Convert $(1+i) ^ {1+i}$ to polar form

Question

Can someone please help me understand the exponent/logarithm relationships to get through this problem? Thank you.

Answer 1

We will need the logarithm function. In the complex plane this involves a choice, and the argument is multi-valued. We choose the principal one, which is the one that we usually expect. For $z=re^{it}$, we have $\log z=\log |z|+it$.

We need to write $1+i$ in polar form: $1+i=\sqrt2\,e^{i\pi/4}$.

By definition, \begin{align}(1+i)^{1+i}&=\exp((1+i)\log(1+i))=\exp((1+i)(\log\sqrt2+i\tfrac\pi4)\\ \ \\ &=\exp(\tfrac12\,(1+i)(\log2+i\tfrac\pi2)=\exp(\tfrac12\,(-\tfrac\pi2+\log 2 + i(\tfrac\pi2+\log 2))\\ \ \\ &\exp(-\tfrac\pi4+\tfrac12\log 2 + i(\tfrac\pi4+\tfrac12\log 2))\\ \ \\ &=e^{-\pi/4}\sqrt2\,\exp\left[i\left(\tfrac\pi4+\log \sqrt2\right)\right]. \end{align}

Answer 2

Hints to explore

• Can you write $1+i$ in polar form $re^{i\theta}$ with $r$ being a positive real?

• Can you write $1+i$ in the form $e^{s+i \theta}$ with $s=\log(r)$ being a real number?

• Then consider $\left(e^{s+i \theta}\right)^{1+i} = e^{(s+i\theta)(1+i)} = e^{(s-\theta)+i(s+ \theta)}$

• Then remember that $e^{x+iy} = e^x \cos(y) + i e^x\sin(y)$

That will get you one value. But complex powers are not so conveniently defined and $re^{i\theta} = re^{i(\theta+2n\pi)}$ for all integers $n$. So you should consider what different values of $n$ do to your earlier answer