2 The limit of a function

The limit of w = f ( z ) as z → z 0 is a number ℓ such that | f ( z ) − ℓ | can be made as small as we wish by making | z − z 0 | sufficiently small. In some cases the limit is simply f ( z 0 ) , as is the case for w = z 2 − z . For example, the limit of this function as z → i is f ( i ) = i 2 − i = − 1 − i .

There is a fundamental difference from functions of a real variable: there, we could approach a point on the curve y = g ( x ) either from the left or from the right when considering limits of g ( x ) at such points. With the function f ( z ) we are allowed to approach the point z = z 0 along any path in the z -plane; we require merely that the distance z − z 0 decreases to zero.

Suppose that we want to find the limit of f ( z ) = z 2 − z as z → 2 + i along each of the paths (a), (b) and (c) indicated in Figure 1.

Figure 1

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  1. Along this path z = x + i (for any x ) and z 2 − z = x 2 + 2 x i − 1 − x − i

    That is: z 2 − z = x 2 − 1 − x + ( 2 x − 1 ) i .

    As z → 2 + i , then x → 2 so that the limit of z 2 − z is 2 2 − 1 − 2 + ( 4 − 1 ) i = 1 + 3 i .

  2. Here z = 2 + y i (for any y ) so that z 2 − z = 4 − y 2 − 2 + ( 4 y − y ) i .

    As z → 2 + i , y → 1 so that the limit of z 2 − z is 4 − 1 − 2 + ( 4 − 1 ) i = 1 + 3 i .

  3. Here z = k ( 2 + i ) where k is a real number. Then

    z 2 − z = k 2 ( 4 + 4 i − 1 ) − k ( 2 + i ) = 3 k 2 − 2 k + ( 4 k 2 − k ) i .

    As z → 2 + i , k → 1 so that the limit of z 2 − z is 3 − 2 + ( 4 − 1 ) i = 1 + 3 i .

In each case the limit is the same.

Task!

Evaluate the limit of f ( z ) = z 2 + z + 1 as z → 1 + 2 i along the paths

  1. parallel to the x -axis coming from the right,
  2. parallel to the y -axis, coming from above,
  3. the line joining the point 1 + 2 i to the origin, coming from the origin.
  1. Along this path z = x + 2 i and z 2 + z + 1 = x 2 − 4 + x + 1 + ( 4 x + 2 ) i . As z → 1 + 2 i , x → 1

      and z 2 + z + 1 → − 1 + 6 i .

  2. Along this path z = 1 + y i

      and z 2 + z + 1 = 1 − y 2 + 1 + 1 + ( 2 y + y ) i . As z → 1 + 2 i , y → 2 and z 2 + z + 1 → − 1 + 6 i .

  3. If z = k ( 1 + 2 i ) then z 2 + z + 1 = k 2 + k + 1 − 4 k 2 + ( 4 k 2 + 2 k ) i . As z → 1 + 2 i , k → 1

      and z 2 + z + 1 → − 1 + 6 i .

Not all functions of a complex variable are as straightforward to analyse as the last two examples. Consider the function f ( z ) = z ̄ z . Along the x -axis moving towards the origin from the right

z = x and z ̄ = x so that f ( z ) = 1 which is the limit as z → 0 along this path .

However, we can approach the origin along any path. If instead we approach the origin along the positive y -axis z = i y then

z ̄ = − i y and f ( z ) = z ̄ z = − 1 , which is the limit as z → 0 along this path .

Since these two limits are distinct then lim z → 0 z ̄ z does not exist .

We cannot assume that the limit of a function f ( z ) as z → z 0 is independent of the path chosen.

2.1 Definition of continuity

The function f ( z ) is continuous as z → z 0 if the following two statements are true:

  1. f ( z 0 ) exists;
  2. lim z → z 0 f ( z ) exists and is equal to f ( z 0 ) .

As an example consider f ( z ) = z 2 + 4 z 2 + 9 . As z → i , then f ( z ) → f ( i ) = i 2 + 4 i 2 + 9 = 3 8 . Thus f ( z ) is continuous at z = i .

However, when z 2 + 9 = 0 then z = ± 3 i and neither f ( 3 i ) nor f ( − 3 i ) exists. Thus z 2 + 4 z 2 + 9 is discontinuous at z = ± 3 i . It is easily shown that these are the only points of discontinuity.

Task!

State where f ( z ) = z z 2 + 4 is discontinuous. Find lim z → i f ( z ) .

z 2 + 4 = 0 where z = ± 2 i ; at these points f ( z ) is discontinuous as f ( ± 2 i ) does not exist.

lim z → i f ( z ) = f ( i ) = i i 2 + 4 = 1 3 i . It is easily shown that any polynomial in z is continuous everywhere whilst any rational function is continuous everywhere except at the zeroes of the denominator.

Exercises
  1. For which values of z is w = 1 z − i defined? For these values obtain u and v and evaluate w when z = 1 − 2 i .
  2. Find the limit of f ( z ) = z 3 + z as z → i along the paths
    1. parallel to the x -axis coming from the right,
    2. parallel to the y -axis coming from above.
  3. Where is f ( z ) = z z 2 + 9 discontinuous?. Find the lim z → − i f ( z ) .
  1. w is defined for all z ≠ i w = 1 x + y i − i = 1 x + ( y − 1 ) i × x − ( y − 1 ) i x − ( y − 1 ) i = x − ( y − 1 ) i x 2 + ( y − 1 ) 2 .

    ∴ u = x x 2 + ( y − 1 ) 2 , v = − ( y − 1 ) x 2 + ( y − 1 ) 2 .

    When z = 1 − 2 i ,   x = 1 , y = − 2  so that u = 1 1 + 9 = 1 10 , v = 3 10 , z = 1 10 + 3 10 i

    1. z = x + i , z 3 + z = x 3 + 3 x 2 i − 2 x . As z → i ,    x → 0   and   z 3 + z → 0
    2. z = y i , z 3 + z = − y 3 i + y i . As   z → i , y → 1   and   z 3 + z → − i + i = 0 .
  3. f ( z ) is discontinuous at z = ± 3 i . The limit is f ( − i ) = − i − 1 + 9 = − 1 8 i .