Question Number 224739 by thetpainghtun_111 last updated on 30/Sep/25
$$\mathrm{z}\:=\:\mathrm{r}\:\left(\mathrm{cos}\:\theta\:+\:\mathrm{i}\:\mathrm{sin}\:\theta\right),\:\mathrm{find}\:\frac{\mathrm{z}}{\overset{−} {\mathrm{z}}}\:+\frac{\overset{−} {\mathrm{z}}}{\mathrm{z}}. \\ $$
Answered by Frix last updated on 30/Sep/25
![z=re^(iθ) ⇔ z^ =re^(−iθ) (z/z^ )+(z^ /z)=e^(2iθ) +e^(−2iθ) = [e^x +e^(−x) =2cosh x] =2cosh (2iθ) =2cos 2θ](https://www.tinkutara.com/question/Q224744.png)
$${z}={r}\mathrm{e}^{\mathrm{i}\theta} \:\Leftrightarrow\:\bar {{z}}={r}\mathrm{e}^{−\mathrm{i}\theta} \\ $$$$\frac{{z}}{\bar {{z}}}+\frac{\bar {{z}}}{{z}}=\mathrm{e}^{\mathrm{2i}\theta} +\mathrm{e}^{−\mathrm{2i}\theta} =\:\:\:\:\:\:\:\:\:\:\left[\mathrm{e}^{{x}} +\mathrm{e}^{−{x}} =\mathrm{2cosh}\:{x}\right] \\ $$$$=\mathrm{2cosh}\:\left(\mathrm{2i}\theta\right)\:=\mathrm{2cos}\:\mathrm{2}\theta \\ $$