ' = ( - i ) n , ( b ) ( 1 - i 1 n = 2 '

If x,=

"osfr*isinfr,

provethatxizxg

"' o=-7'

lf a, B are the roots or rn" equation *' X + 1= 0' prove that

on +Fn =zn*l.cosT.

If a and B are the _"o^pt", cube roots of unity,show that oo*/*o'f'=o',

I

RSM-91 1-P3-MA-Complex Number ' t 5

In general, let 21, 22, zs, be the three vertices of a triangle ABC described in the counter-clock wise sense. Draw OP and OQ parallel and equal to AB and AC respectively. _{Then the point P is z2- 21 and} Q i s z 3 - z 1 a n d z z - z t O Q . = -- (cos 0. + isin cr) z z - z t O P ' C A , - l z ^ - z , l = ' o ' o _ | o - l I _{a l o} B A _{l z z - z t |} : 2 - : = A . ' ' n ' n - " i 2 n t n

z t - 0

_{l = r l}

+ zz= 21 gi2"ln Similarfy, \ = "-i'ntn + zn = z,e-i2ntn z1

Note that arg. (zs- zt) - arg(22- zi = ct is the angle through _{which OP must be rotated in the} anti-clockwise direction so that it becomes parallelto Oe.

Here we can write z::+ =3:?!"-i(2n-a) _{also. In this case we are rotating op in ctockwise} z z - z t _{l z z - z t I}

direction by an angle (2n - a). Since the rotation is in clockwise direction, we are taking negative sign with angle (2n - a)

lllustration 9.

Consider an 'n' sided regular polygon with the origin as it's centre. tf zr be the complex number representing _{a vertex 41, of the polygon, find the complex number associated with} the vertex that is adjacent to 41.

Solution:

Vertex adjacent _{to 41 is either 42 or An. Let z2 and zn represent the vertices & and An} respectively. Considering the rotation about the origin, we get

2nln

lllustration 10.

Complex numbers 21, 22, 23 are the veftices A, B, C respectively of an isoscetes right angled triangle with right angle at C. Show that (21- z)' = 2(zr - zs)(zs- z).

Solution:

In the isosceles triangle ABC, AC = BC and BCJ-AC. lt means that AC is rotated through angle nl2 to occupy the position BC.

Hence we have. Z, - =t = e*inr2 = +i z l - z z

> zZ

+ zl -zzrz, = -G?

+ zA

-2z,zr)

) zz -zs = +i(zt - zz) + zl + z| - 22,2, = 22,23 + 2zrz, - 22,2, - 2z! = 2(4 - zr)(2, - zr) = (21 - zr)' = 2(zt - zr)(2, - zr)

EI

1 6 RSM-9 1 1 -P3-MA-ComPlex Number

lllustration 11.

Let zr and zz be the roots of the equation z' * pz + q = 0, where the coefficients p and q may be complex number. Let A and B represent 21 and z2 in the complex plane. lf ZAOB = a * 0 and OA = OB, where O is the origin, prove that

' . " , 9 ) . p- = 4q 6os_ [ Z,r Solution: Z t + Z z - - P artdzlz2= q . Also zz = zla'o A(zr)

. . . ( 1 )

. . . ( 2 )

. . . ( 3 )

Now p2 = (zt + z2)2 = (21 + z1et")2 = 212(1 * 2ei" + e''")

= qejo 1'l + 2d" * e''") = q(2 + e-'" * e'")

= q(2 + 2coso)

= +qcos't.

lllustration 12.

lf 21, 22, zs ond 21,, z2', 4' represent the veftices of two similar triangles ABC and PQR'

respectivety

then

_{"v" 'J'}

prove

_{n"tl--L I l':=l.l=+llz'}

;z'l > r '

lz, -ztl I zi I lr, - rtl I zb I

Solution:

Q(zz')

Since

AABC

and

APQR

are similar,

ffi

=;* and IBAC=zQPR

= 0

In A ABC, 23 -zt = 49 "'t ' z z - z t _{A B}

ln

APQR,

' z z - z i

4-4= T"''

_{P Q}

From (1) and (2), ' / - " - \ ' " zs - zt ==? -'i. z z - z r z z - 2 1 - zl(22- zs) + zz'(zs- zt) = zs'(zz- zt) + lz'1(22- zs) + zz'(zs- zll = lzs'(22- z1)l > lz' {zz- ze)l + lz/(zs - z)l2 lzs'(22 - z1)l

I z : , z " - z r l . l z | z s - 4 1 ,.

- | - 1 - Z - J l - , - 1 l z r - z t z i I lzz-zt z i I

- l ' i

l z z - 4 1 | z L I l z z - z t l I z e I

| l = , - , = ' l . . l

' , | l " - " 1 > t

l z

= l

-l2z

-2,

l = ,

- . r l

- |

= L

I lr ' - . 1

. ,

| . t I l r r - = , 1

| z L I

. . . ( 1 ) . . . ( 2 )

RSM-91 !-P3-MA-Complex Number

t) tf zl +z2rxzrzrcos?=0, prove that the points represented by 4, zz and the origin form an isosceles triangle.

ii) lf A,B,C represent the complex numbers 21, 22, zs respectively on the complex plane and the angles at B and C are each equal _{' to !1o -a), then prove that}

2 \ (22- 4)2 = 4(zs- z)(a - z)

sinzal2-Secfion Formula:

Let z1 and z2 be any two complex numbers representing the points A and B respectively in the argand plane. Let C be the point dividing the line segment AB internally in the ratio m : n i.e. 19 = t , and let the complex number associated with point C B C n

be z.

Let us rotate the line BC about the point C so that it becomes parallel to CA. The corresponding equation of rotation will be, z'r z =E1-4"'n= !1-t1

z z - z _{l z z -z l} n ' ) t 1 7 : 1 - n z = - m z 2 + m z + z - n z 1 + m z 2

m + n

Similarfy if C(z) divides the segment AB extemally in the ratio of m:n, then ==AZt -rrrzz ln the specific case, if C(z) is the mid point of AB then z =tt *=='

2 lllustration 13.

lf 21, z2 and zs (in anticlochuise sense) represenfs the veftices of a triangle, find the centroid, incentre, circumcentre and the ofthocentre of the triangle.

