How to show that if $f(z)$ is an analytical function in a region containing $z_0$ then $$f(z)=f(z_0)+f^{\prime}(z_0)(z-z_0)+\eta(z-z_0)$$ where $\eta\rightarrow 0$ when $z\rightarrow z_0$.
How to prove $f(z)=f(z_0)+f^{\prime}(z_0)(z-z_0)+\eta(z-z_0)$?
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complex-analysis
ordinary-differential-equations
derivatives
complex-numbers
analyticity
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0Like calculus approach. – 2017-01-16
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0If there would have been couple of more tags then the tagline would have become longer than the question itself ;) ... – 2017-01-16
1 Answers
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The key here: if it is analytic, then $$ \lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}=f'(z_0). $$
We can define $\eta=\eta(z)$ to make the equation $$ f(z)=f(z_0)+f'(z_0)(z-z_0)+\eta\cdot (z-z_0) $$ true. The question, then, is why must $\eta\to 0$ as $z\to z_0$? Well, we can see that $$ \frac{f(z)-f(z_0)}{z-z_0}-f'(z_0)=\eta; $$ so, the limit above (defining the derivative) completes the problem.