Complex Numbers
4
INTRODUCTION
Ø Consider a simple quadratic equation x2 + 4 = 0. Clearly
there is no solution of this equation in the set of real numbers. To permit the solution of such equations the set of complex numbers is introduced.
Ø The solutions of the above equation are given by x2 = –4
. 2 1 – 2 4 x=± - =± =± i Þ
Ø Swiss Mathematician Euler introduced the symbol i (iota) for positive square root of – 1, i.e., i = - .1
The square root of a negative number is called an imaginary number.
Ø Now for any two real numbers x and y, we can form a new number x + iy. This number x + iy is called a complex number. The set C of complex numbers is therefore defined by C = {x + iy | x Î R, y Î R}
The extension of concept of numbers from real numbers to complex numbers enabled us to solve any polynomial equation.
Ø A complex number is deonoted by a single letter such as z, w etc. Given a complex number z = x + iy, x is called its Real part and y its Imaginary Part and we denote
x = Re(z) and y = Im (z)
If y = 0, then z = x is a purely real number. If x = 0, then z = iy is a purely imaginary number. The complex number 0 = 0 + i0 is both purely real as well as purely imaginary.
Ø Two complex numbers z1 = x1 + iy1 and z2 = x2 + iy2 are said to be equal if and only if x1 = x2 and y1 = y2. AXIOMATIC APPROACH TOWARDS THE COMPLEX NUMBER SYSTEM
Ø A complex number is defined as an ordered pair (x, y) of real numbers x and y. Thus,
C = {(x, y) | x Î R, y Î R}
If z = (x, y) then x = Re(z) and y = Im (z), z = (x, y) is equivalent to z = x + iy.
Thus, i is equivalent to (0, 1) and z = (x, –y). IMPORTANT RESULTS TO BE MEMORISED ABOUT i
i (greek letter iota) represents positive square root of
–1, so, i= -1. It is called imaginary unit. We have 1. i2=-1,i3 =-i,i4=1,i5 =i,...etc.
Thus for any integer k,
i4k = 1, i4k+1 = i, i4k+2 = –1, i4k+3 = –i.
That is if power of i is m, mÎ , then divide m by 4N and find the remainder.
If the remainder is zero, then im = 1
If the remainder is one, then im = i
If the remainder is two, then im = –1
If the remainder is three, then im = –i
2. The sum of four consecutive powers of i is zero, for example, i12 + i13 + i14 + i15 = 0
3. For any two real numbers a and b, a´ b= ab is true only if at least one number is non negative or zero. If both a and b are negative then a´ b¹ ab In fact if a > 0 and b > 0 then
1 1
a b a b i a i b ab
- ´ - = - ´ - = ´ =
-OPERATIONS ON COMPLEX NUMBERS
Suppose that z1 = (x1, y1) and z2 = (x2, y2) be two complex numbers, that is, z1 = x1 + iy1 and z2 = x2 + iy2
(i) Equality : z1 = z2, if x1 = x2 and y1 = y2
(ii) Addition : z1 + z2 = (x1 + x2, y1 + y2) or equivalently z1 + z2 = (x1 + x2) + i (y1 + y2)
(iii) Subtraction : z1 – z2 = (x1 – x2, y1 – y2) or equivalently z1 – z2 = (x1 – x2) + i (y1 – y2) (iv) Multiplication : z1z2 = (x1x2 – y1y2, x1y2 + x2y1) or equivalently z1z2 = (x1x2 – y1y2) + i (x1y2 + x2y1) (v) Division : If z2 ¹0 then ÷ ÷ ø ö ç ç è æ + -+ + = 2 2 2 2 1 2 2 1 2 2 2 2 2 1 2 1 2 1 y x x y x y , y x y y x x z z or equivalently ÷÷ ø ö ç ç è æ + -+ ÷ ÷ ø ö ç ç è æ + + = 2 2 2 2 1 2 2 1 2 2 2 2 2 1 2 1 2 1 y x x y x y i y x y y x x z z
(vi) Multiplication by a real number : If z = (x, y) and m Î R then mz = (mx, my) or equivalently if z = x + iy then mz = mx + imy
ALGEBRA OF OPERATIONS
If z1, z2 and z3 belong to set C of complex numbers, then following properties hold.
(1) Closure Property : z1 ± z2; z1z2 and
2 1 z z , z2 ¹0 all also belong to C.
(2) Commutative Property : z1 + z2 = z2 + z1 and z1z2 = z2z1 (3) Associative Property : z1 + (z2 + z3) = (z1 + z2) + z3 and
z1 (z2z3) = (z1z2)z3
(4) Cancellation Property : z1 + z3 = z2 + z3 Þ z1 = z2 and z1z3 = z2z3 Þ z3 = 0 or z1 = z2
(5) Distributive Property : z1 (z2 + z3) = z1z2 + z1z3 (6) Existence of Identity : 0 = (0, 0) is additive identity, i.e. 0
+ z = z + 0 = z " z Î C
1 = (1, 0) is multiplicative identity, i.e. 1(z) = (z)1 = z " z Î C (7) Existence of Inverse : For every complex number z = (x, y), we may get a unique number–z = (–x, –y) such that z + (–z) = (–z) + z = 0. (–z) is Additive Inverse.
For every complex number z = (x, y), z¹ we may get a0 unique number z–1 or z 1 = ÷÷ ø ö ç ç è æ + + 2 2 2 2 y x y – , y x x such that 1 z z 1 z 1 z ÷ = ø ö ç è æ = ÷ ø ö ç è æ . ÷ ø ö ç è æ z 1 is multiplicative inverse.
[NOTE : A set with two operations on it satisfying all above properties is called a Field.]
(8) The order relations 'greater than' and 'less than' are not defined for non real complex numbers. The inequalities like –2i < 0; 1 + 2i > 1; i – 1< i are meaning less.
The Modulus and the conjugate of a Complex Number Let z = x + iy be a complex number. Then the modulus (absolute value) of z, denoted by | z | is defined as follows :
| z | = a2+b2 = {Re ( )}z 2+{Im ( )}z 2
Clearly, modulus of a complex number is a real number. Again let z = x + iy be a complex number. Then the complex number x – iy is called the conjugate of z and is denoted by
z or z*.
Thus, we have Re (z ) = Re (z) and Im ( z ) = – Im (z). Note :
1. The additive inverse of ‘z’ is ‘– x – iy’ and conjugate of ‘z’ is ‘x – iy’.
2. The multiplicative inverse of a non-zero complex number ‘z’ can be given by
1 2 2 2 1 | | x iy z z x iy x y z - = = - = + + or z z =| |z 2
IMPORTANT RESULTS TO BE MEMORISED ABOUT CONJUGATE (i) (z)=z (ii) z1±z2=z1±z2 (iii) n n 2 1 2 1z zz (z ) (z) z = Þ = , nÎN (iv) (iz)=-iz (v) , z 0 z z z z 2 2 1 2 1 = ¹ ÷÷ ø ö çç è æ
(vi) z+z=2Re(z), which is a purely real number (vii) z-z=2iIm(z), which is a purely imaginary number.. (viii) z= if and only if z is purely real.z
(ix) z=-z if and only if z is purely imaginary..