1 7

Solution:

Let G be the centriod and let the line joining A and G meet the line BC at the point D. We have, BD= DC

D = Z z l z g

2

G divides AD internally in ratio 2 : 1 2 ( z z + z s ) _{_ .} = G = 2 ' - 1 - 2 1 + 2 2 + 2 3 2 + 1 3 B(zr) D (2, + zr\

l z

)

Let I be the incentre and let the line connecting A and I meet the line BC at Dr. We have B D r _{= A B _ 1 2 , , - z r l}

D r C A C lzt -zs I

']fItCC

I RSM-91 1 -P3*MA-ComPlex Number 1 8 and + D r -Dg. + D e A l _ AB+AC -tDr BC l z , - z r l + l z r - z t l l z z - z s I z t l z t - z r l + z r l z t - z z l l z . , - z z l + l z , - z t I o1 (l zr - zs | + | zr - zz l) + z, I z, -zs I a n o l = I z, - zz | + | zt - zs | + | zz - zs B(zz) Dr C(2.) c(2.)

Let 'O' be the circum-centre and let the line connecting A and O meet the line BC at Dz'

A(zr) _ z, I z, - z, | +zz I zt - zs | +zs I zt -.zz I - I z t - z z l + l z t - z s l + l z z - z t l BD" sin2C we have or"= *r28 . AO sin28 + sin2C ano oD2 = -- s''n 2A + ,.^ = zzsin2B+zssin2C ^ ^ < vz sin 28 + sin 2C

D^ (sin 28 + sin 2C) + z', sin2A 2., sin 2A + z, sin 29 + zt sin 2C

u -

m 2 C

s i n 2 A + s i n 2 B + s i n 2 c

Let 'P' be the orthocentre and let the line connecting the points A and P meet the line BC and B(zz = 0 . BD. tan C we have' Dp = tanB , AP tanC + tanB ano PDe tan A zrlanB + zrtanC tanB + tanC

D. (tanB + tanC)+ z,' tanA

a n o P = ' . ; .

t a n A + t a n B + t a n C zn tan A + z, tanB + z. tan C

t a n A + t a n B + t a n C

Condition for Collineariry:

ff there are three real numbers (other than 0) /, m and n such that tz'r + rfizz + fizt = 0 and I + m + n = 0 then complex numbers z1,7:2eltld z3 will represent collinear points'

Equation of a straight line:

r E q u a t i o n o f s t r a i g h t | i n e - w i t h t h e h e | p o f c o o r d i n a t e g e o m e t r y :

W r i t i n g * = t 1 2 , y = + e t c . i n Y - Y r = *-*' _{" n d} r e - a r r a n g i n g t e r m s , w e f i n d t h a t 2 ' ' 2 i Y z - Y t x z - x t

the equation of the line through zl and z2 is given by

1 1 1

l z z

z - z t = t-=, o, l=n z

zz -zt

2, -4

lq*,

zj.,

r Equation of a straight line with the help of rotation formula: Let A(zr) and B(22) be any two points lying on any

line and we have to obtain the equation of this line. For this purpose let us take any point C(z) lying on this line.

/ - _{- \}

S i n c e a r q _- | ' - " | = o o r n , l z z - z r )

z - z , t _{= -}z - z ' t z z - z t 2 z - 4

This is the equation of the line that

. . . ( 1 ₎ passes through

r Equation of Perpendicular Bisector:

Consider a line segment joining A(21) and B(zz). Let the line 'L' be it's perpendicular bisector.

lf P(z) be any point on the 'L', we have p A = p B tlz-zl = l z - z 2 l

* l z - z i 2 = lz-zrl'

+ ( z - . r ) ( Z - . ' t ) = ( z - 2 2 ) ( Z - z r )

= ZZ. - ZZ1 - 2,,2 + 2,2, = ZZ - ZZ, - Zr2 + Zr2, = z ( 2 , - 4 ) * 2 ( = , - 2 , ) + z ' t 4 - z z z z = 0

A(zr)and B(22). After rearranging the

A(a)

l z z

terms, it can also be put in the following _{form,lz, 4} lzz 22

1 - 0 .

General equation of the line:

From equation _{(1) we get, z(2, -4) - =r2" + z,t4 =2t=, - zr) - 42, + zr4} + z(2, - A\ *2(zr - zr) + 42, - 2,,2, = Q

Here 42, -ztZz is a purely imaginary number _{as 42" -ztzz =2ilm(4=r).} L e t 4zr-ztZz= ib, b e R

= z(2, - A) * Z(zt - zr) + ib = 0 =+ = i (4 - z2) + A (2, - z1 ) + b = 0 Let a = i(22-z) = a :i(a -zr)

* z a + z a + b : 0

This is the general equation of a line in the complex plane. Slope of a given line :

Let the given line bez6+-za+b = 0 . Replacing z by x + iy, we get ( x + i y ) a + ( x - i y ) a + b = 0 + ( a + a ) x + i y ( e - a ) + 6 = g .

a + a 2 R e ( a ) _ Re(a) ItS SiOpe iS =:;---=i =

--r(a - a) 2i2 h(a) lm(a) '

Equation of a line parallel to the line za +-za+ b = 0 is za + -za +)" = 0 (where l, is a real number).

Equation of a line perpendiculartothe line za +-za+b =0isza --za+ i l, =0 (where 2v is a real number).

RSM-"91 1 "-P3-MA-ComPlex Number 20

I Distance of a given point from a given line:

Let the given line be {a +2a + b : 0 , and the given point be z"' SaY z" = Xc + iYc'

Replacing z by x+iy, in the given equation, w e g e t , x ( a + a ) + i y ( a - a ) + 6 = I

Distance of (x", y") from this line is,

_ lz.a +2".a +bl _. 2 l a l

r arg (z-zd = 0 rePresents a line positive direction of x-axis) lllustration 14.

tf lzl = 1, then prove that the points represented by pe rPe n d i cul ar stra ight I i ne s.

