Show that $\sinh(z^{2})$ is entire using the Cauchy Riemann equations

Question

I was wondering if anyone could assist me with the above problem, I'm struggling to get the function in the form $u(x,y)+iv(x,y)$ to then calculate the partial derivatives.

I understand that I need to use the chain rule, but would I first replace $z$ with $z=re^{i\theta}$ $\implies z^{2}=r^{2}e^{2i\theta}$. Then we would have that $f(z)=\sinh(r^{2}e^{2i\theta})$. I'm stuck here, any help would be great.

Answer 1

Recall the identity $$\sinh z = \frac{e^z - e^{-z}}{2}.$$ Then $$\sinh z^2 = \frac{e^{x^2 - y^2} e^{2xy i} - e^{-(x^2-y^2)} e^{-2xy i}}{2}.$$ Then isolate the real and imaginary components from $$\frac{e^{x^2 - y^2}}{2} (\cos 2xy + i \sin 2xy) - \frac{e^{-(x^2-y^2)}}{2} (\cos (-2xy) + i \sin (-2xy)).$$ The rest is straightforward.

Answer 2

Consider \begin{align} \sinh(it)&=\frac{e^{it}-e^{-it}}{2}=i\frac{e^{it}-e^{-it}}{2i}=i\sin t \\ \cosh(it)&=\frac{e^{it}+e^{-it}}{2}=\cos t \end{align} Then, if $z=x+iy$, $z^2=(x^2-y^2)+2ixy$ and \begin{align} \sinh(z^2) &=\sinh(x^2-y^2)\cosh(2ixy)+\cosh(x^2-y^2)\sinh(2ixy)\\ &=\sinh(x^2-y^2)\cos(2xy)+i\cosh(x^2-y^2)\sin(2xy)\\ \end{align} Thus you have \begin{align} u(x,y)&=\sinh(x^2-y^2)\cos(2xy)\\ v(x,y)&=\cosh(x^2-y^2)\sin(2xy) \end{align}