TT1 Problem 1 (night)

[Victor Ivrii]

Show that
\begin{equation*}
|\sin (z)|^2 = \sin^2 (x)+ \sinh^2 (y)
\end{equation*}
 for all complex numbers $z = x+yi$.

[ZhenDi Pan]

My solution:
\begin{align*}
| \sin(z)^{2}| & =[\sin (x) \cos(iy)]^{2}+[\cos (x)\sin(iy)]^{2}\\
 & = [\sin^{2}(x)\cosh^{2}(y)]+[\cos^{2}(x)\sinh^{2}(y)] \\
 & = [\sin^2(x)(1+\sinh^2(y))]+[(1-\sin^2(x))\sinh^2(y)] \\
 & = \sin^2(x)+\sinh^2(y)-\sin^2(x)\sinh^2(y)+\sin^2(x)\sinh^2(y) \\
 & = \sin^2(x)+\sinh^2(y)
\end{align*}
Please reformat (see how I changed your first line)