(x) If f(z) is a polynomial in a complex variable z, then ) z ( f ) z (
f = [where f means the complex coefficients are replaced by their conjugate is]
(xi) z1z2+z1z2=2Re(z1z2)=2Re(z1z2)
(xii) If 3 2 1 3 2 1 3 2 1 c c c b b b a a a z= then 3 2 1 3 2 1 3 2 1 c c c b b b a a a z=
where ai, bi, ci (i = 1, 2, 3) are complex numbers. ILLUSTRATIVE EXAMPLES
1. The value of the sum
å
+ = 7 n 4 1 k k i is (a) 0 (b) 1
(c) –1 (d) i or –i, depending on n is even or odd
Sol. We have, 1 0 1 7 n 4 4 k k 3 2 7 n 4 1 k k = + + +
å
= - - + =-å
+ = + = i i i i i i i[Note that, k = 4 to 4n + 7 contains 4n + 4 terms, a multiple
2. If ,x,y R 3 y ) 3 2 ( 3 2 x ) 1 ( Î = -+ -+ + -+ i i i i i i i , then (a) x = 3, y = –1 (b) x = –3, y = 1 (c) x = 3, y = 1 (d) x = –3, y = –1 Sol. We have i i i i i i i = -+ -+ + -+ 3 y ) 3 2 ( 3 2 x ) 1 (
Multiplying both sides by (3 + i) (3 – i), we get
[(1 + i)x – 2i] (3 – i) + [(2 – 3i)y + i] (3 + i) = (3 + i) (3 – i)i Þ (4x + 9y – 3) + (2x – 7y – 3)i = 10i
Þ 4x + 9y – 3 = 0 and 2x – 7y – 13 = 0
Solving these equations, we get x = 3, y = –1 Answer (a) 3. If (cos x sin x)(cos y sin y) A B
(cot u )(1 tan v) + + = + + + i i i i i , then
(a) A = sinu cosv cos (x + y – u – v) (b) B = sinu cosv sin(x + y – u – v) (c) A = cosu sinv cos(x + y – u – v) (d) B = cosu sinv sin(x + y – u – v) Sol. We have, (cos x sin x)(cos y sin y)
(cot u )(1 tan v) + + + + i i i i ÷ ø ö ç è æ + ÷ ø ö ç è æ + + + = v cos v sin 1 u sin u cos ) y sin y )(cos x sin x (cos i i i i ) v sin v )(cos u sin u (cos ) y sin y )(cos x sin x (cos v cos u sin i i i i + + + + =
=sinu[cos(cosvu[cos(+vx)++iysin()+uisin(+vx)]+y)]´[cos([cos(uu++vv))--iisin(sin(uu++vv)])] ) v u ( sin ) v u ( cos )] v u y x sin( ) v u y x [cos( v cos u sin 2 2 + + + -+ + -+ = i ) v u y x cos( v cos u sin + -
-= +isinucosvsin(x+y-u-v) ) v u y x cos( v cos u sin A= + - -\
and B=sinucosvsin(x+y-u-v) Answers (a, b) 4. If z1z2z3 are there complex numbers then the value of
z1 Im (z2z3)+z2 Im(z3z1)+z3Im(z1z2)is (a) Re (z1z2z3) (b) Im (z1z2z3) (c) Re (z1 + z2 + z3) (d) 0 Sol. Let z1=x1+iy1, z2=x2+iy2, z3=x3+iy3 1 1 Im( 2) ( 1 1) Im ( 2– 2)( 3 3) Then z z z = x +iy éë x iy x +iy ùû 1 1 2 3 2 3 2 3 3 2 (x iy) Im (x x y y ) i x y( –x y ) = + éë + + ùû 1 1 2 3 3 2 (x iy )(x y –x y ) = + 1( 2 3– 3 2) 1( 2 3– 3 2) x x y x y iy x y x y = + Similarly, z2Im(z3z2)=x2(x3y1–x1y3)+iy2(x3y1–x1y3) and z3Im(z1z2)=x3(x1y2–x2y1)+iy3(x1y2–x2y1) Clearly z1Im(z2z3)+z2Im(z3z1)+z3Im(z1z2)=0
Answer (d) 5. If (x + iy)1/3 = a + ib, where x, y, a, b Î R, show that
2 2 2( ) x y a b a- = -b + Sol. (x + iy)1/3 = a + ib Þ x + iy = (a + ib)3
i.e., x + iy = a3 + i3b3 + 3iab (a + ib)
= a3 – ib3 + i 3a2b – 3ab2 = a3 – 3ab2 + i (3a2b – b3) Þ x = a3 – 3ab2 and y = 3a2b – b3 Thus x a = a 2 – 3b2 and y b = 3a 2 – b2 So, x y a-b = a 2 – 3b2 – 3a2 + b2 = – 2a2 – 2b2 = – 2(a2 + b2).
6. Solve the equation z2 = z , where z = x + iy
Sol. z2 = z Þ x2 – y2 + i2xy = x – iy Therefore, x2 – y2 = x ... (1)
and 2xy = – y ... (2)
From (2), we have y = 0 or x = – 1 2 When y = 0, from (1), we get
x2 – x = 0, i.e., x = 0 or x = 1. When x = – 1 2, from (1), we get y2 = 1 1 4+ or y2 2 = 3 4, i.e., y = 3 2 ± .
Hence, the solutions of the given equation are
0 + i0, 1 + i0, – 1 2 + i 3 2 , – 1 2 – i 3 2
4.1
Solve following problems with the help of above text andexamples. 1. The value of 1 i i i i i i i i i i 574 576 578 580 582 584 586 588 590 592 -+ + + + + + + + is (a) 2 (b) –2 (c) 1 (d) –1 2. 1 + i2 + i4 ... + i2n is
(a) Positive (b) Negative
(c) 0 (d) Can’t be determined 3. The value of
å
= + + 13 1 n 1 n n ) i i ( , where i = 1 - equals (a) i (b) i – 1 (c) – i (d) 04. The least positive integer n such that n i 1 i 2 ÷ ø ö ç è æ + is a positive integer is (a) 2 (b) 4 (c) 8 (d) 16
5. The multiplicative inverse of (6+5i)2 is (a) i 61 60 61 11 - (b) i 61 60 61 11 + (c) i 61 60 619 - (d) None of these
6. The multiplicative inverse of i 5 4 i 4 3 -+ is (a) i 25 31 258 + - (b) i 25 31 258 -(c) i 25 31 258 -- (d) None of these
7. The conjugate complex number of (1 2i)2
i 2 is (a) 252 2511÷i ø ö ç è æ + ÷ ø ö ç è æ (b) 252 2511÷i ø ö ç è æ -÷ ø ö ç è æ (c) 25 i 11 25 2 ÷ ø ö ç è æ + ÷ ø ö ç è æ - (d) i 25 11 25 2 ÷ ø ö ç è æ -÷ ø ö ç è æ -8. The value of çèæ - ÷øö + ÷ ø ö ç è æ + + - 2 4i i 4 3 i 1 3 i 2 1 1 is (a) i 4 9 4 1 - (b) i 4 9 4 1 + (c) i 4 9 4 1 + - (d) i 4 9 4 1 -9. The value of - q+ q sin i 2 cos 1 1 is (a) + q q -q + 5 3cos 2 tan 2 i cos 3 5 1 (b) q + q + q + 5 3cos 2 tan 2 i cos 3 5 1 (c) + q q -q + 5 3cos 2 cot 2 i cos 3 5 1 (d) None of these 10. If 3a-ib=x-iy,then 3 + = ib a (a) x+iy (b) x-iy (c) y+ix (d) y-ix
GRAPHICAL REPRESENTATION OF COMPLEX NUMBERS N (0, y) P (x,y) O M (x,o) y Q (x,–y) x
Ø Every complex n umber x+iy can be r epr esen ted geometrically as a unique point P (x,y) in the xoy plane with x-coordinate representing its real part and y-coordinate representing its imaginary part.
Ø The Point (x, 0) on the x-axis represents the purely real number x. As such x-axis is called the real axis. Similarly, the point (0, y) on the y-axis represents purely imaginary number iy. Therefore, y-axis is called the imaginary axis. Ø The plane having a complex number assigned to each of its points, is called the complex plane or Argand plane or Guassian plane. This representation of complex numbers as points in the plane is known as Argand diagram.
Ø The distance from the origin to the point P(x, y) is defined as the MODULUS (or absolute value) of the complex number z = x + iy, denoted by | z |, thus | z | = x2+y2 Ø The conjugate z of complex number z is represented by
the point Q, which is the mirror image of P on the x–axis.
IMPORTANT RESULTS ABOUT MODULUS 1. |z|³ , 0 |z|=0Ûz=0.
2. zz=|z|2 ( majority of the complex equations are solved using this property)
3. |z|=|z|=|-z|=|-z| 4. |z1z2|=|z1||z2|Þ|zn|=|z|n 5. |z | | z | z z 2 1 2 1 = , z2 ¹0
6. |zz| =1Þ|zz|, is a unimodular complex number (z¹ ).0 7. |z1+z2|2=|z1|2+|z2|2 +2Re(z1z2) 8. |z1-z2|2=|z1|2+|z2|2-2Re(z1z2) 9. |z1+z2|2+|z1-z2|2=2(|z1|2+|z2|2) 10. 2 2 2 2 2 2 1 2 1 2 1 2 | az +bz | +| bz -az | =(a +b )(| z | +| z | ), where a, b Î R 11. If z1,z2¹ then 0 |z1+z2|2=|z1|2 +|z2|2 2 1 z z Û is purely imaginary. ANSWER KEY 1. (b) 2. (d) 3. (b) 4. (c) 5. (a) 6. (c) 7. (d) 8. (b) 9. (c) 10. (a)
POLAR FORM (OR TRIGONOMETRICAL FORM) OF COMPLEX NUMBERS P (z) O M y X X q r = |z | y Let P r epresents the n onzer o
complex number z = x+iy. Let the directed line regment OP be of length r and makes an angle q with the positive direction of the x–axis (q in radians)
The point P is uniquely determined by the ordered pair of real numbers (r, q) called the polar coordinates of the point P. Clearly,
x = r cos q, y = r sin q, r= x2+y2 , tanq =
x y
Thus z=r(cosq + isin q) is the poar from of z. r is the modulus of th e n umber z an d q is called the ARGUMENT (or AMPLITUDE) of the number z, denoted by arg (z) or amp (z) Hence. 2 2 y x | z | r= = + and x y tan ) z ( arg = -1 = q
z = r(cosq + i sinq) is also written as r cis(q)
Ø Note that q is not defined uniquely, In fact q is the solution of simultaneous equations 2 2 y x x cos + = q and 2 2 y x y sin + = q If z = 1 + i, r= x2+y2 = 12+12 = 2; 4 1 x y tanq= = Þq= p Polar from of ÷ ø ö ç è æ p+ p = 4 sin i 4 cos 2 z
Clearly the possible arguments of the number z=1+i are the following angles : I Î p + p p + p p k , k 2 4 .., ,... 2 4 , 4
Any two arguments of a complex number differ by a number which is a multiple of 2 p.
The unique value of q, such that -p<q£p is called the principal value of the Argument.
