Cauchy Riemann equations
(b) using the Cauchy Riemann equations
that if , then for any if exists then .
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.
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 and . Hence does not depend neither on nor , and we are done.
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 (Rofl).
Originally Posted by Enrique2
Ok, I see...
Originally Posted by ThePerfectHacker
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!
Let f(z), , be such that f(z) is always real. Then 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 . 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 . 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.