Complex exercise Level 1 asked in the mathematics teacher group

Exercise:

Show that, $\cos \left( i\ln \left( \frac{a+ib}{a-ib} \right) \right)=\frac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}$ where $a,b \in \mathbb{R}$

Solution: Let ${{z}_{1}}=a+ib\,\,\And \,\,{{z}_{2}}=a-ib$ and notice that ${{z}_{2}}$ is the conjugate of ${{z}_{1}}$ hence ${{z}_{2}}={{\bar{z}}_{1}}$

Thus $if\,\,z=a+ib=r{{e}^{i\theta }}$ then $\bar{z}=a-ib=r{{e}^{-i\theta }}$ so $\frac{a+ib}{a-ib}=\frac{r{{e}^{i\theta }}}{r{{e}^{-i\theta }}}={{e}^{i\left( \theta +\theta  \right)}}={{e}^{i2\theta }}$

We have $a+ib=r\left( \cos \theta +i\sin \theta  \right)$ so $\cos \theta =\frac{a}{r}=\frac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$ and $\sin \theta =\frac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$

L.H.S$=\cos \left( i\ln \left( \frac{a+ib}{a-ib} \right) \right)=\cos \left( i\ln {{e}^{i2\theta }} \right)=\cos \left( i\left( i2\theta  \right)\ln e \right)=\cos \left( -2\theta  \right)=\cos 2\theta$

        $={{\cos }^{2}}\theta -{{\sin }^{2}}\theta =\frac{{{a}^{2}}}{{{a}^{2}}+{{b}^{2}}}-\frac{{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}=\frac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}=$R.H.S