WORKING RULE FOR FINDING PRINCIPAL ARGUMENT
Let z = x + iy has image P on the argand plane and
x y tana= ,x¹0, y¹0 y y¢ a q = P(z) x x O x¢ y
Following cases may arise
Case I : If x > 0, y > 0, then the point P lies in the first quadrant and then q=argz=a
For example, if z=1+i, then arg(z) = 4 p
Case II : If x < 0, y > 0, then the point P lis in the second quadrant and then q=argz=p-a
For example, if z=-1+ 3i, then 3 1 3 tan = -= a so 3 2 3 ) z arg( =p-p= p y¢ q P x O x¢ y a
Case III : If x < 0, y < 0 then the point P lies in the third quadrant and then
p -a = a -p -= = q argz ( ) y¢ q P x O x¢ y a For example, if z=-1-i then 1 1 1 tan = -= a so 4 3 ) 4 ( ) z arg( =-p-p =- p
Case IV : If x > 0, y < 0 then the point P lies in the fourth quadrant and thenq=argz=-a
For example, if z= 3-i, then 3 1 3 1 tana= - = y¢ q P x O x¢ y a so, 6 ) z arg( =-p
Case V : If y = 0, then z is purely real and P lies on real axis, and z = x,
so arg (z) = 0 if x > 0; arg (z) = p if x < 0 For example arg(3) = 0 and ÷=p
ø ö ç è æ -2 1 arg
Case VI : If x = 0, then z = iy is a purely imaginary number and P lies on imaginary axis
So, arg (z) = 2 p if y > 0 and arg (z) = 2 p - if y < 0 For example, 2 ) i 2 arg( = and p 2 ) i 100 arg(- =-p
IMPORTANT RESULTS ABOUT ARGUMENT 1. arg(z)=-arg(z)
2. arg(z1z2)=arg(z1)+arg(z2)+2kp
3. ÷÷= - + p ø ö çç è æ k 2 ) z arg( ) z arg( z z arg 1 2 2 1 4. arg z 2 arg(z) 2k z æ ö = + p ç ÷ è ø 5. arg(zn)=narg(z)+2kp
Where k = 0, –1 or 1 would be taken so that the value comes out into the principal value region
6. If ÷÷=q ø ö çç è æ 1 2 z z arg the ÷÷= p-q ø ö çç è æ k 2 z z arg 2 1 , kÎI 7. arg(z)-arg(-z)=±p 8. arg(z) 2 ) iz arg( =p+ 9. |z1+z2|=|z1-z2|Ûarg(z1)-arg(z2)=p/2 10. |z1+z2|=|z1|+|z2|Ûarg(z1)=arg(z2) EULER'S NOTATION
It can be shown that eiq =cosq+isinq, e-iq=cosq-isinq \ ez =ex+iy =ex.eiy =ex(cosy+isiny)
Also r(cosq+isinq)=reiq Again, i 2 e e sin and 2 e e cos i i i iq -q q- -q = q + = q
LOGARITHM OF A COMPLEX NUMBER Let z=x+iy=r(cosq+isinq)=reiq
Then loge(z)=loge(reiq)=loger+iq So, loge(z)=loge|z|+iarg(z)
As such the argument of a complex number is not unique, the log of a complex number cannot be unique. In general,
)], z arg( k 2 [ i | z | log ) z ( loge = e + p+ kÎI
For example, log(i) logei /2 i 2k 2÷, kÎI ø ö ç è æ p + p = = p So, 2 ip is one of the values of log (i)
Also, log(logi)=log(ip2)=logi+logçèæ p2øö÷=i2p+logæ pèç2ö÷ø [Taking principal value only]
7. If z1 and z2 are any two complex numbers then | z z z | | z z z | 1+ 12- 22 + 1- 12- 22 is equal to (a) |z1| (b) |z2| (c)|z1+z2| (d) |z1+z2|+|z1-z2| Sol. (Trick ) The nature of the problem suggests at once that
we shold use the formula
) | z | | z (| 2 | z z | | z z | 1+ 2 2+ 1- 2 2= 1 2+ 2 2 We have(|z1+ z12-z22 |+|z1- z12-z22)2 | z z z | 2 | z z z | | z z z | 1+ 12- 22 2+ 1- 12- 22 2+ 12- 12+ 22 = 2 2 2 2 2 2 1 2 1| | z z | 2|z | z | 2êëé + - úûù+ = = 2(|z1|2 +|z2|2)+2|z12-z22| =|z1+z2|2+|z1-z2|2 +2|z1+z2||z1-z2| = (|z1+z2|+|z1-z2|)2 \|z1+ z12-z22 |+|z1- z12-z22 | | z z | | z z | 1+ 2 + 1- 2 = Answer (d)
8. The smallest positive integer n for which 1 i – 1 i 1 n = ÷ ø ö ç è æ + is (a) 4 (b) 3 (c) 2 (d) 1 Sol. i 2 i 2 i – 1 i 2 i 1 i 1 i 1 i – 1 i 1 i – 1 i 1 2 2 = = + + = + + ´ + = + 1 i 1 i – 1 i 1 n n = Þ = ÷ ø ö ç è æ + \
Clearly the smallest value of n is 4. Answer (a) 9. If a+ib =1. the simplified form of
ai b 1 ai b 1 -+ + + is (a) b + ai (b) a + bi (c) (1 + b)2 + a2 (d) ai Sol. As a+ib =1Þa2+b2=1 = ) ai b 1 )( ai – b 1 ( ) ai b 1 ( ai – b 1 ai b a 2 + + + + + = + + + 2 2 2 2 a ) b 1 ( ai ) b 1 ( 2 a – ) b 1 ( + + + + + = = b 2 ) b a ( 1 abi 2 ai 2 b 2 b ) a – 1 ( 2 2 2 2 + + + + + + + b 2 1 1 abi 2 ai 2 b 2 b b2 2 + + + + + + = = b ai b 1 i ) b 1 ( a ) b 1 ( b b 1 abi ai b b2 = + + + + + = + + + + Answer (a) 10. The argument of the complex number 1+sina-icosa is
(a) 4 2 p -a (b) 2 a (c) 4 2 p + a (d) 2 2 p + a Sol. Let 1+ sina – icosa = r(cosq + isinq)
Then r cos q = 1 + sina and r sinq = – cosa r2 = (1 + sina)2 + (–cosa)2 = 2 + 2 sina =
2 2 sin 2 cos 2 ÷ ø ö ç è æ a+ a r = ÷ ø ö ç è æ a+ a 2 sin 2 cos 2 or ÷ ø ö ç è æp-a 2 4 cos 2
Also, tan q = ÷ ø ö ç è æp-a + ÷ ø ö ç è æp-a -= a + a -2 cos 1 2 sin sin 1 cos
[Note the step]
= ÷ ø ö ç è æp-a ÷ ø ö ç è æp-a ÷ ø ö ç è æp-a -2 4 cos 2 2 4 cos 2 4 sin 2 2 ÷ ø ö ç è æa-p = ÷ ø ö ç è æp-a -= 4 2 tan 2 4 tan \ q = 4 2 p -a Hence, Modulus = ÷ ø ö ç è æp-a 2 4 cos 2 and argument = 4 2 p -a . Answer (a)
11. Convert the complex number 16
1 i 3
-+ into polar form. Sol. The given complex number 16
1 i 3 -+ = 16 1 i 3 -+ × 1 3 1 3 i i -= 16 (1 23) 1 ( 3) i i - -- = 16 (1 3) 1 3 i - -+ = – 4 (1-i 3) = – 4 + i4 3 q O Y X Y' X' P( 4, 4 3)
-Let – 4 = r cos q, 4 3 = r sin q By squaring and adding, we get
16 + 48 = r2 (cos2 q + sin2 q) which gives r2 = 64, i.e., r = 8
Hence, cos q = – 1 2, sin q = 3 2 q = p – 3 p = 2 3 p
Thus, the required polar form is 8 cos 2 sin2
3 i 3
p p
æ + ö
ç ÷
è ø
12. If | z2 – 1| = | z |2 + 1, then show that z lies on imaginary axis.
Sol. Let z = x + iy. Then | z2 – 1| = | z |2 + 1 Þ | x2 – y2 – 1 + i2xy | = | x + iy |2 + 1 Þ (x2 – y2 – 1)2 + 4x2y2 = (x2 + y2 + 1)2 Þ 4x2 = 0 i.e., x = 0
Hence z lies on y-axis.
4.2
Solve following problems with the help of above text and examples.
