How do I solve $|\sinh(x+iy)|^2 = (\sin(y))^2+(\sinh(x))^2$

Question

How do I solve this ?

$|\sinh (x+iy)|^2 = ( \sin (y))^2+ ( \sinh (x))^2$

I'm not sure how to solve the left hand side.

Answer 1

As $\displaystyle\sin(iA)=i\sinh A$ and $\cos(iB)=\cosh B$

$\displaystyle\sinh(x+iy)=\frac{\sin[i(x+iy)]}i=\frac{\sin(ix-y)}i=\frac{\sin(ix)\cos y-\cos(ix)\sin y}i$

$\displaystyle=\frac{i\sinh x\cos y-\cosh x\sin y}i=\sinh x\cos y+i\cosh x\sin y $

$$\implies|\sinh(x+iy)|^2=\sinh^2x\cos^2y+\cosh^2x\sin^2y$$

$$=\sinh^2x(1-\sin^2y)+(1+\sinh^2x)\sin^2y=\cdots$$

( utilizing $\displaystyle\cos^2y=1-\sin^2y$ and $\displaystyle\cosh^2x=1+\sinh^2x$ to eliminate $\cos y,\cosh x$)

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Alternatively, using $\displaystyle2\sinh A=e^A-e^{-A},$

$$2\sinh(x+iy)=e^{x+iy}-e^{-x-iy}=e^xe^{iy}-e^{-x}e^{-iy}$$

then using Euler's formula,

$$2\sinh(x+iy)=e^x(\cos y+i\sin y)-e^{-x}(\cos y-i\sin y)$$ $$=\cos y(e^x-e^{-x})+i\sin y(e^x+e^{-x})$$

$$\implies|2\sinh(x+iy)|^2=[\cos y(e^x-e^{-x})]^2+[\sin y(e^x+e^{-x})]^2$$

$$=(1-\sin^2y)(e^x-e^{-x})^2+\sin^2y(e^x+e^{-x})^2$$

$$=(e^x-e^{-x})^2+\sin^2y[(e^x+e^{-x})^2-(e^x-e^{-x})^2]$$

$$=(2\sinh x)^2+\sin^2y[4(e^x)(e^{-x})]$$