Evaluation for Complex number

Exercise:

Evaluate, $2\cos \left( i\ln 4 \right)+2{{\sin }^{2}}\left( i\ln 2 \right)+i\sin \left( i\ln 4 \right)$ where $i\in \mathbb{C}$ such that ${{i}^{2}}=-1$

Solution: we know that $\ln 4=\ln {{2}^{2}}=2\ln 2$ and ${{\sin }^{2}}\theta =\frac{1-\cos 2\theta }{2}$

So $2\cos \left( i\ln 4 \right)+2{{\sin }^{2}}\left( i\ln 2 \right)+i\sin \left( i\ln 4 \right)$ $=2\cos \left( i\ln 4 \right)+\left( 1-\cos \left( 2\left( i\ln 2 \right) \right) \right)+i\sin \left( i\ln 4 \right)$

$=1+2\cos \left( i\ln 4 \right)-\cos \left( 2\left( i\ln 2 \right) \right)+i\sin \left( i\ln 4 \right)$ $=1+2\cos \left( i\ln 4 \right)-\cos \left( i\ln 4 \right)+i\sin \left( i\ln 4 \right)$

$=1+\cos \left( i\ln 4 \right)+i\sin \left( i\ln 4 \right)$ $=1+{{e}^{i\left( i\ln 4 \right)}}=1+{{e}^{-\ln 4}}=1+{{e}^{\ln {{4}^{-1}}}}=1+\frac{1}{4}=\frac{5}{4}$