Answer to Question #313820 in Complex Analysis for Jyo

Question

Show that 𝑠𝑖𝑛𝑖𝑦 = 𝑖 𝑠𝑖𝑛hy

Answer 1

Solution in the first way. We apply the following formulas

$\sin z=\frac{e^{iz}-e^{-iz}}{2i}$ and $\sinh z=\frac{e^{z}-e^{-z}}{2}$. Then $\sin iy=\frac{e^{i(iy)}-e^{-i(iy)}}{2i}=\frac{e^{-y}-e^{y}}{2i}=-\frac{1}{i}\frac{e^{y}-e^{-y}}{2}=i\sinh y$, as required.

Solution in the second way. We apply the following formulas

$\sin z=\sum\limits_{n=0}^{+\infty}\frac{(-1)^n}{(2n+1)!}z^{2n+1}$ and $\sinh z=\sum\limits_{n=0}^{+\infty}\frac{1}{(2n+1)!}z^{2n+1}$. Then

$\sin iy=\sum\limits_{n=0}^{+\infty}\frac{(-1)^n}{(2n+1)!}(iy)^{2n+1}=\sum\limits_{n=0}^{+\infty}\frac{(-1)^n}{(2n+1)!}i^{2n+1}y^{2n+1}$

$=\sum\limits_{n=0}^{+\infty}\frac{(-1)^n}{(2n+1)!}i(-1)^ny^{2n+1}=i\sum\limits_{n=0}^{+\infty}\frac{1}{(2n+1)!}y^{2n+1}=i\sinh y$, as required.