# Evaluating $\log(2+i)$

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.

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For $z\in\mathbb C$, perhaps try expressing $z$ in polar form? i.e. $z = re^{i\theta}$ for $r,\theta$ real. Then using usual properties of logs ... – Eugene Shvarts Nov 14 '12 at 11:38
Indeed, plotting $2+i$ as the point $(2,1)$ makes the hypotenuse of length $\sqrt{5}$ and the angle $\arctan 1/2$. – coffeemath Nov 14 '12 at 11:46

## 2 Answers

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

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

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

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