CSLS Chapter 01

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Welcome to the presentation

on

Limits and continuity of a

function of complex variable

Presented by :

Mohammed Nasir Uddin

Assistant Professor Dept. Of ICT

Faculty of Science and Technology(FST)

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Learning Out comes

Objectives:

Introduction

Discussion Points

Functions

Limits

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To understand functions of Complex Number To understand functions of Complex Number

To know Limits To know Limits

Be familiar with the Branch Points and Branch Lines Be familiar with the Branch Points and Branch Lines

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Single & Multiple Valued Functions Single & Multiple Valued Functions

Branch Points

Limits

Infinity

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Home Work

Ch-2 / MRS

Solved Problem:

1, 6,7, 14,19, 23,24,25,26,29,30

Supplementary Problems:

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FUNCTIONS

Functions:

If to each value which a complex variable z can assume their corresponds one or more values of a complex variable w,

we say that w is a function of z and write w = f(z) or w=G(z) etc.

Variable:

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Single valued functions

Single valued functions:

If only one value of w corresponds to each value of z, we say that w is a single valued function of z or that f(z) is single valued.

Example: If w=z2 , then to each value of z there is only one

value of w.

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Multiple valued functions

If more than one value of w corresponds to each value of z, we say that w is a multiple valued function. A multiple valued function can be consider as a collection of single valued function.

Example: If w=z1/2 , then to each value of z there is two

values of w.

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If w = f(z),then we can also consider z as a function of w , written z = g(w)=f-1(w) .

The function f-1 is often called the inverse function

corresponding to f.

Thus w=f(z) and w =f-1(z) are inverse functions of each

other.

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If w = u + iv is a single valued function of z = x + iy, we can write u + iv = f(x + iy).

By equating real and imaginary parts this is seen to be equivalent to,

u = u (x,y) , v =v (x,y)………(1)

Thus given a point (x,y) in the z plane such as p in fig-1,

there corresponds a point (u,v) is the w plane, say p/ in fig-2.

The set of equation (1) is called a transformation.

We say that point p is mapped or transformed into point p/ by means of the transformation and call p/ the image of p.

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TRANSFORMATIONS

Z-Plane

y

x P

Q

W-Plane

v

u Q/

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Example (MRS/41/1):

Let w = f(z) = z2. Find the values of w which correspond to

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Algebraic Functions

If w is a solution of the polynomial equation P₀(z) wn + P

1(z)wn-1 + … … + Pn-1(z)w + pn(z) = 0

Where, P₀  0, P₁(z), … … P(z) are polynomials in z and n is a positive integer,

Then w = f(z) is called algebraic function of z

Example: Let us consider a polynomial equation is

w2 – z = 0 … (1)

 w2= z

w= z1/2

So w= z1/2 is a solution of equation (1).

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Let the polynomial equation

P₀(z) wn + P

1(z)wn-1 + … … + Pn-1(z)w + pn(z) = 0 … (1)

Where, P₀  0, P₁(z), … … P(z) are polynomials in z

and n is a positive integer.

Any function which cannot be expressed as a solution of (1) is called a transcendental function.

The logarithmic, trigonometric and hyperbolic functions and their corresponding inverses are examples of transcendental functions.

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Branch Points and Branch Lines

Let us consider a function

w = z1/2 … (1)

Suppose we allow z to make a complete circuit (counter clockwise) around the origin starting from point A (Fig-1).

A

O B

z plane

=₁

Branch line or

Branch cut

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We have z = r ei ,

w = r ei/2 [ by equation (1)]

w = r ei1/2 , at A, = 1

After a complete circuit back to A, then  = ₁ + 2 and

w = r ei(1+2)/2 = -r ei1/2 [ since, cos(1+2)/2 + isin (1+2)/2

= cos(₁/2+) + isin (₁/2+) = -cos(₁/2) - isin (₁/2)

= -[cos(₁/2) + isin (₁/2)= - ei1/2 ]

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However, by making a second complete circuit back to A, i.e.  = ₁+4

w = r ei(1+4)/2 = r ei1/2

[ since, cos(₁+4)/2 + isin (₁+4)/2 = cos(₁/2+2) + isin (₁/2+2)

= cos(₁/2) + isin (₁/2) = ei1/2

Now we have achieved the same value of ‘w’ with which we started.

We can describe the above by stating that if 0   < 2 we are on one branch of the multiple-valued function z1/2_,

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It is clear that each branch of the function is single-valued.

In order to keep the function single-valued, we set up an artificial barrier such as OB where B is at infinity (although any other line from O can be used) which we agree not to cross.

This barrier (drawn heavy in the figure) is called a branch line or

branch cut , and point O is called branch point.

It should be noted that a circuit around any point other than z=0

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Example (MRS/43/6):

Let w5=z and suppose that corresponding to the particular

value z=z₁ we have w=w₁.

(a)If we start at the point z₁ (see fig. 1) and make one complete circuit counter click wise around the origin, show that the value of w on returning to z₁ is w₁e2i/5.

(b)What are the values of w on returning to z₁ , after 2, 3,

… complete circuits around the origin?

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Home Work:

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Let f(z) be defined and single valued in a neighbourhood of z = z with the possible exception of z = z itself. We₀ ₀

say that the number l is the limit of f(z) as z approaches z₀

and write

If for any positive number ε we can find some positive number δ such that |f(z)-l |< ε Whenever 0 <| z - z |<δ.₀

0

lim ( )

zz f z l

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RIGHT HAND LIMIT:

If the values of f(z) can be made as close as possible to L by

taking values of z sufficiently close to ‘a’ (but greater than a), then we can write Lim f(z) = L

z→a+

Which is read “the limit of f(z) as z approaches a from the right is L.”

