# Complex show differentiable only at z=0

• saraaaahhhhhh
In summary, the function f(z) = zRez is only differentiable at z=0, as shown by taking the limit as z_0 approaches 0 of [f(z+z_0) - f(z)]/(z_0) and simplifying. When z_0 is purely real, the limit is 2x+iy, while when z_0 is purely imaginary, the limit is x. Therefore, the limit does not exist and the function is not differentiable everywhere. However, at z=0, when both x and y are equal to 0, the limit simplifies to 0, showing that the function is differentiable at this point. The issue with getting different values for the real and
saraaaahhhhhh

## 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?

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?

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!

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

## 1. What does it mean for a complex function to be differentiable only at z=0?

Being differentiable at a point means that the function has a well-defined derivative at that point. In the context of complex functions, this means that the function is smooth and has a unique tangent line at that point. When a complex function is differentiable only at z=0, it means that it is not differentiable at any other point in the complex plane except for z=0.

## 2. Why is it important for a complex function to be differentiable?

The concept of differentiability is important in complex analysis because it allows us to determine the behavior of the function at a particular point. A differentiable function is also continuously differentiable, meaning that it has derivatives of all orders. This allows us to use techniques such as power series to approximate the function and make calculations easier.

## 3. What are some examples of complex functions that are only differentiable at z=0?

One example is the function f(z) = |z|. This function is not differentiable at any point in the complex plane except for z=0. Another example is the function f(z) = Re(z), which is the real part of a complex number. This function is only differentiable at z=0 because its derivative is equal to 1 for all values of z.

## 4. Can a complex function be differentiable at more than one point?

Yes, a complex function can be differentiable at multiple points in the complex plane. In fact, most complex functions are differentiable at infinitely many points. However, there are some functions, such as the examples given in question 3, that are only differentiable at a single point.

## 5. How can we determine if a complex function is differentiable only at z=0?

To determine if a complex function is differentiable only at z=0, we can use the Cauchy-Riemann equations. These equations state that a function is differentiable at a point if and only if it satisfies the Cauchy-Riemann conditions, which are a set of two partial differential equations. If the function satisfies these conditions only at z=0, then it is differentiable only at that point.

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