Solution:

Since lzl = 1. z lies on a unit circle having centre at the origin'

( l + z \ _{n . 3 n}

" t s I r _ r J

= ' r ' o ' * Z

= 1+ z = keiil. or kei3dz

1 - z

where k is a real parameter and its value dePends upon _{the Position} of z'

Letcr

=

lE+,,

= fi einr+or.

fi

"'too

= a lies on one or other of the two perpendicular lines' Illustration

ir*oz =c, b *0 be a tine inthe comprex prane, where 6 istne complexconiugate of b. lf zt is the reflection of a point zz through the tine, then show that c = Zrb + zz| '

D + Z a + b = 0 za

passing through zo with slope tanO' (making angle 0 with the

lie on one or other of two fixed

tu

1 - z I I ,

r . \

\ 1 + z ' . \ 1-z.r" / 1 / --/ B ( 1 , 0 ) 5 T / Z .P'(z) A(-Solution: T h e g i v e n l i n e b z + b Z = c , " " ' ( 1 ) b + 0 is the perpendicular bisector of the line joining _{the Points} 21 ?trd 22.

Then lz - ziz = lz - zrlt (z-z) (2 - 2,) =(z-zr) (Z - Zz)

z(22 -Z)* 2(22 - z',1+ l.,l' - l=rl'=0 '..(2) (1) and (2) are the same. Therefore

b _- b _{= - - : - = k ( s a y )}

2, -2, zz - zt l.rl' -l=,1"

= c , b + 0

Q ( z r )

l x . (a + a ) + iy "

( a - a ) + o l

( a + 5 ) 2 -( " - a ) ' , l l z c a + z c a + o l +(ne(a;)'z + a(im(a))'z 1 + z 1 - z

RSM-91 1 *P3-MA*Complex Number 2 1

N o w b 4 + 6 z z - c = k ( z z - z i Z t + l c z z ( Z z - 2 , ) - k ( l z r l ' - l z l ' )

= nl=yo

-lr,l' *lrrl' - zz4

-lrrl' *l=,1'l

= 0 = b4 +62,

= c,

.

i) _{lf lzl = 2, then show that the porlfs representing the complex numbers -1 + 5z} lie on a circle.

i i ) _{l f z = 2 * 1 * i t [ g - 1 2 , where t is reatand ( < s, show thatthe modulus of '*7.} z - 1 is independent of t. Also show that the locus of the points z for different values of t is a circle and finds its centre and radius.

Equation of a Circle:

Consider a fixed complex number zs and let z be any complex number which moves in such a way that it's distance fromzs is always equals to'r'. This implies z would lie on a circle whose centre is zs and radius r. And it's equation _{would be lz -zol = 7.}

= lz -zsl2 = f + e - zs) (2 - zo) = 12 = z2 - zzo - zzl + zozo - 12 = 0 . L e t - a = Z o o l t d z o z o - r 2 = b = zz+ aZ + az+b : 0.

It represents the general equation of a circle in the complex plane.

Remarks:

t zz.+aZ+az+b =0represents _{a circle whose centre is -a and radius is Jaa-b.} T h u s z Z + a 2 + a z + b = 0 , ( b e R ) r e p r e s e n t s a r e a l c i r c l e i f a n d o n l y if a a - b > 0 .

r Now let us consider a circle described on a line.segment AB, (A(21), B(22)) as its diameter.

Let P(z) be any point on the circle. As the angle in the semicircle is nl2. IAPB = nl2 ( =n - z) .^ A(zr) :+ afg \ z z - z ) = t-" is purely imaginary = t-=t * z - z z z - z z + (z - .r)(Z -2r) + (z - zr) (2 -21) = 0

r _{Let z1 and z2 be two given complex numbers and z be any} , , : j L l =

o , *her" ae(0, ru). complex _{number such that, arg |}

-\ z - z z )

Then 'z' would lie on an arc of segment of a circle oo 2122, containing angle cr. Clearly if cr, e (0, nl2),2 would lie on the major arc (excluding the points 21 and z) and if a e (n12, 7t). 'z' would lie on the minor arc(excluding the points z1 and z2).

Note:

The sign of o determines the side of zl 22otl which the segment lies. Thus a is positive in fig. (1) and negative in fig. (2)

:=L =o

Z - Z z B(zz) f i s . ( 1 ) A(zz) A(zz) B(zr) B(zr) ris. (2)

r Let ABCD be a cyclic quadrilateral such that A(21), B(22), C(23) and D(z+) lie on a circle' Clearly l A + l C = n ( z o - 2 , , \ ( = , - . " \

--

*sl#,)+ars\-:---.

)=

n

_ aro

?o

-=,\(._z_:z)=

*

- \zz -21)\24 -4 ) > 24 -z't .22 -23 is purely real. z z - z ' t z + - z l T h u s p o i n t s A ( z r ) , 8 ( 2 , ) , C ( z a ) a n d D ( z + ) ( t a k e n i n o r d e r ) w o u | d b e c o n c y c | i c i f (=o - =,)(., - zs) ,^ -- _{'o Pur€lY} r€al' ( z z -z ' r ) \ z + - z s )

Equation of tangent to a given circle at a given point:

Let lz - zsl =r be the given circle ano we have to obtain the tangent at A(zr)' Let us take any point P(z) on the tangent line at A(21)'

Clearly IPAB= nl2

ng(

=-=''

I=,n,, = iaa ispurelvimasinary.

