complex expression to the power of a complex expression

Question

I have a math exam tomorrow, and i am not sure with my solution for a exercise. can you please tell me if i am right.
Question is: $$(1+i)^{(1-i)}$$
My solution is: $$\sqrt{2} e^{(i {\pi\over4})^{(1-i)}} = \sqrt{2} e^{i {\pi\over4} - i² {\pi\over4}} = \sqrt{2}e^{i{\pi\over2}}$$ Is this correct?

Thanks in advance.

EDIT: changed \pi\over2 to \pi\over4 due to comments, thank you :)

Answer 1

Follow the comments, because otherwise you shoud elevate $\sqrt{2}$ to the power $(1 - i)$ too.

$ \left( \sqrt{2} e^{(i \frac{\pi}{4})} \right)^{(1-i)} = (\sqrt{2})^{(1-i)} e^{i \frac{\pi}{4}(1-i)}$

Answer 2

There are infinitely many solutions when raising a number to complex power. For example: \begin{equation} i^i=(e^{(2n+1)\pi i})^i=e^{-(2n+1)\pi} \end{equation} where $n$ is any integer.

$1-i$ can be also written as $\sqrt2e^{-\pi i/4}$, because \begin{equation} \sqrt2e^{-\pi i/4} =\sqrt2\left(\cos\frac{\pi}{4}-i\sin\frac{\pi}{4}\right) =\sqrt2\left(\frac{1}{\sqrt2}-i\frac{1}{\sqrt2}\right)=1-i. \end{equation} Actually, it's also any of $e^{(2n-1/4)\pi i}$, where $n$ is any integer, because $e^{2n\pi i}=1$.

Therefore, \begin{eqnarray} (1-i)^{1-i}&=&(1-i)(1-i)^{-i} \\ &=& (1-i)(\sqrt2e^{(2n-1/4)\pi i})^{-i} \\ &=& (1-i)e^{-(2n-1/4)\pi} 2^{-i/2} \\ &=& (1-i)e^{-(2n-1/4)\pi} e^{-i(\ln2)/2} \\ &=& (1-i)e^{-(2n-1/4)\pi} \left( \cos\frac{\ln2}{2}-i\sin\frac{\ln2}{2} \right)\\ &=& (1-i)e^{-(2n-1/4)\pi} \left( \cos\frac{\ln2}{2}+\sin\frac{\ln2}{2} \right)\\ \end{eqnarray}