computing $\ln[(1+i)^{2}]$ How to compute Natural logarithm of complex numbers?
and how to verify our answer?
in example: $\ln[(1+i)^{2}]$
 A: Solution 1
The point corresponding to the complex number $1+i$ is $\sqrt{2}$ units from the origin, located $\pi/4$ radians (i.e. 45°) counter-clockwise from the real axis, and therefore it can be represented in polar coordinates as
$$1+i=\sqrt{2}e^{i\pi/4}$$
This representation is not unique, however, because you can add any multiple of $2\pi$ radians (360°) to the angle.  So more generally
$$1+i = \sqrt{2} e^{i\pi/4 + 2n\pi i}$$
Squaring this is easy:
$$(1+i)^2 = 2e^{i\pi/2 + 4n\pi i}$$
Therefore the logarithm is:
$$\ln(1+i)^2 = \ln(2) + \frac{i \pi}{2} + 4n\pi i$$
Notice that this is not a single value, but a collection of infinitely-many different values corresponding to different choices of $n$.  This is because the $\ln$ function on the complex plane is a multi-valued function; see here for more information.
Solution 2
Note that $(1+i)^2 = 2i$.  So really the question is asking us to compute $\ln(2i)$.  Since $i=e^{i\pi/2}$, this is $\ln(2e^{i\pi/2})= \ln(2) + \ln(e^{i\pi/2}) = \ln(2)+\frac{i\pi}{2}$.  In contrast, this approach gives only the principal branch of the logarithm function.
A: To compute $\ln[(1+i)^{2}]$ :
first we should answer : $(1+i)^{2}$ as follows :
$\sqrt{2} cis\dfrac{\pi}{4}=\begin{cases}  r=\sqrt{1^{2}+1^{2}}=\sqrt{2} \\ \tan\theta=\dfrac{1}{1}=1 =>\theta=\dfrac{\pi}{4} \end{cases}$
$[\sqrt{2} cis\dfrac{\pi}{4}]^{2}=2Cis\dfrac{\pi}{2}
$
so we reach to this point : $\ln(2Cis\dfrac{\pi}{2})$ :
$
\ln(2Cis\dfrac{\pi}{2})=\ln(2)+i(\dfrac{\pi}{2}+2k\pi)$ 
assuming ($k=0$) => $0.693147181 + 1.57079633 i$
but to proof our answer :
$
e^{\ln(2)+i(\dfrac{\pi}{2}+2k\pi)}=\sqrt{2cis(\dfrac{\pi}{2}+2k\pi)}=\sqrt{2} cis\dfrac{\pi}{4}
$
A: $$\ln\left(\left(1+i\right)^2\right)=\ln\left(\left(|1+i|e^{\arg(1+i)i}\right)^2\right)=$$
$$\ln\left(\left(\sqrt{2}e^{\frac{\pi i}{4}}\right)^2\right)=\ln\left(2e^{\frac{\pi i}{2}}\right)=\frac{\pi i}{2}+2i\pi\lfloor{\frac{-\frac{\pi}{2}+\pi}{2\pi}}\rfloor+\ln(2)=\ln(2)+\frac{\pi i}{2}$$