1. For any two complex number z1, z2 2 2 1 2 2 1z | |z z | z 1 | - - - is equal to : (a) (1+|z1|2)(1+|z2|2) (b) (1-|z1|2)(1-|z2|2) (c) (1+|z1|2)(1-|z2|2) (d) (1-|z1|2)(1+|z2|2)
2. If z1, z2 and z3, z4 are two pairs of conjugate complex
numbers, then arg ÷÷
ø ö çç è æ + ÷ ÷ ø ö ç ç è æ 3 2 4 z z arg z z1 equals (a) 0 (b) 2 p (c) 2 3p (d) p 3. If |b| = 1, then 1b--aab is equal to (a) 0 (b) ½ (c) 1 (d) 2
4. Let z1 and z2 be complex numbers such that z1¹z2 and
|z1| = |z2|. If z1 has positive real part and z2 has negative
imaginary part, then
2 1 2 1 z z z z -+ may be (a) real and positive
(b) zero
(c) real and negative
(d) either zero or purely imaginary
5. If 1 2 z 7 z 5
is purely imaginary number then
2 1 2 1 z 3 z 2 z 3 z 2 -+ is equal to (a) 7 5 (b) 9 7 (c) 49 25 (d) None
6. If z1 and z2 be two non-zero complex numbers such that | z z | | z z
| 1+ 2 = 1- 2 then arg (z1) – arg (z2)= (a) 2 p (b) 0 (c) 2 p - (d) 4 p
7. The Argument of the complex number z =
) 3 i – 1 ( i 4 ) 3 i 1 ( + 2 is (a) 6 p (b) 4 p (c) 2 p (d) None of these 8. The modulus of the complex number
z = ) sin i – (cos ) i – 1 ( 2 ) sin i (cos ) 3 i – 1 ( q q q + q is (a) 2 1 (b) 2 2 1 (c) 3 1 (d) None of these 9. If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then
(a2 + b2) (c2 + d2) (e2 + f ) (2 g + 2 h ) =2
(a) A2 – B2 (b) A2 + 2
B
(c) A4+B4 (d) A4–B4
10. If q is real, then the modulus of
q + q +cos ) isin 1 ( 1 is (a) 2 1 sec 2 q (b) 2 1 cos 2 q (c) sec 2 q (d) cos 2 q DE MOIVRE THEOREM
These are two statements of De Moivre Theorem 1. (cosq+i sinq)n = cos nq + i sin nq, nÎI
2. If nÎQ ÷÷ ø ö çç è æ Î ¹ = ,q 0,p,q I q p
n then cos nq+i sinnq
is one of the values of (cosq+ i sinq)n
IMPORTANT RESULTS
1. Take care that (sinq +i cos )q ¹n (sin nq +i cos n )q in fact
n ) cos i (sinq+ q = n 2 sin i 2 cos ú û ù ê ë é ÷ ø ö ç è æp-q + ÷ ø ö ç è æp-q = ÷ ø ö ç è æ p- q + ÷ ø ö ç è æ p- q n 2 n sin i n 2 n cos
2. (cosa +i sin )(cosa b +i sin )(cosb g +i sin )....g = cos(a + b + g +...) i sin(+ a + b + g +...) 3. cos( ) isin( ) sin i cos sin i cos = a-b + a-b b + b a + a
ROOTS OF A COMPLEX NUMBER
If z=r(cosq+isinq) and n is a positive integer, then n 1 n 1 )] sin i (cos r [ ) z ( = q+ q
[
]
n 1 n cos(2k ) isin(2k ) r p+q + p+q = 1 k = 0, 1, 2, ..., n – 1 ANSWER KEY1. (b) 2. (a) 3. (c) 4. (d) 5. (d) 6. (a) 7. (c) 8. (a) 9. (b) 10. (a)
ú û ù ê ë é ÷ ø ö ç è æ p+q + ÷ ø ö ç è æ p+q = \ n k 2 sin i n k 2 cos r ) z ( n 1 n 1
Putting k = 0, 1, 2, ..., n – 1, we get n values which represent nth roots of complx number z
PROPERTIES
1. These n roots always form a G. P. with common ratio ei 2p/
n
2. Their images on argand diagram lie on a circle of radius
n / 1
r and centre origin.
3. The points representing these roots from the vertices of a regular polygon of n sides
y¢ P1 x O y P2 P3 Pn q 2q
CUBE ROOTS OF UNITY
3 k 2 sin i 3 k 2 cos ] k 2 sin i k 2 [cos ) 1 ( z= 1/3= p+ p1/3= p+ p, k = 0, 1, 2 So, three cube roots of unity are 1,
3 2 sin i 3 2 cos p+ p and 3 4 sin i 3 4 cos p+ p or 1, 2 3 i 1+ and 2 3 i 1 -. As ÷÷ ø ö ç ç è æ- -= ÷ ÷ ø ö ç ç è æ- + 2 3 i 1 2 3 i 1 2 and 2 1 i 3 1 i 3 2 2 æ- - ö =æ- + ö ç ÷ ç ÷ ç ÷ ç ÷ è ø è ø
So, we denote the non-real roots by w and w2 we write mostly 2 3 i 1+ -= w and 2 3 i 1 2= - -w ALGEBRAIC METHOD Let z = (1)1/3 0 1 z3- = Þ Þ(z-1)(z2+z+1)=0, which gives the roots of z as z = 1, the real root and
2 3 i 1
z=- ± ,the non-real roots , i.e., w and w2. Clearly, we can always write
) z )( z )( 1 z ( ) 1 z ( 3- = - -w -w2 IMPORTANT CHARACTERISTICS OF w 1. 1+w+w2=0 2. w=w2 and (w )2 =w, so 1 3= w = w w 3. |w|=|w2|=1 4. 3 2 )
arg(w = p and arg( 2) 4 3 p w = or 3 2p
-5. For nay positive integer k, w3k =1, w3k+1=w and 2
2 k
3 =w
w +
6. For any real a, b, c; a+bw+cw2 =0 Þa=b=c 7. The cube roots of unity lie on a unit circle and divide the
circumference into three equal parts
8. The points represented by cube roots of unity form the vertices of an equilateral triangle.
9. |wz|=|w||z|=|z| 10. arg(z) 3 2 ) z arg(w = p+ –1 1 w w2 i –i x x¢ y y¢
SOME USEFUL IDENTITIES (i) x2+y2=(x+iy)(x-iy) (ii) x3+y3=(x+y)(x+wy)(x+w2y) ) y x )( y x )( y x ( + w+ w2 w2+ w = (iii) x3-y3 =(x-y)(x–wy)(x–w2y)
(iv) x2 + xy + y2 = (x – yw) (x – yw2), in particular,
x2 + x + 1 = (x – w) (x – w2) (v) x2 – xy + y2 = (x + yw) (x + yw2), in particular, x2 – x + 1 = (x + w) (x + w2) (vi) x2 + y2 + z2 – xy – xz – yz = (x + yw + zw2) (x + yw2 + zw) (vii) x3+y3+z3–3xyz ) zx – yz – xy – z y x )( z y x ( + + 2+ 2+ 2 = = (x+y+z)(x+yw+zw2)(x+yw2+zw) nTH ROOTS OF UNITY Since 1=cos0+isin0, therefore
n / 1 n / 1 ) 0 sin i 0 (cos ) 1 ( = + =(cos 2kp +i sin 2k )p1/ n cos2 k i sin2 k n n p p = + ; k = 0, 1, 2, ..., (n – 1) i2k n e p = ; k = 0, 1, 2, ...., (n – 1) =1,e(i2p/n),e(i4p/n),...,e[i2(n-1)p/n] =1,a,a2,a3,....,an-1 where a=e(i2p/n) SQUARE ROOTS OF A COMPLEX NUMBER
Let z = x + iy and let the square root of z be the complex number a + ib. Then
ib a iy
x+ = + or (x+iy)=(a+ib)2 =(a2-b2)+(2ab)i Equating real and imaginary part, we get
2 2 b a x= - …(1) and y = 2ab …(2) Now, a2+b2= (a2-b2)2+4a2b2 = x2+y2 …(3) Solving the equations (1) and (3), we get
÷÷ ÷ ø ö çç ç è æ + + ± = 2 x y x a 2 2 ; ÷÷÷ ø ö çç ç è æ + -± = 2 x y x b 2 2
From (2), we can determine the sign of ab. If ab > 0, then a and b will have same sign. Thus
ú ú ú û ù ê ê ê ë é ÷÷ ÷ ø ö çç ç è æ + -+ ÷÷ ÷ ø ö çç ç è æ + + ± = + 2 x y x i 2 x y x iy x 2 2 2 2 If ab < 0, then ú ú ú û ù ê ê ê ë é ÷÷ ÷ ø ö çç ç è æ + -÷÷ ÷ ø ö çç ç è æ + + ± = + 2 x y x i 2 x y x iy x 2 2 2 2
Thus, square roots of z = a + ib are :
ú ú û ù ê ê ë é -+ + ± 2 a | z | i 2 a | z | for b > 0 and ú ú û ù ê ê ë é -+ ± 2 a | z | i 2 a | z | for b < 0 For example :
(i) Square root of i is ÷÷ ø ö çç è æ + ± 2 i 1 , as x = 0, y = 1 > 0 and | i | = 1
(ii) Square root of -i is ÷÷ ø ö çç è æ -± 2 i 1 , as x = 0, y = -1 < 0 and | i | = 1
(iii) Squar e root of -3 + 4i is ÷÷
ø ö ç ç è æ - -+ -± 2 ) 3 ( 5 i 2 3 5 , or ±(1+2i), As x = –3 and y = 4 > 0.
(iv) Squar e root of 1- 3i is ÷÷ ø ö ç ç è æ -+ ± 2 1 2 i 2 1 2 or ÷ ÷ ø ö ç ç è æ -± 2 i 3 ; As y=- 3<0 13. 20 15 20 15 ) i 1 ( ) 3 i 1 ( ) i 1 ( ) 3 i 1 ( + -+ -+ is equal to (a) 132 (b) 64 (c) – 64 (d) none
Sol. (Trick :Any complex number z with |Re(z) | : | Im(z) | = 1 : 3 or 3:1 can be expressed in terms of w and w2)
Let z = 20 15 20 15 ) i 1 ( ) 3 i 1 ( ) i 1 ( ) 3 i 1 ( + -+ -+ -=
[
]
[
2]
10 15 15 10 2 15 15 ) i 1 ( 2 3 i 1 2 ) i 1 ( 2 3 i 1 2 + ÷ ÷ ø ö ç ç è æ- -+ -÷ ÷ ø ö ç ç è æ- + 5 2 10 3 5 5 2 5 3 5 10 15 2 15 10 15 15 ) i ( ) ( 2 ) i ( ) ( 2 ) i 2 ( ) ( 2 ) i 2 ( ) ( 2 + w = w + w -w = 64 2 25- 5= -= Answer (c)[You can also convert the numbers into polar form and apply De Moivre theorem but above approach is better] 14. The value of (2 – w) (2 – w2) (2 – w10) (2 – w11) is (a) 49 (b) 16 (c) 16w (d) 49w2 Sol. w10 = w9.w = w, w11 = w9.w2 = w2 \ (2 – w) (2 – w2) (2 – w10) (2 – w11) = (2 – w)2 (2 –w2)2 Now x3 – 1 = (x – 1) (x – w) (x – w2), dividing by x – 1 \ x2 + x + 1 = (x – w) (x – w2) or (x2 + x + 1)2 = (x – w)2 (x – w2)2.
