Imaginary and real part of $\sin\left(iz^2\right)$

Question

I really can’t wrap my head around this question. I want to find the imaginary and the real part of $\sin\left(iz^2\right)$ where $z= x+iy$. I thought of starting from $$\sin\left(iz^2\right) = \sinh\left(z^2\right),$$ but got a way too long working out. If anyone got a shortcut to answer questions like these, I’d be very thankful.

Answer 1

Hint: $$\sin(iz^2) \rightarrow \sin (i(x+yi)^2) \rightarrow \sin (i(x^2 + 2xyi - y^2)) \rightarrow \sin (ix^2 - 2xy -iy^2)$$

Then you have $$\sin (-2xy + i(x^2 - y^2))$$

Also $$\sin (u+iv) = \sin u \cosh iv + i \cos u \sinh iv \rightarrow \sin u \cosh v + i \cos u \sinh v$$

Can you finish?

Answer 2

Write: $$ \sinh{z^2} = \frac{e^{z^2} - e^{-z^2}}{2} $$ If $z = x + iy$, then: $$ e^{z^2} = e^{x^2 - y^2 + 2xyi} = e^{x^2 - y^2}(\cos(2xy) + i\sin(2xy)) $$ Can you do the same for $e^{-z^2}$?