' ' ' \ = o - = . ) z o -z ' r

-

zo -z'r

'-='' * !:-1- =g

zo - 4

-

(z - zi (4 -A) * G -4)Go - 4) = o

- = ( 4 - 2 r ) * 2 ( = o - 2 . ) + z ' , 4 - z t Z o + 2 , 2 , -4 z o = 0 + z(4 -Z)*2(.o -zr)+2lz',12 -ztzo -Zzo =0

ln particular if given circle is lzl = r, equation of A + z t = z l z l 2 = 2 r 2 .

b<planation:

Let A(21) and B(22) be two given complex numbers and P(z) be any arbitrary complex number' Let PAr and P& be internal and extemal bisectors of angle IAPB respectively. Clearly lA2PAl = nl)'

Ap l=-=.1 lZ_Ztl=

l" (say)

t r t o w '

_{g p}

=

V q =

l . - r r l

Thus points Ar and Az would divide AB in the ratio of l' : Hence P(z) would be lying on a circle with Ar& being it's

D(a

A(zt

. ftlz-=.|=^ (^ e R*, r + 1), where 21 end 22 are given complex numbers and z is an l z - z z l

arbitrary complex number then z would lie on a circle'

the tangenl al z = Zt would be,

1 internally and externally respectively' diameter. ,. B1a) I I l

', az

f|lllcc nt*t

RSM-91 1-P3-MA-Complex Number 23

Note:

I lf we take 'C' to be the mid-point of A2A1, it can be easily proved that CA.CB = (CAr)2 i.e. lz7 zollzz- =ol = f , where the point C is denoted by zs and r is the radius of the circle.

r _{lf l, =1 > lz - z,tl = lz - zrl hence P(z) would lie on the right bisector of the line A(21) and} B(22). Note that in this case 21 and 22 e,re, the miffor images of each other with respect to the right bisector.

lllustration 16. z - i

lf q = :---, show that, when z lie.s above the real axis, a will lie within the unit circle which

z + l

has centre at the origin. Find the locus of a as z travels on the real axis from - mto +o.

I

j

Solution:

From figure, it is clear that lz -il < lz +il (as z lies above the real axis).

lz-il = l a l = H . t

l z + t l

> a lies within the unit circle which has centre at the origin.

Now if z is travelling on the real axis lm(z) = 0 and Re(z) L e t z = x + i 0 x - i I G = ---x + l l x - i l + l a l = - = 1 ( a s l x - i l = l x + i l V x e R . l x + t l

= o moves on the unit circle which has centre at the origin lltustration 17.

lf 21, 22, ft Qra complex numbers such that 2 -- l- * J- , show that the points represented 21 22 23

by 21, 22, 4lie on a circle passrng through the origin. Solution:

Since P(21), Q(zz), R(zs) and S(za) are concyclic points, ZpSe = ZpRe

varies from -oo to +m.

+ a r g . 2 2 - 2 4 = a r g 2 2 - z g z t - z + z , t - z g f / \ / \ - l

= " r s l l = ' - ' o

_{1 1}

4 - z s ll=o

L \ z r - z q ) \ z z - z s ) ) = \=, - =o) .(z', - =t) = ," , \ z t - z c ) l z z - z e ) l f 4=g+i0,then z z . z t - z t - r " ^ 1 2 1 z Z - z S w e h a v e 2 = 1 * 1 _{f r o r w h i c h z " -} z ' t z z z1 z2 23 . 22, -2, . . . ( 1 ) . . . ( 2 )

z1

-From (1) and (2),, r * 22, - z, = real

z1 z'rzt

z . -

-' 2 2 , - 2 ,

Z c - Z t , n

= ---* = reat + I = real, which is true.

- 2(2, -2,) 2

Therefore z1,22,z3and the origin are concyclic' Alternative Solution: 2 = 1 * 1 = 1 _ 1 = 1 _ 1 z1 z2 23 21 z2 zg 21 Z t - z ' t z t - z s - - = -ztzz ztzz z c - z t 2 2 = :] = --z l - --z t 2 3

' ^ - ^ ( " - 1 ) = " , . o

( - z \ = *

( z ' \

- ^ r n l T . , )

- ( zs)

. " * l . ; , , l = o r = n - F > * + p = t r

= Points A, B, C, D are concYclic' Illustration 18.

Plotthe resion represented bv

:, "-(#)=? 'n

the Arsand ptane'

Solution:

Let us take arg(4) _{' \ z - 1 )} = 2n13, clearly z lies on the minor arc of the circle passing through (1, 0) and C1' 0)' similarly,

"g f 1+l \ z - 1 ) = n/3 means that'z' is lying on the major arc of the circle passing through (1, 0) and (-1' 0)' Now if we take any point in the region included between the two arcs, say P{zr),

. I t ( z + 1 \ _{' 2 n}

w e s e t

_{A s a r s I z t J = T}

Thus I <

"rof Z11l . ] ,"pr".ents the shaded

region (exctuding points (1, 0) and C1, 0)). . . . " -

3 _ _ _ \ z _ j ) 3 Illustration 19.

lf zt and 22 dre two complex numbers, then prove that lz,-zrl' s(1+k)lzl2 + 1t+K1)lzrl2 vk e R'' Solution: l(k4+zr)12 >0 + (kz1+7r)(k2, +2r)>-0 = ( r , * p l 1 r , z r + 2 r ) , - 0 a s k > 0 = k l z , l ' * | t = r l 2 + 2 , 2 r + 2 , 2 r > 0 \ . ' k ) + ( 1 + k ) l z t 12 + ( 1 + k - 1 _{) l z , l ' > l z t 1 2} + lz,12 -2,2, -Ztzz + (1 + k) | zt 12 +(1 + k-1 _{)l zz 12} >l z1 - z2 l2'

RSM-91 1-P3-MA-Complex Number 2 5

i)

ii]

Consider a square OABC in the Argand plane, wftere 'O' is origin and A= A(2il, then show that equation of the circle that can be inscribed in this

| _{- t " , i \ l}

square isZlz-'#i=lzol _{Pertices of square are given in anticlackwise)}

r( =)(=) = k, then shaw that points A(2,), B{22), c(3, 0) and D(2, a} \ 2 - , , ) \ 3 - r , )

