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How would I show that $\sinh(z)$ is analytic in the whole complex plane? Would I separate $\sinh(z)$ into exponentials or use that $\sinh(x+iy) = \sinh(x)\cosh(iy) + \cosh(x)\sinh(y)$?

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  • $\begingroup$ What definition of $\sinh$ are you working with? And what definition of "analytic"? $\endgroup$ – 5xum Sep 24 '18 at 7:51
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Analytic is a function that is locally given by a convergent power series. So, if you have $$\sinh z = \frac{1}{2}(e^z-e^{-z}),$$ and the exponential has an everywhere convergent power series $$e^z=\sum\limits_{n=0}^{\infty}\frac{z^n}{n!},$$ so the hyperbolic sine is analytic on the whole plane: $$\sinh z = \frac12\left(\sum\limits_{n=0}^{\infty}\frac{z^n}{n!}-\sum\limits_{n=0}^{\infty}\frac{(-z)^n}{n!}\right)=\sum\limits_{n=0}^{\infty}\frac{z^{2n+1}}{(2n+1)!}$$

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Verify that your definition of $\sinh$ gives $\sinh z=\frac {e^{z}-e^{-z}} 2$. Since $e^{z}$ is analytic in the whole complex plane so is $\sinh z$.

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