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Math Help - Cauchy Riemann equations

  1. #1
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    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.
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  2. #2
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    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.
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  3. #3
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    Quote Originally Posted by Enrique2 View Post
    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 .
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  4. #4
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    Quote Originally Posted by ThePerfectHacker View Post
    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!
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  5. #5
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    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.
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