blestimd4pz

2022-03-30

Prove that $\mathrm{sin}\left(z\right)=\frac{{e}^{iz}-{e}^{-iz}}{2i}=\mathrm{sin}\left(x\right)\text{cosh}\left(y\right)+i\mathrm{cos}\left(x\right)\text{sinh}\left(y\right)$
For complex z=x+iy. How to prove that
$\mathrm{sin}\left(z\right)=\frac{{e}^{iz}-{e}^{-iz}}{2i}=\mathrm{sin}\left(x\right)\text{cosh}\left(y\right)+i\mathrm{cos}\left(x\right)\text{sinh}\left(y\right)$
by using the power series definition
$\mathrm{sin}\left(z\right)=z-\frac{{z}^{3}}{3!}+\frac{{z}^{5}}{5!}-\dots$
and Euler's formula
${e}^{z}={e}^{x}\left(\mathrm{cos}\left(y\right)+i\mathrm{sin}\left(y\right)\right)$?

### Answer & Explanation

Ali Price

Is it really necessary to use the power series definition of sine? The left equality is easier to prove right-to-left with the exp. power expansion.
$\mathrm{sin}z=\sum _{n=0}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{n}}{\left(2n+1\right)!}{z}^{2n+1}=\sum _{n=0}^{\mathrm{\infty }}\frac{{i}^{2n}}{\left(2n+1\right)!}{z}^{2n+1}$
$=\frac{1}{i}\sum _{n=0}^{\mathrm{\infty }}\frac{{i}^{2n+1}}{\left(2n+1\right)!}{z}^{2n+1}=\frac{1}{i}\sum _{n=0}^{\mathrm{\infty }}\frac{{\left(iz\right)}^{2n+1}}{\left(2n+1\right)!}$
Consider $\frac{1-{\left(-1\right)}^{k}}{2}$ which equals 1 for odd k and 0 for even k.
$\mathrm{sin}z=\frac{1}{i}\sum _{k=0}^{\mathrm{\infty }}\left(\frac{1-{\left(-1\right)}^{k}}{2}\right)\frac{{\left(iz\right)}^{k}}{k!}=\frac{1}{2i}\sum _{k=0}^{\mathrm{\infty }}\frac{1-{\left(-1\right)}^{k}}{k!}{\left(iz\right)}^{k}$
$=\frac{1}{2i}\sum _{k=0}^{\mathrm{\infty }}\frac{{\left(iz\right)}^{k}}{k!}-\frac{1}{2i}\sum _{k=0}^{\mathrm{\infty }}\frac{{\left(-iz\right)}^{k}}{k!}$
$=\frac{{e}^{zi}-{e}^{-zi}}{2i}$
$=\frac{{e}^{\left(x+yi\right)i}-{e}^{-\left(x+yi\right)i}}{2i}$
$=\frac{{e}^{-y+\xi }-{e}^{y-\xi }}{2i}$
$=\frac{{e}^{-y}{e}^{\xi }-{e}^{y}{e}^{-\xi }}{2i}$
$=\frac{{e}^{-y}\left(\mathrm{cos}x+i\mathrm{sin}x\right)-{e}^{y}\left(\mathrm{cos}x-i\mathrm{sin}x\right)}{2i}$
$=\frac{i}{i}\cdot \mathrm{sin}x\cdot \frac{{e}^{y}+{e}^{-y}}{2}-\frac{1}{i}\cdot \mathrm{cos}x\cdot \frac{{e}^{y}-{e}^{-y}}{2}$
$=\mathrm{sin}x\text{cosh}y+i\mathrm{cos}x\text{sinh}y$

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