Showing that $\sinh(\mathrm{e}^z)$ is entire

Question

I am attempting to show that $\sinh(\mathrm{e}^z)$, where $z$ is a complex number, is entire. The instructions of the problem tell me to write the real component of this function as a function of $x$ and $y$, which I used algebra to do; this function is $u(x, y)=\cos(\mathrm{e}^x \sin(y))\cosh(\mathrm{e}^x \cos(y))$. The instructions then say to "state why this function must be harmonic everywhere", i.e., that the sum of the second derivatives $u_{xx}, u_{yy}$ is zero.

This is where I'm stuck: Even the first derivative looks like a nightmare to compute, and the language of the problem's instructions seems to suggest that I shouldn't need to, that I should simply be able to glance at the function and know that it's harmonic, and state why in a simple sentence. Why should this function be clearly harmonic?

Answer 1

You should have a theorem in your text to the effect of "the composition of two entire functions is entire" and "the sum of two entire functions is entire".

Note that if $\sinh(z)$ is entire, then $\sinh(e^z)$ must be entire since this would be the composition of entire functions.

Note that $$ \sinh(z) = \frac 12 \left(e^z - e^{-z}\right) $$ Is the sum of entire functions.

Answer 2

You have real and imaginary part $ u(x,y )$ and $v(x,y)$, so you can control if the Cauchy-Riemann equations are verified: $$u_{x} = v_{y} \ \ \ u_{y} = -v_{x}$$

Answer 3

Hint: $z\mapsto e^z$ is entire. $z\mapsto\frac12(z-\frac1z)$ is holomorphic on $\mathbb{C}\setminus\{0\}$. $e^z$ is never zero.