computing $\ln[(1+i)^{2}]$

Question

How to compute Natural logarithm of complex numbers?

and how to verify our answer?

in example: $\ln[(1+i)^{2}]$

Answer 1

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} $

Answer 2

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.

Answer 3

$$\ln\left(\left(1+i\right)^2\right)=\ln\left(\left(|1+i|e^{\arg(1+i)i}\right)^2\right)=$$ $$\ln\left(\left(\sqrt{2}e^{\frac{\pi i}{4}}\right)^2\right)=\ln\left(2e^{\frac{\pi i}{2}}\right)=\frac{\pi i}{2}+2i\pi\lfloor{\frac{-\frac{\pi}{2}+\pi}{2\pi}}\rfloor+\ln(2)=\ln(2)+\frac{\pi i}{2}$$