Evaluating $\log(2+i)$

Question

Could someone please help me understand why $\log(2+i)=\log\sqrt{5}+i\tan^{-1}(1/2)$. I don't know how to evaluate logarithms involving complex numbers.

Answer 1

$$ \begin{eqnarray} \log(2+i)&=&\log\left(\sqrt{(\Re{(2+i)})^2+(\Im{(2+i)})^2 }e^{i\tan^{-1}\left(\frac{{\Im{(2+i)}}}{{\Re{(2+i)}}}\right)}\right)\\ &=&\log\left(\sqrt{2^2+1^2 }e^{i\tan^{-1}(\frac12)}\right)\\ &=&\log\left(\sqrt{2^2+1^2 }\right)+\log\left(e^{i\tan^{-1}(\frac12)}\right)\\ &=&\log\sqrt{5}+i\tan^{-1}(1/2)\\ \end{eqnarray} $$

Answer 2

$log(Z)=ln|Z| +iarg(Z)$

where $|Z|^2=(Re(Z))^2 + (Im(Z))^2$ and $arg(Z)$ is the angle that Z makes with the positive real axis.