(taken in clackwise senseJ is lie an a circle for k > A

Some Importanf Resulfs to Remember:

r The triangle whose vertices are the points represented by complex numbers 21,22, z3is e q u i l a t e r a l 6 - _ 1 * 1 * - - 1 = 6 i . e . i t z l + z f + z l = z t z z + 2 2 2 3 + : . 3 : . 1 .

z z - z s z l - z t z t - z z

t _{lz -zl + lz -z2l = 1. , represents an ellipse if lz1 -z2l < 1", having the points z1 and 22 as its} focii. And if lz1-z2l = 1", then z lies on a line segment connecting \ end 22.

t _{l z - z l - l z - z r . l = l . , r e p r e s e n t s a h y p e r b o l a i f l z r - z z l > l . , h a v i n g t h e p o i n t s z l a n d z 2 a s i t s} focii. And it lz1-z2l = 1,, z lies on the line passing through 21 ?nd z2 excluding the points between 21 a,nd 22.

Illustration 20.

Let z1 and zz be the complex roots of the equation 3l + 3z+ b = 0. tf the origin, together with the points represented by z1 and z2 form an equilateral triangle then find the vatue of b. Solution:

21, z:2, 0 will be the vertices of equilateral triangle if z ' f + z z 2 + 02 = ztzz + Oz2+ Oz1+ 1 -? =! + n = t.

3 3

i) lf far a carnplex nwmber _{z, I z * 1l + lz + 1l = 2, then sftour that z lies on a line} segrnenf.

ii) _{If z be any complex number such that l3z * 2l + l3z + 2l = 4, then shaw that} loeus af z is a line-segment.

RSM-91'! -P3-MA-ComPlex Number 26

Exercisel.

i ) . ( a ) 0 = n n

ii) 0

iii) (a) t (1 - 3i)

Exercise 2.

iii). -160.

Exercise 3.

i X a ) . n = 4 k , k e l ( b ) n = 0

Exercise 5.

ii) centre is (2, 0) and radius = J5

o = n n t f , w h e r e n e l

, ( x v , i )

=[t - t- t)

(b)

(b)

RSM-91 1 -P3-MA-Complex Number 27

GOTGTPT

TORITUII

[I GIITGT

Properties of Modulus and Argument c _{lz1+22+ ... +znl < lzil + lzrl + ... + lznl} . _{lzt - z2l> llal - lz2ll}

. _{l z t + z 2 l ?} = (zt + zr) (2t + 2z)=lz.,l2 _{+ ltl'*.,72+} z z Z t c _{arg(z4z) = arg(4) + ary(22)}

. arg (z1lz2) = arg(z) - arg(22)

De Moivre's Theorem :lf z= r(cos 0 + isin 0), and n is a positive integer, then 1 t ^ . , ^ f 2 k n + 0 2 k r c + 0 1 .

2 " " = t " " lcos-+isin-"-- " l , k = 0 , 1 , 2 , . . . n _ 1 . L N N J

The nth Root of Unity

L e t x b e t h e n t h r o o t o f unity. T h e n x n = 1 = c o s 2 k n + i s i n 2 k n ( w h e r e k isan integer) 2kn 2kr

3 X = c o s -" " + i s i n -' " " k = 0 , 1 , 2 , . . . . , n - 1 . n n

Let cr = cos2-!+ isin?I.tnen the nth roots of unity are crt n n

( t = 0,1,2,....,n - 1), i.e. t h e n t h r o o t s o f u n i t y a r e 1 , a , a 2 , . . . a n - 1 . sum of the Roots

4 n n-1 n .r

1 + a + o ' * . . . . + o n - 1 = 1 - - o = 0 =)I"or2kn=0 _{a n d l r i n ? E = O}

1 - a f t n f t n

Thus the sum of the roots of unity is zero. Product of the Roots

1 , a . a 2 . . . c r n - 1 = ( - 1 ) " ( - 1 ) = C1)n*,

Concept of Rotation : lf 21,22, zs, era the three vertices of a triangle ABC described in the counter-clock wise sense. then

z q - z t O Q , C A , ^ l z , - z n l i ^ -__:_________.j_ = _(COSc[ + tStnc) e'" = :___l____-l-r. e'u z z - z t O P ' B A _{l z z - z t l}

Note that arg (zs- z.) - arg(22- zi = o is the angle through which OP must be rotated in the anti-clock-wise direction so that it becomes parallelto OQ.

Geometrical Applications: Condition for Collinearity

lf there are three real numbers (otherthan 0) l, m and n such that lz1 + fizz* 1123 = Q and l+ m + n = 0 then complex numbers zl,22etldz2willbe collinear.

Equation of a Straight Line

. _{Equation of a straight line with the help of rotation formula:}

Let A(21) and B(22) _{be any two points lying on any line and we have to obtain the} equation of this line. For this purpose let us take any point C(z) on this line

/

. { z - z t \ _^ z - 2 , Z - 2 " s l n c e a r g l - l = u o r r - ; ' = = - - - : - . \ z z - z t , ) z z - z t Z z - Z t

,,.,P*,

E=

General equation of the line:

From equation (1) we get za+-za+b = 0 ' w h e r e e = i ( 4 - l ) a n d i b = Z z 2 - 2 . 2 r , b e R

This is the general equation of a line in the complex plane' Slope of a given line:

ne(a) lI za +-za+ a = 0 is the given line then its slope = - _t("')

Equation of a line parallel to the line za +-za+ b = 0 is zd +-za + l' = 0

(where l, is a real number)' _{- _7a+ i1. = 0 .}

Equation of a line perpendicular to the line za +-za+b = 0 is za (where l, is a real number)'

Equation of Perpendicular Bisector:

consider a line segment joining A(21) and B(22)' Let the line 'L', be it's perpendicular bisector' lf P(z) be any point on the'L', we have PA = PB'

= l z - z l = l z - z 2 l

+ z(2, - a\ +z(2, - zt) + 2.4 - zr2, = o Distance of a given point from a given line:

