The computation of $(1+i)^{1+i}$

Question

Being bound to compute $(1+i)^{1+i}$, I guess I should somehow use the complex logarithm.

But how can I firstly write the stated number in the form $re^{i\theta}$?

Answer 1

Considering $$A=(1+i)^{1+i}$$ you could start taking the logarithms $$\log(a)=(1+i)\log(1+i)$$ Now, using the representaion of complex logarithms, you have $$\log(1+i)=\frac{\log (2)}{2}+i\frac{ \pi }{4}$$ which makes $$\log(A)=\left(\frac{\log (2)}{2}-\frac{\pi }{4}\right)+i\left(\frac{\log (2)}{2} +\frac{\pi }{4}\right)$$

I am sure that you can take from here and arrive to the desired result.

Answer 2

Using $1 + i = \sqrt{2} \, e^{\pi i/4}$ and $a = e^{\ln a}$ then \begin{align} (1+i)^{1+i} &= (\sqrt{2} \, e^{\pi i/4})^{\sqrt{2} \, e^{\pi i/4}} \\ &= e^{\sqrt{2} \, \ln(\sqrt{2}) \, e^{\pi i/4}} \, e^{\sqrt{2} \, \pi i \, e^{\pi i/4}/4 } \\ &= e^{(1+i) \, \ln(2)/2} \, e^{(i-1) \, \pi/4} \\ &= e^{(2 \, \ln(2) - \pi)/4} \, e^{i \, (2 \ln(2) + \pi)/4} \\ &= e^{(2 \, \ln(2) - \pi)/4} \, ( \cos((2 \ln(2) + \pi)/4) + i \, \sin((2 \ln(2) + \pi)/4) ) \end{align}