Cauchy-Riemann conditions derivative of a complex function How do you check using Cauchy Riemann equation $e^{\cosh z}$ is analytic. I have struggled with it for some time .Would appreciate a little assistance.
 A: If you have to use the Cauchy-Riemann equations, the first thing is to separate the real and imaginary parts, of both, function and argument.
Let us start with the inner function,
$$\cosh (x+iy) = \cosh x \cosh (iy) + \sinh x \sinh (iy)$$
by the addition theorem. Now use $\cosh (iy) = \cos y$ and $\sinh (iy) = i\sin y$ to obtain
$$\cosh (x+iy) = \cosh x \cos y + i \sinh x\sin y.$$
Next, we have $e^{u+iv} = e^u(\cos v + i\sin v)$, so
$$\begin{align}
e^{\cosh (x+iy)} &= e^{\cosh x\cos y}\bigl(\cos (\sinh x\sin y) + i \sin (\sinh x\sin y)\bigr)\\
&= \underbrace{e^{\cosh x\cos y}\cos (\sinh x\sin y)}_g + i\, \underbrace{e^{\cosh x\cos y}\sin (\sinh x\sin y)}_h.
\end{align}$$
Start to differentiate.
$$\frac{\partial g}{\partial x}(x+iy) = e^{\cosh x\cos y}\left(\cos(\sinh x\sin y)\cdot \sinh x\cos y - \sin(\sinh x\sin y)\cdot \cosh x\sin y\right).$$
And so on. Not so pretty (but not impossible).
Does it look better if we try to avoid the ugly special case and abstractly verify that the composition of two functions satisfying the Cauchy-Riemann equations also satisfies the Cauchy-Riemann equations?
Let $f = g+ih$ and $w = u+iv$ be two complex differentiable functions which can be composed. Then, writing everything in real form,
$$(f\circ w)(x,y) = f(u(x,y),v(x,y)) = \underbrace{g(u(x,y),v(x,y))}_a + i\, \underbrace{h(u(x,y),v(x,y))}_b,$$
differentiating gives us
$$\begin{align}
\frac{\partial a}{\partial x}(x,y) &= \frac{\partial g}{\partial u}(u(x,y),v(x,y))\cdot \frac{\partial u}{\partial x}(x,y) + \frac{\partial g}{\partial v}(u(x,y),v(x,y))\cdot \frac{\partial v}{\partial x}(x,y),\\
\frac{\partial b}{\partial y}(x,y) &= \underbrace{\frac{\partial h}{\partial u}}_{-\frac{\partial g}{\partial v}}(u(x,y),v(x,y))\cdot \underbrace{\frac{\partial u}{\partial y}}_{-\frac{\partial v}{\partial x}}(x,y) + \underbrace{\frac{\partial h}{\partial v}}_{\frac{\partial g}{\partial u}}(u(x,y),v(x,y))\cdot \underbrace{\frac{\partial v}{\partial y}}_{\frac{\partial u}{\partial x}}(x,y),
\end{align}$$
and that is at least not worse than the special case. The other equation is verified similarly, not too bad. Then it remains to check the Cauchy-Riemann equations for $w\mapsto e^w$ and $z\mapsto \cosh z$, which is much less of a chore.
It gets better if you use the Wirtinger derivatives
$$\frac{\partial}{\partial z} = \frac{1}{2}\left(\frac{\partial}{\partial x} - i\frac{\partial}{\partial y}\right),\qquad \frac{\partial}{\partial \overline{z}} = \frac{1}{2}\left(\frac{\partial}{\partial x} + i\frac{\partial}{\partial y}\right)$$
and write the Cauchy-Riemann equations in complex form,
$$\frac{\partial f}{\partial\overline{z}} = 0.$$
Then, after finding the chain rule in terms of the Wirtinger derivatives, you get
$$\frac{\partial (f\circ g)}{\partial \overline{z}} = \frac{\partial f}{\partial w}\circ g\cdot \frac{\partial g}{\partial\overline{z}} + \frac{\partial f}{\partial\overline{w}}\circ g\cdot \frac{\partial \overline{g}}{\partial\overline{z}}.$$
Then if $f$ is complex differentiable, the second term vanishes, and if $g$ is complex differentiable, the first term vanishes, so it is easily seen that the composition of complex differentiable functions is complex differentiable.
