3
$\begingroup$

Can someone show me how to find the limit of a complex function?

Example:

z1 = 3 + 4*I

some_function[z] = z * z1

Set-up:

Limit[some_function[z],z->z1]

The result should be

(4*I + 3)*z

but I get

Limit[some_function[z], z -> 3 + 4 I]

What am I doing wrong? I tried to search in google and in the help files in the section about complex analysis.

Thank you for your attention.

$\endgroup$
0

2 Answers 2

3
$\begingroup$

can't use _ in names. the _ is used for patterns. Also your function definitions are not correct. Need to use delay assignment :=

May be this will work for you:

ClearAll[z, z1];
z1 = 3 + 4*I;
someFunction[z_] := z*z1;
Limit[someFunction[z], z -> z1]

  (*-7 + 24 I*)

see DefiningFunctions.html and what-are-the-most-common-pitfalls-awaiting-new-users

$\endgroup$
0
1
$\begingroup$

Taking a sample complex function as example with z->1+i

eq = z^2 - 2 z + 1;

Separating complex and Real part,

eqt = ComplexExpand[eq //. {z -> x + I y}]

1 - 2 x + x^2 - y^2 + I (-2 y + 2 x y)

Taking real part,

Cases[eqt, x_ /; FreeQ[x, I]] /. List -> Plus

1 - 2 x + x^2 - y^2 Calculating limit for real part as,

Limit[Limit[1 - 2 x + x^2 - y^2, x -> 1], y -> 1]

-1

Taking Complex part,

Complement[eqt, 1 - 2 x + x^2 - y^2]

I (-2 y + 2 x y) Taking limit as,

Limit[Limit[(-2 y + 2 x y), x -> 1], y -> 1]

0

So, by theorems of Complex Analysis its limit is -1. Use Example 2.17 on this Page

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.