# Show that $\sinh(z)$ where $z = x + iy$ is analytic in the whole complex plane

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

• What definition of $\sinh$ are you working with? And what definition of "analytic"? – 5xum Sep 24 '18 at 7:51

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: $$\sinh z = \frac12\left(\sum\limits_{n=0}^{\infty}\frac{z^n}{n!}-\sum\limits_{n=0}^{\infty}\frac{(-z)^n}{n!}\right)=\sum\limits_{n=0}^{\infty}\frac{z^{2n+1}}{(2n+1)!}$$
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 $$\sinh z$$.