Put x = 2, (2 – w)2 (2 – w2)2 = (7)2 = 49 Answer (a) 15. If w is a complex cube root of unity, then the value of
2 2 2 2 a c b c b a b a c c b a w + w + w + w + + w + w + w + w + is (a) –1 (b) 1 (c) a + b + c (d) 0 Sol. 2 2 2 2 a c b c b a b a c c b a w + w + w + w + + w + w + w + w + 2 2 2 2 a c b c b a b a c c b a w + w + ÷ ø ö ç è æ + + w w w + w + w + ÷÷ ø ö çç è æ + w + w w = = 1 a c b ) c b a ( b a c ) c b a ( 2 2 2 2 2 2 -= w + w = w + w + w + + w w + w + w + + w + w w = ú ú û ù ê ê ë é w = w w = w . etc 2 3 1 Answer (a) 16. If cos6x =
å
= 6 0 r r rcos x a , then (a)å
= = 6 0 r r 1 a (b) a 0 = 0 (c) a6 = 1 (d) a1 = a3 = a5Sol. [To find the value of cos nx and sin nx in ascending
powers of cos x or sin x we expand (cosx+isinx)n using
De Moivre theorem and Binomial theorem then equate real and imaginary parts from two expansion to get the required identity as done for above example]. We have (cosx + i sinx)6 = cos6x + isin6x ...(i)
(Using De Moivre theorem) Also (cos x + isinx)6
= x sin i C x sin i x cos C x sin i x cos C x sin i x cos C x sin i x cos C x sin i . x cos C x cos C 6 6 6 6 5 5 5 6 4 4 2 4 6 3 3 3 3 6 2 2 4 2 6 5 1 6 6 0 6 + + + + + +
(Using Binomial theorem)
= 6C0cos6x-6C2cos4xsin2x+6C4cos2xsin4x-6C6sin6x
+ i[6C1cos5xsinx-6C3cos3xsin3x+6C5cosxsin5x] ...(ii)
From (i) & (ii), equating the real parts we get Cos 6x = x sin C x sin x cos C x sin x cos C x cos C 6 6 6 4 2 4 6 2 4 2 6 6 0 6 -+ -= 3 2 2 2 2 2 4 6 ) x cos 1 ( ) x cos 1 ( x cos 15 ) x cos 1 ( x cos 15 x cos -+
= 32 cos6x – 48 cos4x + 18 cos2x – 1= a cos6x 6 0 r r
å
= We get, a0 = – 1, a1 = 0, a2 = 18, a3 = 0, a4 = – 48, a5 = 0, a6 = 32 Clear ly a a0 a1 a2 a3 a4 a5 a6 1 6 0 r r = + + + + + + =å
= .17. If sin 5x =
å
= 5 0 r r rsin x a , then (a)å
= = 5 0 r r 1 a (b) a 1 = a3 = a5 (c) a0 = a2 = a4 (d) a1+a3+a5=a0+a2+a4 Sol. Solve as above, expand (cos x + i sinx)5 using two theoremsand equate imaginary parts. Answer (a, c)
18. If sin6x =
å
= 6 0 r rcosrx a . Then (a) a0 = 0 (b) a1 = a3 = a5 (c) a2 + a6 = 2 1 (d) 2a0 + a1 + 3a2 + a3 + a4 + a5 + a6 = 0Sol. To express cosnx or sinnx as a series of multiple angles
of cos and sin, we use Euler's representation cosx =
2 e eix+ -ix and sinx = i 2 e eix- -ix
and expand cosnx =
n ix ix 2 e e ÷ ÷ ø ö ç ç è æ +
by binomial theorem etc. as for the above example. sin6x = 6 ix ix i 2 e e ÷ ÷ ø ö ç ç è æ -
-[
]
ix 6 6 6 ix 5 ix 5 6 ix 4 ix 2 4 6 ix 3 ix 3 3 6 ix 2 ix 4 2 6 ix ix 5 1 6 ix 6 0 6 6 6 e C e e C e e C e e C e e C e . e C e C i 2 1 -+ -+ -+ -= =[
]
3 6 ix 2 ix 2 2 6 ix 4 ix 4 1 6 ix 6 ix 6 0 6 6 C ) e e ( C ) e e ( C ) e e ( C 2 1 -+ + + -+ -] C C [Qn r =n n-r=
[
2cos6x 6 2cos4x 15 2cos2x 20]
64 1 -´ + ´ = 8 5 x 2 cos 32 15 x 4 cos 16 3 x 6 cos 32 1 + -+ - =å
= 6 0 r rcosrx a We get a0 = 8 5 , a1 = 0, a2 = 32 15 - , a 3 = 0, a4 = 16 3 , a5 = 0 and a6 = 32 1 - .Only (d) is satisfied with these values. Answer (d) 19. If a, b, g are the cube roots of a negative number p then for any three real numbers x, y, z, the value of ba++ gb++ ag
z y x z y x is (a) 2 i 3 1 -(b) (x + y + z)i (c) ip (d) x2 2yi÷p ø ö ç è æ + Sol. Let q = –p, q > 0 Cube root of p = p1/3 = (–q)1/3 = –q1/3. (1)1/3
Cube roots of p are – q1/3, –q1/3w, –q1/3w2
Let a = –q1/3, b = –q1/3w, g = –q1/3w2 \ ) q ( z ) w q ( y ) w q ( x ) w q ( z ) w q ( y ) q ( x z y x z y x 3 / 1 2 3 / 1 3 / 1 2 3 / 1 3 / 1 3 / 1 -+ -+ -+ -+ -= a + g + b g + b + a = w 12 3i z yw xw ) z yw xw ( w z yw xw zw yw x 2 2 2 2 2 2 - -= = + + + + = + + + + 1 3i or 2 æ- + ö ç ÷ è ø Answer (a)
4.3
Solve following problems with the help of above text andexamples. 1. If z = 50 17 ) i – 1 ( ) i 3 ( +
, then amp (z) is equal to
(a) 3 2p (b) – 3 p (c) 3 2p - (d) None 2. Arg 2 6 ) 3 i – 1 ( 4 ) i 3 ( i + is equal to (a) – 6 p (b) 6 p (c) 3 10 p (d) 10 5p
3. For z = cos q + i sin q, then the value of 2n n 2 z 1 z 1 + is (a) –i tan n q (b) i tan n q (c) tan n q (d) i 4. If x = a + b, y = aw + bw2 and z = aw2 + bw, then x3+y3+z3
=
(a) 3 (a3 + b3) (b) (a + b)3 (c) a3 + b3 – a2 b – ab2 (d) None of these 5. If 1, w, w2 are the cube roots of unity then
(1 – w + w2)5 + ( 1 + w – w2)5 =
(a) 32 (b) 0
(c) 32w (d) 16w2
6. If 1, w, w2 are the cube roots of unity then
(3 + 3w + 5w2)6 – (2 + 6w + 2w2)3 =
(a) 0 (b) 64 (c) 36 (d) – 36
7. xÎR, then square roots of x+i x4+x2+1 are
(a) ÷ ø ö ç è æ + + + - + ± x x 1 i x x 1 2 1 2 2 (b) ÷ ø ö ç è æ + + - - + ± x x 1 i x x 1 2 1 2 2 (c) ± èæç x -x+1+i x +x+1ö÷ø 2 1 2 2 (d) ÷ ø ö ç è æ - + - + + ± x x 1 i x x 1 2 1 2 2
8. If w 1 is a cube root of unity, then the roots of the¹ equation (x+aw)3 + a3 = 0 are :
(a) a, aw2, – 2aw (b) a, –2aw2, aw
(c) –2a, aw2, aw (d) a, – 2aw2, –aw 9. Let a=z1+z2,b=wz1+w2z2, g=w2z1+wz2 where
w is a complex cube root of unity, then : (a) a2+b2 +g2 =8z1z2
(b) ab+bg+ga=3z1z2 (c) a3+b3+g3 =3(z13+z23) (d) abg=2(z13+z23)
10. The polynomial x3m+x3n+1+x3k+2, is exactly divisible by x2+x+1 if
(a) m n, k are rational (b) m, n, k are integers
(c) m, n, k are positive integers (d) none of these. 11. If i= -1,then 334 365 1 i 3 1 i 3 4 5 3 2 2 2 2 æ ö æ ö + - +çç ÷÷ + - +çç ÷÷ è ø è ø is equal to (a) 1-i 3 (b) -1+i 3 (c) i 3 (d) -i 3
12. Ifw(¹1) be a cube root of unity and (1+w)7 =A+Bw, then A and B are respectively the numbers
(a) 0, 1 (b) 1, 1 (c) 1, 0 (d) –1, 1
GEOMETRY OF A COMPLEX NUMBER
O Y
X P(z) As in vectors, we represent a
point by the position vector of the p oi n tOP= with respect tor orgin O. Similarly the point P can be r epr esen ted by a complex n u mber z, su ch t h at l en g t h OP= z
and ÐXOP=arg(z). The point P is called the IMAGE of the complex number z and z is said to be AFFIX or complex coordinate of the point P.
