1. ## Cauchy Riemann equations

Show that
(a) directly
(b) using the Cauchy Riemann equations
that if $f(z) \in \mathbb{R}$ $\forall z \in \mathbb{C}$, then for any $z \in \mathbb{C}$ if $f'(z)$ exists then $f'(z)=0$.

I don't know how to do this. For part (a) I have tried using the definition of the derivative and wasn't sure how to do it. I am not sure how to use the Cauchy Riemann equations, we have just started on them. Thanks for any help.

2. a) Directly, you can use that a non costant holomorphic mapping is open. This makes imposible the range being real (the real line has emty interior in C!)

b) Using Cauchy Riemann, if f=u+iv, the assumption implies v=0, then $u_x=v_y=0$ and $u_y=-v_x=0$. Hence $u$ does not depend neither on $x$ nor $y$, and we are done.

3. Originally Posted by Enrique2
a) Directly, you can use that a non costant holomorphic mapping is open. This makes imposible the range being real (the real line has emty interior in C!)
This is complete overkill. Obviously, this is a first week of a complex analysis course. The problem is intended to be solved by writing out the limit. You might as well use the Picard theorem to completely overkill this problem .

4. Originally Posted by ThePerfectHacker
This is complete overkill. Obviously, this is a first week of a complex analysis course. The problem is intended to be solved by writing out the limit. You might as well use the Picard theorem to completely overkill this problem .
Ok, I see...

Anyway, I don't see the way of giving a "direct proof" without using neither Cauchy Riemann equations nor deeper properties of holomorphic functions. But you are right, this "direct proof" is clearly out of the first week of a complex analysis course. I thought it could be a problem for a global exam, in my country (Spain) in setember there are global exams to pass subjects that one failed in june!

5. Directly:

Let f(z), $z\in C$, be such that f(z) is always real. Then $f'(z)= \lim_{h\to 0}\frac{f(z+h)- f(z)}{h}$ where h can go to 0 along any path in the complex plane.

In particular, suppose h= ik where k is a real number (that is, h goes to 0 along the imaginary axis). Then $f'(z)= \lim_{h\to 0}\frac{f(z+h)- f(z)}{h}= \lim_{k\to 0}\frac{f(z+ik)- f(z)}{ik}$ $= i\lim_{k\to 0}\frac{f(z+ik)- f(z)}{k}$. Since f is real valued, that last limit is real and so f'(z) is a pure imaginary number.

Now, do the same thing taking h= k where k is a real number (that is, h goes to 0 along the real axis). Then $f'(z)= \lim_{h\to 0}\frac{f(z+h)- f(z)}{h}= \lim_{k\to 0}\frac{f(z+k)- f(z)}{k}$ $= \lim_{k\to 0}\frac{f(z+ik)- f(z)}{k}$. Since f is real valued, that last limit is real and so f'(z) is a real number.

If f' exists, then, it must be 0.

Of course, that is the kind of argument that is used to arrive at the Cauchy-Riemann equations.