Why is $e^{z^2}$ holomorphic?

Question

I read somewhere that $e^{z^2} $ is holomorphic function, but I can't think of an easy way to prove that. I thought of using Cauchy Riemann equations, but that's probably overkill. Is there a simple approach to show that this function is holomorphic?

Answer 1

Of course it's a composition of holomorphic function, and thus it's holomorphic. But you can also use Cauchy-Riemann equation observing that $$e^{z^2}=e^{(x+iy)^2}=e^{x^2-y^2+2ixy}=e^{x^2-y^2}\cos(2xy)+ie^{x^2-y^2}\sin(2xy).$$

Answer 2

It's because you can obtain $e^{z^2}$ composing the exponential function with the function $z\mapsto z^2$, both of which are holomorphic.

Answer 3

With substitution $z^2$ in series expansion of $e^z$ we have $$e^{z^2}=\sum_{n=0}^\infty\dfrac{z^{2n}}{n!}$$ which shows $e^{z^2}$ is holomorphic.