DISTANCE BETWEEN TWO POINTS
If two points P and Q have affices z1and z2 respectively then
1 2–z z PQ= PQ=z2–z1= Affix of Q – Affix of P.. SECTION FORMULA n m QRPR = P (z )1 Q (z )2 R (z)
If a point R (z) divides the join of two points P (z1) and Q (z2) in ratio m:n, then n m nz mz z 2 1 + + = (internally) and n m nz mz z 2 1 -= (externally)
Ø Mid Point of PQ is given by 2
z z1+ 2 ANSWER KEY
ANGLE BETWEEN TWO LINES (CONCEPT OF ROTATION) ) z ( R 3 ) z ( P 1 ) z ( Q 2 1 q q 2 q
(i) Consider three points P(z1), Q(z2) and R(z3) Then angle between PQ an PR (counter clockwise)
1 2-q q = q =arg(PR)–arg(PQ) =arg(z3–z1)–arg(z2 –z1) 1 2 1 3 z – z z – z arg = q \
If P(z1), Q(z2) and R(z3) be collinear Points, then q = 0 or p, i.e. 1 2 1 3 z – z z – z = 0 1 z z 1 z z 1 z z z – z z – z 3 3 2 2 1 1 1 2 1 3 Þ =
(A complex number z is purely real if z= )z If PR is perpendcular to PQ.Then arg 1 2 1 3 1 2 1 3 z – z z – z 2 z – z z – z Þ p ± = ÷÷ ø ö çç è æ is puerly imaginary That is, 1 2 1 3 1 2 1 3 z – z z – z z – z z – z -=
(A Complex number z is purely imaginary if z=-z) (ii) ) z ( R 3 P(z) 1 ) z ( Q 2 S(z4) q
Consider Four Points P, Q, R and S with affices respectively
3 2 1,z ,z
z and z4. As in (i) the angle between SR and QP..
÷÷ ø ö çç è æ = q 2 1 4 3 z – z z – z arg If SR and QP be perpendicular, 2 p ± = q Then 2 1 4 3 z – z z – z is purely imaginary.. 2 1 4 3 2 1 4 3 z – z z – z – z – z z – z = Þ
or alternatively ik,forsomek {0} z – z z – z 2 1 4 3 = ÎR -) z – z ( ik z – z3 4= 1 2 Þ
(iii) Multiplying a complex number z by i is equivalent to rotating the image of z in Argand plane by 90º about origin,
anticlockwise, as |z| = |iz| and arg
2 ) i ( arg z iz p = = ÷ ø ö ç è æ O P (z) Q (iz) p/2
(iv) Multiplying a complex number z by w is equivalent to rotating the image of z in Argand plane by 120º (or 240°) about origin anticlockwise, for |z| = |wz| (|w| = 1) and
arg wz arg (w) 2 or 4 z 3 3 p p æ ö= = æ ö ç ÷ ç ÷ è ø è ø O P (z) Q (wz) 3 2p ANGLES OF A TRIANGLE ) z ( C 3 ) z ( A 1 ) z ( B 2
If z1 z2 and z3 be the affices of vertices A, B and C of a tringle ABC described in counterclockwire sense. Then
) A sin i A (cos BA CA z – z z – z 1 2 1 3 = + or iA 1 2 1 3 e BA CA z – z z – z =
IMPORTANT RESULTS ABOUT TRIANGLES If z1, z2 , z3 are the vertices of a triangle then
1. centroid z is givn by 3 z z z z= 1+ 2+ 3 2. Incentre I (z) of the DABC is given by
c b a cz bz az z 1 2 3 + + + + = 3. Circumcentre O(z) of theDABC is given by
C 2 sin B 2 sin A 2 sin ) C 2 (sin z ) B 2 (sin z ) A 2 (sin z 1 z z 1 z z 1 z z 1 z | z | 1 z | z | 1 z | z | z 1 2 3 3 2 3 2 2 2 1 2 1 3 2 3 2 2 2 1 2 1 + + + + = =
4. Orthocentre H(z) of DABC is given by
1 z z 1 z z 1 z z 1 z | z | 1 z | z | 1 z | z | 1 z z 1 z z 1 z z z 3 2 3 2 2 2 1 2 1 3 2 3 2 2 2 1 2 1 3 2 3 2 2 2 1 2 1 + = OR 1 2 3 1 2 3
(tan A)z (tan B)z (tan C)z z
tan A tan B tan C
(a sec A)z (b sec B)z (c sec C)z a sec A b sec B csec C
+ + = + + + + = + +
5. The centroid G lies on the segment joining the orthocentre H and the circumcentre O of the triangle and divides
internally in ratio 2 : 1, i.e
1 2 OG HG = H O G
6. Ar ea of the DABC is given by the modulus of
1 1 2 2 3 3
z
z 1
1
z
z
1
4 z z 1
7. Triangle ABC is equilateral if and only if
0 z z 1 z z 1 z z 1 2 1 1 3 3 2 = -+ -+ -2 1 1 3 3 2 2 3 2 2 2 1 z z z z z z z z z + + = + + Û 0 z z 1 z z 1 z z 1 2 1 1 3 3 2 = Û
EQUATION OF STRAIGHT LINE THROUGH TWO POINTS Z1 AND Z2
Let variable point z be a point on this line then
0 1 z z 1 z z 1 z z 2 2 1 1 = or z(z1-z2)+z(z2-z1)+z1z2-z1z2=0 0 ) z z z z ( i i ) z z ( z i ) z z ( z 1- 2 + 2- 1 + 1 2- 1 2 = Þ
(i) Let (z2-z1)i=a, a con stant complex number then a i ) z z ( 2- 1 =
- Also i(z1z2-z1z2) is purely real constant number, say b then the above equation is az+az+b=0 It is called the general equation of a straight line
(ii) The complex equation za+za+b=0 represents a straight line in complex plane where 'a' is a complex number and 'b' is a real number. The complex slope of the line is given
by a a - .
(iii) The equation of the perpendicular bisector of the line segment joining the points A(z1) and B(z2) is
2 2 2 1 2 1 2 1 z ) z(z z ) |z | |z | z ( z - + - = -IMPORTANT RESULTS
1. The complex slope of line joing points A (z1) and B(z2)
is define as 2 1 2 1 z z z z -= m
2. Two lines with complex slopes m and 1 m are parallel2 if m1=m2 and perpendicular if m1+m2 =0
3. The length of perpendicular from a point A(a to the) line az+az+b=0 is given by | a | 2 | b a a | p= a+ a+ EQUATION OF A CIRCLE : P (z) r ) z ( C 0
(1) Consider a circle with centre C having affix z0 and radius r.
For any point P(z) on this circle CP = r i.e. z–z0 = ...(1)r
1 4 tan 76 y 12 x 14 y x 36 y 6 2 2 = p = + -+ -Þ Þ x2 + y2 – 14x – 18y + 112 = 0 ... (i)
which is a circle with centre (7, 9)
and radius = 72+92-112= 49+81-112= 18 Answer (b) Note : The equation (i) may be converted to complex form as
following : x2 –14 + y2 – 18y + 112 = 0 Þ (x2 – 14x + 49) + (y2 – 18y + 81) + 112 – 49 – 81 = 0 Þ (x – 7)2 + (y – 9)2 = 18 Þ |(x – 7) + i (y – 9)|2 = 18 Þ |(x + iy) – (7 + 9i)|2 = 18 Þ |z – (7 + 9i)|2 = 18 Þ |z – (7 + 9i)| = 18 , which is equivalent to |z – z0| = r
Hence centre of the circle is z0 = 7 + 9i = (7, 9) radius of the circle is r = 18
21. Consider the complex number z satisfying |z – 5i| £ 3, then (a) Value of z having the least modulus is z =2i
(b) Value of z having the greatest modulus is z = 8i (c) Value of z having the least positive argument is
z = (3 4i) 5
4 +
(d) Value of z having the greatest positive argument is
z = ( 3 4i) 5
4 - +
Sol. The inequation | z – 5i | £ 3 represents all the complex numbers lying inside or on the circle
| z – 5i | = 3 ...(1)
Clearly the circle (1) has centre at (0, 5) and radius = 3 Let us plot this circle on the xy – plane.
The given inequation represents the points inside or on the circle shown in figure.
O B D C E y x A (5i) a b f
Note that the modulus of a complex number is represented by the distance of the image of the complex number from origin. Clearly the point at the least distance from origin is D and the point at the greatest distance from origin is E. Hence the affices of D and E give us the complex numbers of the least and the greatest moduli respectively.
Ø
z–z0 < represents the region inside the circle given byr (1)Ø
z–z0 > represents the region outside the circle givenr by (1).(2) General equation of a circle
Consider the equation of a circle |z-z0|=r
0 r | z | z z z z z z r ) z z )( z z ( - 0 - 0 = 2Þ - 0- 0+ 0 2- 2= Þ
Let -z0=a, a constant complex number and , b r | a | r | z
| 0 2 - 2= 2 - 2 = a constant real number then the above equation becomes zz+az+az+b=0,
It is called the general equation of circle .
Hen ce th e complex equation zz+az+az+b=0 reprectents a circle in complex plane where 'a' is a complex number and 'b' is a real number. The centre of the circle is a point with affix '–a' and the radius is given by a2 –b.
For the existence of the circle a2 –b>0.
(3) (z – z )(z – z )1 2 +(z – z ) (z – z )2 1 =0 represents a circle in the complex plane which is described on a line as diameter having extremities z1 and z2.
