Complex show differentiable only at z=0

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Homework Help Overview

The problem involves analyzing the differentiability of the complex function f(z) = zRez, specifically showing that it is differentiable only at z=0 and finding the derivative at that point.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the limit definition of differentiability, attempting to evaluate the limit as z_0 approaches 0 for both real and imaginary inputs. There are questions about the results obtained for the limits and the elimination of terms during simplification.

Discussion Status

The discussion is ongoing, with participants sharing their attempts and questioning the validity of their calculations. Some express confusion about the differing results for real and imaginary parts, while others suggest revisiting the expansion and simplification steps. There is no explicit consensus on the correct approach yet.

Contextual Notes

Participants are grappling with the implications of their findings, particularly regarding the existence of limits and the conditions under which differentiability is determined. There is an acknowledgment of potential errors in calculations, but no resolution has been reached.

saraaaahhhhhh
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Homework Statement


Show that f(z) = zRez is differentiable only at z=0,
find f'(0)


The Attempt at a Solution



This should be easy. I find the limit as z_0 approaches 0 of [f(z+z_0) - f(z)]/(z_0) for this function...expand it out, simplify, and find what the limit is when z_0 is purely imaginary vs purely real. I did this, and I got x for the real part and x_0 + 2x + iy for the imaginary part. They're different, so it's not differentiable everywhere.

But apparently it's differentiable only at z=0. So I tried plugging in z=0 and resolving, and I got different values for z_0 real and imaginary.

I get, for the limit when z_0 is only real: 1
When z_0 only imaginary: x_0.

What's going on? Is there some other way to differentiate a complex function that's not using the limit?
 
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saraaaahhhhhh said:
This should be easy. I find the limit as z_0 approaches 0 of [f(z+z_0) - f(z)]/(z_0) for this function...expand it out, simplify, and find what the limit is when z_0 is purely imaginary vs purely real. I did this, and I got x for the real part and x_0 + 2x + iy for the imaginary part. They're different, so it's not differentiable everywhere.

Try your expansion and simplification again. When z_0 is real you should find that \lim_{z_0\to 0} \frac{f(z+z_0) - f(z)}{z_0}=2x+iy and when z_0 is imaginary you should find that \lim_{z_0\to 0} \frac{f(z+z_0) - f(z)}{z_0}=x
 
Last edited:
I mixed up my original message. I actually got x for the imaginary part and x_0 + 2x + iy for the real part. I still don't see how the x_0 was eliminated in your version of the expansion, above.

But the main issue is the fact that I get different values in the second part, evaluating the limit at z=0. I get different values for the real and imaginary parts, which means the limit doesn't exist. But according to the problem it should exist! I must just be doing the math wrong.

Am I at least on the right track with the method of finding the limit as z_0 goes to 0, setting z=0, and solving it all out and seeing if the real and imaginary parts of z_0 both go to the same value?

Thanks again for the help!
 
saraaaahhhhhh said:
I still don't see how the x_0 was eliminated in your version of the expansion, above.

Doesn't x_0 go to zero when z_0 goes to zero?:wink:

But the main issue is the fact that I get different values in the second part, evaluating the limit at z=0. I get different values for the real and imaginary parts, which means the limit doesn't exist. But according to the problem it should exist! I must just be doing the math wrong.

You are doing the math wrong!:smile:

At z=0, x and y are both zero too aren't they?...Last time I checked 0=0 was a true statement:wink:
 

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