LEFT HAND LIMIT:

If the values of f(z) can be made as close as possible to L by taking Values of z sufficiently close to a (but less than a), then we can write Lim f(z) = L

z→a

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Infinity:

By means of the transformation w = 1/z the point z = 0 is mapped into w =, called the point at infinity in the w plane. Similarly we denote by z =  the point at infinity in the z plane.

To consider the behavior of f(z) at z = , it suffices to let z = 1/w and examine the behavior of f(1/w) at w = 0.

We say that, or f(z) approaches l as z approaches infinity, if for any ε > 0 we can find M > 0 such that |f(z)-l|< ε whenever |z|> M.

We say that, =  or f(z) approaches infinity as z

approaches z , if for any N > 0 we can find δ > 0 such that |f(z)| > N ₀

whenever 0 < |z - z |< δ₀

l z f

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Proof:

we must show that given any ε > 0 we can find δ (depending in general on ε) such that |z2_{– z}2

0| < ε whenever 0< |z – z0| < δ.

If δ 1,then 0< |z – z₀| < δ implies that |z2_{– z}2

0| = |z – z0| |z + z0|

< δ |z – z₀ + 2z₀| [since 0< |z – z₀| < δ ] < δ {|z – z₀|+ |2z₀|}

< δ (1+ |2z₀|)

Take δ as 1 or ε / (1+2|z₀|) ,whichever is smaller. Then we have |z2 – z2

0| < ε whenever |z – z0| < δ,

and the required result is proved.

Problem: MRS / 50/23(a)

If f(z) = z2 , prove that lim f(z) = z2 0

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Problem: MRS / 50/24(a)

Geometrical interpretation of 23(a)

The equation w = f(z) = z2_{defines a transformation or mapping of points of the z}

plane into points of the w plane.

In particular let us suppose that point z₀ is mapped into w₀= z2 0 .

z plane .z x y  Z₀ .w w plane v u  W₀

In problem 23(a) we have that given any ε > 0 we can fine δ > 0 such that |w – w₀| < ε whenever |z – z₀| < δ.

Geometrically this means that if we wish w to be inside a circle of radius ε we

must choose δ (depending on ε) so that z lies inside a circle of radius δ. According to problem 23(a) this is certainly accomplished

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Theorems on Limits

Ch-2/SL/51/26

    0 0 0

1 2 1 2

1 0

2 0

lim ( ) , .

Pr :

We must show, lim ( ) lim ( ) , .

, 0 , We find 0 such that

/ 2 0

z z

z z z z

If f z exists provethat it must beunique

oof

f z l and f z l thenl l

By hypotesis given any

f z l when z z

                               

1 2 1 2

1 2

2 2

( )

Then

l l l f z f z l

l f z f z l

l l proved

        

   

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Ch-2/MRS/51/25

 

4 3 2

4 3 3 2 2

3 2 8 2 5

Pr lim 4 4

Pr :

we must show that 0 , we find 0 s.t.

3 2 8 2 5

4 4 when, 0

sin ,

3 2 8 2 5

3 3 3 3 5 5 5 5

z i

z z z z

ove that i

z i oof

z z z z

i z i

z i

ce z i we can write

z z z z

z i

z z i z i z z zi zi

        _{ }        _ _ _ _{  }                             





3 2 2

2 2 2 2

3 3 5 5 2 2

z iz iz z

z i

z z i z i z i z z i i z i z z i iz z i z i

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 





                             

3 2 2

3 2

4 3 2

3 2

2

3 2 3 5 2 5

3 2 8 2 5

4 4

3 2 3 5 2 5 4 4

3 2 3 5 2 4

3 6 2 1 4

3 12 2 10 6

z i z z iz z iz i

z i z i z i z i

z z z z

Now i

z i

z i z i z i i

z i z i z i

z i z i z i

z i z i i i z i i i

z i z i i z i i

            _      _                                                 





  2

3 12 2 10 6

3 13 12

28

1 / 28 ,

.(proved)

z i i z i i

Taking as the smaller of and the required result follows

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Theorems on Limits

Ch-2/SL/53/30

0

Pr .

Pr :

limit int 0.

Let z 0along the x axes,

Then y 0, and z x iy= x and x so that the required

lim

z

ove that does not exist

z

oof

If the is to exist it must beindependent of the manner in which z approaches the po

z x iy           0 0 limit is lim 1

Let z 0along the y axes,

Then x 0, and z x iy = and , so that the required limit is

lim 1

Since the two approaches do not give the same answer so the limit doe

x

y

x x

iy z x iy iy

iy y              _{ }

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Theorems on Limits

Ch-2/SL/54/36

2 2 2 0 0 0

Pr is uniformly continuous in the region 1.

Pr :

We must show that given any 0 we can find 0such that

when ,

where depends only on and not on the particular point z of

( )

ove that z

oof

z

f z z

z z z

            0 2 2 0 0 0

0 0 0

2 2 0 0 2 2 0 0 0 2 the region. z points in 1,

{ } 2

it follows that 2 .

Choosing / 2

where depends only on and not on z .

Hence (

If z and are any z Then

z z z z

z z z z z z

Thus if z

we see that when z z

z z

z z z

z z f                           

is uniformly continuous in the region.

)

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