Let the given line be za+-za+b= 0 , and the given point be z" then distance of z" from this

. l

l z ^ a + z ^ a + o l l i n e i s - - r : # '

zPl Equation of a Circle

Equation of a circle of radius r and having its centre at zs is lz-zsl= r'

> |z _zo|, = f * ( z - zi (z _^) = f2 > fi. + az +62 +b = 0, where _ ? = 7o and b = zo4 _ 12' It represents the general equation of a circle in the complex plane'

o E q u a t i o n o f a c i r c l e d e s c r i b e d o n a | i n e s e g m e n t A B , ( A ( 2 1 ) , 8 ( 2 2 ) ) a s d i a m e t e r i s ( z - z ) ( 2 - 2 r ) * ( t - z r ) ( 7 - 4 ) = o ' o L e t z l A f i d z z b e t w o g i v e n c o m p | e x n u m b e r s a n d z b e a a n y c o m p | e x n u m b e r s u c h t h a t , ( - - \ a r o l ' - " _- l = a , w h e r e a \ z - z z )

e (0, n). Then'z'will lie on the arc of a circle'

o Let ABCD be a cyclic quadrilateral such that A(21)' B(22)' c(4) and D(a) lie on a circle' Then ( z o - 2 . ) . ( z r - z t )

i s p u r e l y r e a t . ( z r - 2 . ) ( z o - 2 " )

Some lmportant Resultto Remember _{-^--,^-. ^..*hara z.}

r The triangle whose vertices are the points represented by complex numbers z1' 22' zs ts

e q u i l a t e r a l i f l + + - . = + = 0 i ' € ' ' i t z l + z ? ' + z ! : z l z z + 2 2 2 3 + z s z 1 '

z z - z g z s - z t 2 ' t - 7 2

C | z - 2 1 | + | z _ : , 1 2 | = l , , r e p r e s e n t s a n e | | i p s e i f | 2 1 _ 2 2 | < x , h a v i n g t h e p o i n t s 4 a n d z z a s i t s foci. And iI lz1-z2l= 1" then z lies on a line segment connectin Q 21 and z2'

. | z _ 2 . | - | z _ z z | = ^ , , r e p r e s e n t s a h y p e r b o | a i f | 2 . , _ 2 , | > } , , h a v i n g t h e p o i n t s z , t A n d z 2 a S i t s foci. And if lz1-z2l= 1., the z lies on in" tin" passing through zt and z2 excluding the points

between 21 artd 72'

f||?lcc

RSM-91 1-P3-MA-Complex Number 29

s01$D

Pn0Bfirs

I

I

i

,

I

{

Subjective:

Problem 1.

lf iz3 + / - z + i = 0, then show thatlzl =1. Solution:

i z 3 + 22 - z + i = 0

By substituting z = i in the equation, we get 0 = 0 H e n c e z - i i s a f a c t o r o f i z 3 + 2 2 - z + i

+ iz2 1z-i) - 1(z-i) = 0 = (t - 1) (z -i1 = g Either izz -1 = 0 or z- i= 0

W h e n _{z - i = 0 , 2 = i ; . l z l = | 0 +i.11} = J O ' * 12 =1. W h e n i z 2 - 1 = 0 , 2 2 = 1 = - i

. ' l t l = l o - 1 . i 1 = !6'*(lF =1+ l='l=1 or lzl=1

.'. In anY case we have lzl = 1. Problem 2.

tf lzl <1,lwl <1, showthatlz-vn1' <Ml-lwl)t + ( argz-argw)2. Solution:

Let us consider a unit circle with its centre as the origin Let ZAOX =0r dnd IBOX= 0z

.'. arg(z) = 0r ?nd arg(w) = 02 z = O A , w = O B

Now in A oAB; cos

o = t

on f t]_oe t1: tBA l'z

2 t o n t toe l'

t_t2 t_t2 t_t2

= lenl =loAl +loal -2loAl loBl coso

+lz-w l2 =lzl2 * lwl'- 2 lzl lwlcosO =(lzl _{- lwl )' * qEl lwl sin2} (e/2) we know tnat sing = 9 =[9)' > sin2 9

2 2 \ 2 ) 2

= (:i >lzl lwl sin29 [sincelzl , l w l < l , i t i s a u n i t c i r c l e ] H e n c e _{l z - w l ' < ( l z l - lw l ) ' * ( a r g z - a r g} w ) 2

Problem 3.

(i) Prove that the polynomial x3n + i:m+1 + v3k+2 is exactly divisibte by f + x+ 1 if m, n, k are non negative integers.

(ii) Show that the polynomial f sin0- p''1x sinne +pn sin(n-l)0 is divisibte by f-2px cosl+ p2.

I

'l']ntcc

30 RSM-9 1 1 -P3-S[A-€omPlex Number

Solutian:

(i) x2 + x + 1 = (x- rrr) (*-t'), where w is a cube root of unity'

+ We have to prove that x - ro and x -co' are factors of the polynomial' Put x = rrt

0 ) 3 n + o ) 3 m + 1 + r 3 k + 2 = 1 + r o + ( D 2 = 0 = ( x - r r t ) i S a f a C t o r o f t h e p o l y n o m i a l Put x = ro2

( D o n + o o m + 2 + r o k + 4 = 1 + c D 2 + ( D = 0 = x - r o 2 is a l s o a f a c t o r o f t h e p o l y n o m i a l . Hence the polynomial x3n + x3m+1 + x3k+2 is exactly divisible by x2 + x + 1'

(ii) We can write, x'-2pxcosg + p'= (x- p(cosg + ising))(x- p(cosg - ising))

Gase-l x - p(coso + i sin0) is substituted in the given expression the given expression = p n ( c o s n O + isinn0)sinO- p n ( c o s 0 + i s i n O ) s i n n 0 + pn(sinn - 1 e )'

=pn[(cosnosin0+isinnesin0t(cososinn0+i sin0 sinne)+(sinn0cos0 - cosn0 sin0)]=g

Gase- ll Likewise 1= p(cosg - i sino) can be substituted in the expression to show it equalto zero. Hence the result.

