letumsnemesislh

2022-07-04

Solve $2\mathrm{cosh}z+\mathrm{sinh}z=i$ for z

Expert

$2\mathrm{cosh}\left(z\right)+\mathrm{sinh}\left(z\right)=i$
${e}^{z}+{e}^{-z}+\frac{{e}^{z}-{e}^{-z}}{2}=i$
$2{e}^{z}+2{e}^{-z}+{e}^{z}-{e}^{-z}=2i$
$3{e}^{z}+{e}^{-z}=2i$
$3{e}^{z}+\left({e}^{z}{\right)}^{-1}=2i$
Let ${e}^{z}=x$
$3x+{x}^{-1}=2i$
$3{x}^{2}+1=2ix$
$3{x}^{2}-2ix+1=0$
We solve equations of the form $a{x}^{2}+bx+c=0$ with
${x}_{1,2}=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
By plugging in a=3, b=−2i and c=1:
${x}_{1,2}=\frac{2i±\sqrt{-4-4\cdot 3\cdot 1}}{6}$
$=\frac{2i±\sqrt{-4-12}}{6}=\frac{2i±\sqrt{-16}}{6}=\frac{2i±4i}{6}=\frac{i}{3}±\frac{2i}{3}$
The two roots are therefore
${x}_{1}=i,\phantom{\rule{1em}{0ex}}{x}_{2}=-\frac{i}{3}$
Since we required ${e}^{z}=x$, $z=\text{Ln}\left(x\right)$ (Ln being the complex-valued logarithm). We are lazy and throw both values in Wolfram Alpha and get
${z}_{1}=\mathrm{ln}\left(i\right)=\frac{i\pi }{2},\phantom{\rule{1em}{0ex}}{z}_{2}=\mathrm{ln}\left(-\frac{i}{3}\right)=-\mathrm{ln}\left(3\right)-\frac{i\pi }{2}$
And now we have ${z}_{1}$ and ${z}_{2}$, which are the solutions to the initial equation. Notice that the solution to that complex logarithm has arbitrary multiples of $2\pi i$ in it

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