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Suppose $f$ is a real-valued function of a complex variable that is differentiable at every $z \in \mathbb{C}$. Show that $f'(z)=0$ for all $z \in \mathbb{C}.$

My approach: Since $f$ is a real-valued function of a complex variable that is differentiable at all $z \in \mathbb{C}$, we can write that: $$f'(z)=\lim_{\lambda\to0} \frac{f(z+\lambda)-f(z)}{\lambda}= L$$ for some real function $L$.

If this is true, then $f(z+\lambda) - f(z) = \lambda L + g(\lambda)$ such that $\lim_{\lambda\to0} \frac{g(\lambda)}{\lambda}=0$, where $g(\lambda)$ is a real valued function. However, if L is real, that implies $\lambda L$ is complex and arises from the subtraction of two real-valued functions. Since this is not possible, $\lambda L$ has to be real, which implies $L = 0$ or $L=c\bar{\lambda}$, where $c$ is some constant. But since L is real, L cannot be $\bar{\lambda}$. Hence, L has to be zero. This implies $f'(z) = 0$.

Is this proof correct? If not, how should I correct it?

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  • $\begingroup$ In your first paragraph you say that $f $ is differentiable, but in your second you assume that the derivative is constant. Which one is it? $\endgroup$ – Martin Argerami Feb 2 '18 at 3:40
  • $\begingroup$ That is true. Can I assume that the derivative is a real function? $\endgroup$ – Sat D Feb 2 '18 at 3:42
  • $\begingroup$ I have posted an answer. $\endgroup$ – Martin Argerami Feb 2 '18 at 3:54
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The problem with your argument is that you have no control over $g $. Also, "complex" does not preclude "real".

Fix $z $. Since $f'(z) =L$ exists, we may approach along any line. So $$L=\lim_{t\to0}\frac {f (z+t)-f (z)}{t}\in\mathbb R. $$ Also, $$L=\lim_{t\to0}\frac {f (z+it)-f (z)}{it}\in i\mathbb R. $$ So $L $ is both real and imaginary: $L=0$.

As $z $ was arbitrary, $f'(z)=0$ for all $z $.

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  • $\begingroup$ Should that be $f(z+it)$ in the second numerator? $\endgroup$ – G Tony Jacobs Feb 2 '18 at 3:58
  • $\begingroup$ Inded. Corrected. Thanks! $\endgroup$ – Martin Argerami Feb 2 '18 at 4:04

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