20. Let z1 = 10 + 6i and z2 = 4 + 6i. If z is any complex number such that the argument of
2 1 z z z z is 4 p
. Then z must lie on a circle, which has
(a) Centre (0, 0); radius 6 (b) Centre (7, 9); radius 18 (c) Centre ( 2 , 1); radius 18 (d) Centre (7, 9); radius 6 Sol. Let z = x + iy Th en i ) 6 y ( ) 4 x ( i ) 6 y ( ) 10 x ( ) i 6 4 ( ) iy x ( ) i 6 10 ( ) iy x ( z z z z 2 1 -+ -+ -= + -+ + -+ = -i ) 6 y ( ) 4 x ( i ) 6 y ( ) 4 x ( x i ) 6 y ( ) 4 x ( i ) 6 y ( ) 10 x ( -+ -+ -= 52 y 12 x 8 y x i ) 36 y 6 ( 52 y 12 x 8 y x 76 y 12 x 14 y x 2 2 2 2 2 2 + -+ -+ + -+ + -+ = (on simplifying)
Now given that arg
4 z z z z 2 1 = p ÷÷ ø ö çç è æ -; Hence 4 76 y 12 x 14 y x 36 y 6 tan 2 2 1 =p ú ú û ù ê ê ë é + -+ -úû ù êë é + = -x y tan ) iy x ( arg 1 Q
Now OC = 5, DC = 3, Hence OD = 2 and OE = 8 \ Point D, i.e. complex number with least modulus is 2i Point E, i.e. complex number with maximum modulus is 8i Furthermore the argument of a complex number in Argand diagram is given by the angle that the line joining the origin to its image forms with positive x–axis. Clearly the two tangents drawn on the circle from origin, OA and OB represent the least and the greatest values of this angle respectively for any point on or inside the circle. Hence, the affices of points A and B give us the complex numbers with least and greatest arguments respectively.
Now Ð XOA = a = , AOC
2-f f=Ð p and ÐXOB = b = p+f 2 Clearly cos f = OC OA = OC AC OC2- 2 = 5 4 5 3 52- 2 =
[Where AC=radius=3] and sin f = 5 3 Affix of point A is r (cos a+ i sin a) r = OA = 4, cos a = cos çèæp2-f÷øö = sin f =
5 3 ;
sin a = sin çèæp2-f÷øö = cos f = 5 4 . \ A is (3 4i) 5 4 + Also affix of point B is r(cos b + i sinb)
r = OB = OA = 4, cos b = cos ÷ ø ö ç è æp+f 2 = – sin f = –5 3
sin b = sin çèæp2+f÷øö = cos f = 5 4
\ B is ( 3 4i) 5
4 - +
Hence, the complex number with least argument is
) i 4 3 ( 5 4
+ and the complex number with greatest argument is ( 3 4i)
5 4
+
- .
All options (a), (b), (c) and (d) are correct
Note : Student may feel that the solution of this example is lengthy on the contrary the solution is quite simple and direct from the figure only.
22. Suppose that z1, z2, z3 represent the vertices of a triangle taken in order. The triangle is equilateral if and only if
(a) z 1z z 1z z 1z 0 1 3 3 2 2 1 = -+ -+ -(b) 0 z z 1 z z 1 z z 1 1 3 3 2 2 1 = -(c) 3 2 2 1 z z 1 z z 1 -= -(d) z12+z22+z32-z1z2-z2z3-z3z1=0 Sol. Let a = z1-z2, b=z2-z3 and g=z3-z1
Then a + b + g = 0 ....(1)
Clearly a+b+g = 0 ....(2)
Let the triangle be equilateral, then | z1 – z2 | = |z2 – z3 | = | z3 – z1 | = l say That is | a | = | b | = | g | = l Now a l = a Þ l = a = a a 2 2 2 | | Similarly, g l = g b l = b 2& 2 Hence, from (2) la +lb +lg 2 2 2 = 0 g + b + a Þ 1 1 1= 0 1 3 3 2 2 1 z z 1 z z 1 z z 1 -+ -+ -Þ = 0 Conversely let 1 3 3 2 2 1 z z 1 z z 1 z z 1 -+ -+ - = 0 bg a = bg g + b -= a Þ = g + b + a Þ 1 1 1 0 1 from (1) | || || | | | | | | | y 2 3 2=b Þ a =bg Þa = a b g a \ Similarly |b|3=|a||b||g| and |g|3=|a||b||g| Hence |a|=|b|=|g|Þ|z1-z2|=|z2-z3|=|z3-z1|. That is the triangle is equilateral.
Also note that 0
z z 1 z z 1 z z 1 1 3 3 2 2 1 = -+ -0 z z z z z z z z z12+ 22+ 23- 1 2- 2 3- 3 1= Þ Answer (a, d)
23. The centre of a regular hexagon has affix i. The affix of one vertex is 2 + i. The affix z of adjacent vertices are (a) 1 + i (1± 3) (b) i+1± 3 (c) 2+i(1± 3) (d) 1±2i Sol.
Let A be the image of complex number 2 + i, ÐAOB = 3 p
Let B (or F) be z, then ÐAOB = arg (2z+-i)i-i
(refer to angle between two lines in the text)
3 p ± = ÷ ø ö ç è æ p± p = -+ -\ 3 sin i 3 cos AO BO i ) i 2 ( i z = cos 3 sin i 3 p ± p
[
QAO=BO]
(
)
1 3 z i 2 i z i 1 i 3 1 i 1 3 2 2 æ ö \ - = çç ± ÷÷Þ = + ± = + ± è ø Answer (a) 24. Let a,bÎR, such that 0 <a < 1, 0 < b < 1. If the complex numbers z1 = – a + i, z2 = –1 + bi and z3 = 0 form an equilateral triangle, then values of a and b are(a) a = b =2- 3
(b) a = 2- 3, b = 2+ 3 (c) a = 3 , b = - 3 (d) None of these
Sol.
Clearly from the figure 0 z 0 z 1 2 = cos 3 sin i 3 p + p ÷ ÷ ø ö ç ç è æ + = \ 2 3 i 2 1 z z2 1 or,, ÷ ÷ ø ö ç ç è æ + + -= + -2 3 i 1 ) i a ( bi 1 = ÷÷ ø ö ç ç è æ -+ ÷ ÷ ø ö ç ç è æ -- a 2 3 2 1 i 2 3 2 a
Equating real and imaginary parts,
3 2 a 1 2 3 2 a + = Þ = -3 2 b b a 2 3 2 1- = Þ = -Answer (a)
25. Let z1 and z2 be the roots of complex equation z2 + pz + q
= 0. The points represented by z1, z2 and the origin form an equilateral triangle if
(a) p2 > 3q (b) P2 < 3q (c) p 2 = 3q (d) p = 3q Sol. Given z1 and z2 are roots of z2 + pz + q = 0
Hence, z1 + z2 = – p – (i), z1z2 = q –(ii)
We know that if z1, z2, z3 are vertices of an equilateral triangle then z12+z22+z23-z1z2-z2z3-z3z1=0 (see example 16) Here z3 = 0, then z12+z22-z1z2=0 0 q 3 p 0 z z 3 ) z z ( 1+ 2 2- 1 2= Þ 2- =
Þ [from (i) & (ii)]
Answer (c) 26. If the imaginary part of 2 1
1
z iz
+
+ is – 2, then show that the locus of the point representing z in the argand plane is a straight line.
Sol. Let z = x + iy. Then
2 1 1 z iz + + = 2( ) 1 ( ) 1 x iy i x iy + + + + = (2 1) 2 (1 ) x i y y ix + + - + = {(2 1) 2 } {(1 ) } x i y y ix + + - + × {(1 ) } {(1 ) } y ix y ix - -= 2 2 2 2 (2 1 ) (2 2 2 ) 1 2 x y i y y x x y y x + - + - - -+ - + Thus, Im 2 1 1 z iz æ + ö ç + ÷ è ø = 2 2 2 2 2 2 2 1 2 y y x x y y x - - -+ - + But Im 2 1 1 z iz æ + ö ç + ÷ è ø = – 2 (Given) So, 2 2 2 2 2 2 2 2 1 2 y y x x y y x - - = -+ - + Þ 2y – 2y2 – 2x2 – x = – 2 – 2y2 + 4y – 2x2 i.e., x + 2y – 2 = 0, which is the equation of a line. 27. Let z1 and z2 be two complex numbers such that
1 2 0
z +i z = and arg (z1 z2) = p. Then find arg (z1). Sol. Given that z1+i z2=0
Þ z1 = iz2, i.e., z2 = – iz1
Thus, arg (z1z2) = arg z1 + arg (– iz1) = p Þ arg (– iz12) = p Þ arg (– i) + arg (z12) = p Þ arg (– i) + 2 arg (z1) = p Þ 2 - p + 2 arg (z1) = p Þ arg (z1) = 3 4 p
4.4
Solve following problems with the help of above text andexamples.