Problem 4.

Find the complex numbers z which simultaneousty satisfy the equations

l z - 1 z l - s lr - a l - '

| * Bil-

s' lr - 81-

''

Solufion:

t"*

_l=l

=,'

=

l=#l

=' = (x-4)'+

f = (x-8)'

* t' =x = 6

With

x = 6,

l z - 1 2 1 5 | - 6 + i y I s

_{= _ :}

l z - a i l 3 1 6 + i ( y - 8 ) l 3

= 9 ( 3 6 * f ) = 2 5 [ 3 6 + ( y - 8 ) ' ] = f - 2 5 y + 1 3 6

= 0+ y=17,8'

Hence

the required

numbers

are

z = 6 + 17i,6

+ Bi'

Problem 5.

| .tl

tf lzl > 3, then determine the least value of l, * t z l _-l and the corresponding z.

Solufion;

l'.:l- l'',-*l=

I'o-#l=

vt-ttas

rzr>3

Let f(x) = 1- f o r x > 3

f ' ( x ) = 1 * I r o

x-+ f(x) is increasing function x-+ f(x).1n at x =3

l - 1 l -l'-, 1l =3-1=8.

l'

* 21,''

=

_llzl

-

lal,o*

- " -5 - 3'

I I

I

I

I I

I

I

L

X

RSM-9 1 1 -P3-MA-Complex Number 3 1

To find z: Let z = r eio, where r = 3.

rnen

lz

* 1l

= lr"''

* 1"-''l

= l9.o.e

* 9,r,nrl

=

I z l I

r

I 1 3

3

|

/ r o \ 2 , ^ / g ) 2 . , ^ / a ) 2 / t o ) 2 - l - l c o s - U + l - l s l n - 0 = l - l = l - l c o s \ 3 1 \ . 3 / \ 3 / \ 3 / + 0 = 2 ' 2 I

t

'e = l,9)'

_\.3i

= - 3 i .

c o s 2 o

+ c o s ' e = o

+ 2i\ > z= 3-il2 or 1-3i12. Problem 6.

ABCD is a rhombus. lts diagonals AC and BD intersect at the point M and satisfy BD=2AC. I t s p o i n t s D a n d M r e p r e s e n t t h e c o m p l e x n u m b e r s l + i a n d 2 - i r e s p e c t i v e l y . F i n d t h e complex number represented by A.

Solution:

Let A be (x, y)

lt is given that BD = 2AC+ MD = 2AM Also DM is perpendicular to AM > (1-2)' + (1+1)2 _{= 4 l$-2)2} + (y+1)21 ...(1 ₎

" n o

Y * 1 . 1 * l

_{= - 1 + 2(y+1)=x-2}

x - 2 1 - 2 W i t h x - 2 = 2 ( y + 1 ) , ( 1 ) g i v e s ( y + 1 ) 2 = l l 4 = y = - 1 1 2 , - 3 1 2 = x = 3 , 1

+ A represents z = 3 - i12, or 1-3i12 Alternative Sol.

MD = 2AM and AM IDM i.e. ZAMD = nl2

= J:-(2:)-= 3Y

"tT

= +it2 > z-(2

- i1

= 1

( 1 + i ) - ( 2 - i ) M D Problem 7.

iu'

Prove the following inequalities geometrically and analytically : t l

@ l_Z

_4

<1arg

zl

llzl I

Solution:

(b) lz - 1l <llzl - 1l + lzllars

zl.

(a) The quantity zllzllies on a unit circle centred at the

l l

origin.

lt is clear

that

_{l= - I}

i z l I

= AP < arc(AP) = a= argz. lf o is negative, this gives

t l

13 - rl< larg

zl. To show

this

analytically,

put

l l z l I

z = r(cosa

+ isina),

where

t l

r = lzl.rnenlS

- tl = llcoscr

- 1) + i sincrl

l l z l I

RSM-91 1 -P3-MA-ComPlex Number

= [(coso - 1l' * sin2cr]1/2 = l2(1 ' cosu)11/2

= (o"rn'9)t"=

zl.ingl

l ' -

2 )

| 2 l

t l

< 2lil = lcrl

= larg zl.[sin0

< e if e >0]

t 2 l

(b) Referring to the shown figure and using the result of Part (a),

we

have

_{lz- t1=}

AP

< AQ

* nf llzl

- 1l + lz-lzll

= l z - 1 1 <

l l z l -

1 l + lzl li-tl < llzl-

1 l + lzl larszl.

I I Z I I

Analytically, we can get this result as follows : l z - 1 1 = l G - l z l ) + f l z l - , 1 ) l < l z , - l z l l + llzl- 1l

- l z -

1 l <

l l z l -

1 l +

l z l f t r l < tt z t -

1 l

_{+ lzl}

l a r s

z l '

Problem 8.

Find z such thatlz - 2 + 2il < 1 and z has

I f,

1

(i) least absolute value Solution:

(i) Given equation represents a circle with centre (2, -2) and radius 1.

Distance of lz - 2 + 2il = lwill be minimum from origin along the line which is normal to the circle passing through the origin, so it will pass through the centre of the circle also. Op = 1Z,lZ _ 1)

OA = OP cos 45'

and OB = OP sin 45' so that / - \

12"t2

-1) .(zJ| -t\

' l - l

, l z

I v z )

(ii) numerically least amplitude

-(ii) we have to find the complex number z represent by Q where oQ is a tangent to the

circle.

Now

sq= .6Eccf =Ei

; 1' -t

= 1 E *(2)\ = J7 .. lzl=

trt .

zQox

=-ltcox -t coQ'

= -

_{|]-rit-1;!-cl}

- 1,, ^:--1 '' .l = -(l- -rin

, -L)

(

" -,"^,-'

l-l

= -la-sin-'p:2

i ll to

2 1 2 ) >

a m p z =

-

[ a - t ' t ' 2 J z )

= z = lzl {cos(amPz)+ i s i n ( a m P z ) }

- [ t ( *

n 1 ) ] l ( " _ . , . - , + l f l

= Jt

l

*,1_[; - sin-'

dl+

isinl-[z

2J2

)) )

1 \ . . ( n - l 1 ) ' l -

_{2J2 )}

l - l s l n l - - S l n - - - - = I | .

