Solve $z^{1+i}=4$

Question

$\def\Log{\operatorname{Log}}$ I have to solve $z^{1+i}=4$. Is there any easy way?

I'm starting like this:

$$e^{(1+i)\Log z}=e^{2\Log2}$$

Then I solve

$$(1+i)\Log z=2\Log2$$

But I really doubt I can do that like this, since $\Log$ is multivalued...

Edit:

$$(1+i)(\ln|z|+i\arg z+i2k\pi)=2(\ln|2|+i2n\pi), k,n\in\mathbb{Z}$$

Answer 1

Use the definition $\Log z = \ln |z|+i\arg z$. $$e^{(1+i)\Log z}=e^{(1+i)\ln |z|+(1+i)i\,\arg z}=e^{\ln|z|-\arg z}\cdot e^{i(\ln |z|+\arg z)}=4$$ Taking the log-maginitude and argument of the above equation, $$ \ln|z|-\arg z = \ln 4 $$ $$ \ln|z|+\arg z = 2\pi k $$ Solving this system gives you a value of $z$ for each integer $k$.