I want to find a proof that shows the composition of two analytic functions is analytic. I know I should prove this using Cauchy-Riemann equations, but I wasn't able to use them in the proper way in order to solve the problem.
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hmakholm left over Monica
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Saeid Ghafouri
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Do you take analytic to mean "differentiable in an open set" or "expressible locally in a power series?" – Andrew Jul 17 '15 at 07:04
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1By analyctic I mean a such function that is correct in the Cauchy-Riemann equations – Saeid Ghafouri Jul 17 '15 at 07:08
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Please visit the duplicate question to see a proof using the cauchy riemann equations directly. – Sidharth Ghoshal Sep 17 '22 at 17:41
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In comments you have defined "analytic" to mean "satisfies the Cauchy-Riemann equations". But I strongly suspect what you really mean is "has a derivative everywhere", because there are functions that satisfy they Cauchy-Riemann equations everywhere but are not analytic according to the usual meaning of that word. For example, $$ f(z) = \begin{cases} \exp(-1/z^4) & \text{for }z\ne 0 \\ 0 & \text{at }z=0 \end{cases} $$ satisfies the Cauchy-Riemann equations everywhere but is not differentiable at $0$, and any number of important theorems are going to break horribly if we consider it analytic in $\mathbb C$.
So what you really want to prove is that the composition of two differentiable functions is itself differentiable. A natural plan for this would be:
Dig up your old real calculus textbook and find the proof of the chain rule which says that the composition of two differentiable functions is differentiable (and also tells you exactly what the derivative will be).
Copy out the proof of the chain rule for the real case, checking that each step in it still works for complex functions.
You won't use the Cauchy-Riemann equations here, just the plain old $\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}$ definition of differentiation.
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