1. The points z1, z2, z3, z4, are the vertices of a parallelogram (in a complex plane) taken in order if and only if
(a) z1+z4 =z2+z3 (b) z1+z3=z2+z4 (c) z1+z2 =z3+z4 (d) None
2. A square ABCD has its centre at the origin. If A be z1 then the centroid of triangle ABC is
(a) 3 iz1 ± (b) 3 z1 ± (c) ÷ ø ö ç è æ p ± p ÷ ø ö ç è æ 3 sin i 3 cos 3 z1 (d) çæèz31ö÷øæèçcos6p±isin6p÷öø 3. If z1 = 1+2i, z2 = 2 + 3i, z3= 3 + 4i then z1, z2 and z3 represent
the vertices of a/an
(a) equilateral triangle (b) right angled triangle (c) isoceles triangle (d) none of these
4. Circumcentre of an equilateral triangle is at the origin and one of the vertex is r cis a then the other two vertices are at (a) rcis(-a),rcis(2a)
(b) r cis2a,r cis3a
(c) rcisèçæa+23p÷øö,rcisçæèa+43pö÷ø (d) rcisçæèa+p3÷øö,rcisçæèa+23pö÷ø
5. Let z be a complex number satisfying z–5i £ 1 such that amp z is minimum. Then z is equal to
(a) 5 6 2 + 5 i 24 (b) 5 24 + 5 i 6 2 (c) 5 6 2 – 5 i 24 (d) None of these
6. If z–25i £ 15, then max.amp(z)–min.amp(z) =
(a) ÷ ø ö ç è æ 5 3 cos–1 (b) p – 2 ÷ ø ö ç è æ 5 3 cos–1 (c) 2 p + ÷ ø ö ç è æ 5 3 cos–1 (d) ÷ ø ö ç è æ 5 3 sin–1 – ÷ ø ö ç è æ 5 3 cos–1
7. In the Argand plane, the area in square units of the triangle formed by the points 1 + i, 1 – i, 2i is
(a) 2 1
(b) 1 (c) 2 (d) 2
8. If the complex numbers z , 1 z , 2 z are in A.P. , then they3 lie on a
(a) circle (b) parabola (c) line (d) ellipse.
9. Let the complex numbers z , 1 z and 2 z be the vertices3 of an equilateral triangle. Let z the circumcentre of the0 triangle. Then z12 + z22 + z32 – 3z02 equals
(a) 1 (b) – 1
(c) 0 (d) None of these
10. The complex numbers z = x + iy which satisfy the equation
i 5 z i 5 – z + = 1 lie on (a) the x-axis
(b) the straight line y = 5
(c) a circle passing through the origin (d) None of these.
ANSWER KEY
1. (b) 2. (a) 3. (d) 4. (c) 5. (a) 6. (b)
7. (b) 8. (c) 9. (c) 10. (a) SOME LOCI IN COMPLEX PLANE :
1. arg (z) = a, aÎR represents a line starting from the origin (excluding origin) and making an angle a with the real axis. a z Y X O
2. arg(z-z0)=a, aÎR represents a line starting from the
point z0 (excluding the point z0) and making an angle a with the real axis.
a z Y
X
O z0
| z z | | z z
| - 1 = - 2 is the perpendicular bisector of the segment joining A(z1) and B(z2)
4. Th e complex equation z-z1 + z-z2 =2a wh er e ,
z z a
2 > 1- 2 'a' is positive real number represents an
ellipse in complex plane, z1 and z2 are affices of two foci of ellipse. If 2a=|z1-z2|, then |z-z1|+|z-z2|=2a represents the line segment joining z1 and z2
If 2a<|z1-z2|, then the equation does not represent any curve 5. Th e complex equation z-z – z1 1-z2 =2a. wh er e 2 1 z z a
2 < - , and 'a' is positive real number, represents a hyperbola in complex plane, z1 and z2 are affices of two foci of hyperbola.
If 2a=|z1-z2|, |z-z1|-|z-z2| =2a represents the straight line joining A(z1) and B(z2) but excluding the segment AB
A(z )1 B(z )2
6. The complex equation K
z – z z – z 2 1 = represents a circle if 1
K¹ and a straight line if K=1.
7. The complex equationz–z12+ z–z2 2 =K represents a circle if z –z ,K
2 1
K³ 1 22 a real number..
8. Let z1 and z2 be two fixed points and a be a real number such that 0£a£p then
(a) , z z z z arg 2 1 =a ÷÷ ø ö çç è æ -2 , 0<a<p a¹p r epr es en ts a segment of the circle passing through A(z1) and B(z2).
a p(z) B A (b) 2 z z z z arg 2 1 = p ÷÷ ø ö çç è æ
represent a circle with diameter as
the segment A(z1) and B(z2).
p/2 p/2 P(z) P(z) (c) z z 0 z z arg 2 1 = ÷÷ ø ö çç è æ
represents the line segment joining A(z1) and B(z2) A B (d) ÷÷=p ø ö çç è æ -2 1 z z z z
arg represents the straight line joining
A(z1) and B(z2) but excluding the segment AB
A B
28. Let z1 = 6 + i and z2 = 4 – 3i. If z be a complex number
such that arg zz zz 2
2 1 = p ÷÷ ø ö çç è æ -, then z satisfies
(a) | z – (5 – i)| = 5 (b) | z – (5 – i)| = 5 (c) | z – (5 + i)| = 5 (d) |z – (5 + i)| = 5 Sol. C A(z )1 P(z) (z )B2 Given arg z z 2 z z 2 1 = p ÷÷ ø ö çç è æ -...(i) or arg zz zz 2 2 1 =- p ÷÷ ø ö çç è æ
-Which means that
2 APB=- p
Ð . Hence the relation (i) represents the points lying on a semicircle as shown in figure. Centre of this semicircle is 5 – i (mid point of AB)
and radius = 5 çèæ =12AB÷øö. Therefore z satisfies | z – (5 – i)| = 5 [Complex form of equation of circle] ALTERNATE : Let z = x + i y Now zz zz 4x 3iyi x6 iyi 2 1 -+ = -= i ) 3 y ( ) x 4 ( i ) 3 y ( ) x 4 ( i ) 3 y ( ) x 4 ( i ) 1 y ( ) 6 x ( i ) 3 y ( ) x 4 ( i ) 1 y ( ) 6 x ( + + -+ + -´ + -+ -= + -+ -= (4 x)2 (y 3)2 i )] x 4 ( ) 1 y ( ) 3 y ( ) 6 x [( )] 3 y ( ) 1 y ( ) x 4 ( ) 6 x [( + + -+ + -+ +
-Given arg z z z z 2 z z z z 2 1 2 1 -Þ p = ÷÷ ø ö çç è æ is purely imaginary Or Re zz zz 0 2 1 = ÷÷ ø ö çç è æ -0 ) 3 y ( ) 1 y ( ) x 4 ( ) 6 x ( - - - - + = Þ 0 21 y 2 x 10 y x2+ 2- + + = Þ
Which is a circle with centre (5, – 1) and radius = 5 The equivalent form in complex number is
|z – (5 – i)| = 5 Answer (b)
29. The complex number z in the argand diagram satisfying | z –a | = 2 and arg (z) = tan–1
3 4 where a = 3 + 4i, is (a) (3 4i) 5 7 or ) i 4 3 ( 5 3 + + (b) (3 4i) 7 5 or ) i 4 3 ( 3 5 + +
(c) 3 + 4i and 3 – 4 i (d) None of these.
Sol.
| z – a | = 2 is the circle with centre at a = (3, 4) and radius = 2. Clearly 3 4 tan XOC=q= –1 Ð Since arg (z) = 3 4
tan–1 , z must lie along OC and on the circle. It must therefore be either point A or B.
Now, OA = OC – AC = 5 – 2 = 3 OB = OC + CB = 5 + 2 = 7 \ A is 3 (cos q +i sin q ) = (3 4i) 5 3 + B is 7 (cos q +i sin q ) = (3 4i) 5 7 + Answer (a)
30. If z = x + iy such that | z + 1| = | z – 1| and amp
4 1 z 1 z = p + -then (a) x= 2+1, y=0 (b) x=0,y= 2+1 (c) x=0,y= 2-1 (d) x = 2-1,y=0 Sol. Given that | z + 1 | = | z – 1 |Þ (x +1)2 + y2 = (x – 1)2 + y2 Þ x = 0 ...(1) 4 1 z 1 z arg ÷=p ø ö ç è æ + -Þ arg 4 y ) 1 x ( ] iy 1 x [ ] iy ) 1 x [( 2 2 p = ïþ ï ý ü ïî ï í ì + + -+ + -4 y ) 1 x ( iy 2 ) 1 y x ( 2 2 2 2 p = ïþ ï ý ü ïî ï í ì + + + -+ Þ 1 1 y x y 2 , 0 1 y x , 0 y 2 2 2 2 = -+ > -+ > Þ 0 1 y x , 0 y , 0 1 y 2 y x2+ 2- - = > 2+ 2- > Þ
Th er efor e the locus r epresen ted by th e equation
arg zz 11÷= 4p ø ö ç è æ +
-is the arc ABC of the circle x2 + y2 – 2y – 1 = 0. Solving
with x = 0, we get y = 1 2,y 0, y 1 2 2 8 2± = ± > \ = + Answer (b) 31. In argand plane the locus of z¹1 such that
arg 23 2 z z 3 3 z 5 z 2 2 2 p = ú ú û ù ê ê ë é -+ is
(a) the straight line joining the points z = 3/2, z = –2/3 (b) the straight line joining the points z = –3/2, z = 2/3 (c) a segment of a circle passing through z = 3/2, z = –2/3 (d) a segment of a circle passing through z = –3/2, z = 2/3
Sol. ) 1 z )( 2 z 3 ( ) 1 z )( 3 z 2 ( 2 z z 3 3 z 5 z 2 2 2 -+ -= -+ -= asz 1. 3 / 2 z 2 / 3 z . 3 2 2 z 3 3 z 2 ¹ + -= +
-\ The given condition reduces to arg zz 32//23÷=23p ø ö ç è æ +
-This implies that the line joining the points z = 2 3 . and z = 3 2
- subtend a constant angle 3 2p
= at the point z. Thus z describes the segment of a circle through z =
2 3 and z = 3 2
- at which the chord subtends an angle 3 2p
= .