_\4

_2.t2

₎₎

f /

-= vzl*'[i - sin-1

RSM-91 1-P3-MA-Com$ex ilumber 3!l

Problem 9.

Find the common tangents of the curues _{R,(z)= lz - 2aJ and jz - 4aj=9".} Solution:

Put z = x + iy in R" (z) = lz- 2aland lz - 4a[ = 3" > \ f = 4 a ( x - a ) a n d ( x - 4 a ) 2 + f = ( 3 a ) 2 Tangent _{to f = 4a (x - a)}

y = m ( x - a ) * a m

For this to be a tangent to (x - 4a)2 + f = P42

t ^ l

lm(3a1+

el

l '

m l = g a + a = 3 = , m = t *

_'

m 2 - J 3

Also, the line x = a is a common tangent to both the curves Hence the common tangents are

1 2 a - x 2 a 1 = ? ' Y = J d * + J T a n d t = J a - J a o r , R " ( z ) = t , t * ' 2 a z - Z

2 J i -. f = z

z + z + 4^ = P-Z\Ji = 1z

- lJdt

I

(1 *..6 i) z + (1 -J5 i)z + 4a= 0.

( 1 -. F )z+ (1+J5 i)z + 4a = o.

Prablem lA,

21,

Zs Zs

are complex

numbers

and p, q, r are real numbers

such

that

P j , = T - L , . P r o v e t h a t P 2 * Qt * " = 0 .

E;1=

lzs - zrl lzr - zzl

zz - Zs zs - 4 zr - zz

Sotutton:

1rys

6"u"--4-

-

g'

^

=

,t

,,

-- r (say)

-

l=,

- =rl' l=, -.,1' l=,

- =rl

= Q ' - f ' = ,

- (=, - zr)(a -zr) (., -.,)(2" -2,) (r,, - =r)(q -zr)

) n 2 . 2

e

P - =k(r"-Z.),:L=k(Zs

_{- q ) , _ ' _ =k(4-=r)}

z z - z g z s - z l ' ' zt -zz

= - - p ' - - * -9'= * -" - =k(zr-zs+zs-4+4-+)=0.

_{z z - z s z g - z t 7 t - 2 2} I i ) I I I

34 RSM-91 1-P3-MA-Complex Number

Objective :

Problem 1.

It arg (4) = arg(z), then ( A ) z ' = k z ; 1 ( k > 0 ) (C) lztl = lZ t Solution:

Problem 2.

f f a > 0 , a e R , z = a + 2 i a n d z l z l - a z + 1 (A) z is always a positive realnumber (C) z is purely imaginary number

(B) zz= kz1 (k > 0) (D) None of these

= 0 then

(B) z is always a negative real number (D) such a complex z does not exist.

(B) 0

(D) none of these 7 = ztzt =12.,l' 2.,-1 = arg(zi|) = ary(4) = arg(22) = zz= kz',-1 1k > 01

' 2 .

Hence (A) is the correct answer.

Solution:

Putting z = a+ 2i in the given equation and comparing imaginary parts, we get a2 + 4 = a2, which is not possible .

Hence (D) is the correct answer. Problem 3.

tf a is the angte which each side of a regular polygon of n srdes subfends at its centre, then 1 + cosa + cos2a + cos3a ... + cos(n-l)a is equal to

(A) n

(c) 1

Solution:

n-1

f cosro

= nef e; = sum of the n roots

of unity

= 0

n-l irl

lf the imaginary paft of the expression (A) a straight line parallelto x-axis (C) a circle of radius 1

' -,1 * "t' o" zero, then locus of z is e o ' z - 1

(B) a parabola (D) none of these

r=0 r=0

Hence (B) is the correct answer. Problem 4.

Solution:

z - 1 e t ' 1 L e t U = 7 + _ - = - .

e o i z - 1 u

Given that imaginary _Part of u +

t

u rs zero.

RSM-91 1-P3-MA-Complex Number 3 5

f " * 1 )

- ( o * 1 ) = o

(

u / \ .

u /

( u - u ) . H : o + t,-ol(''-*) = t

-t . t 2 = l=l = 1 = lz-11= 1, which isacircle. l e ' ' l

Hence (C) is the correct answer.

, t - # = o =lul2=1

Problem 5.

lf 4 , 22, 23 are the vertices area of the triangle having zt,

l o 1411' , 2 , ' . r 3 J 3 l " / 4 Solution: DE = BD - BE = 2*312= 112 Area of AACD = 1 DE T AC 2

= 1 , 1 J a = € .

2 2 4

of an equilateral triangle inscribed _{in the circle lz | = t then the} - 22, 23, as ifs yertices is

@ 9

₄

(D) none of these

Hence (B) is the correct answer. Problem 6.

The complex numbers z = x + iy which satisfy the (A) the x-axis

(C) a circle passing through the origin

eouationlzriLl=1

_'

fieon

l z + 1 i l

(B) the straight line y - 5 (D) None of these Solutlon:

l z - 5 i l

l - l = 1 =lz-sil =lz+5il lz + 5il

+ z would lie on the right bisector of the line segment connecting the points 5i and - 5i Thus z would lie on the x-axis.

Hence (A) is the correct answer. Problem 7.

The inequality lz-l l< f z-2 f represents the region given by

(A) Re (z) > 0 (B) Re (z) . O (C) Re (z) > 2 (D) None of these Solution: lz - 4l < | z-21 + | z- 4l' < | z - 2l' + ( z-Q Q -\ < ( z-2) (z -2) + z 2 - 4 ( z + 2 ) + 1 6 < z 2 - 2 ( z + 2 ) + 4 B (zz) D(-zz)