Find the Value of z in z^{1+i}=4 using Logarithms

[Terrell]

Homework Statement

Find $z$ in $z^{1+i}=4$. Is my solution correct

Homework Equations

$\log(z_1 z_2)=\log(z_1)+\log(z_2)$ such that $z_1, z_2\in \{z\in\Bbb{C} : (z=x+iy) \land (x\in\Bbb{R}) \land -\infty \lt y \lt +\infty\}$
$re^{i\theta}=r(\cos\theta + i\sin\theta)$

The Attempt at a Solution

Note $4=4(\cos 0 +i\sin 0)$. Thus,
\begin{align}
r^{1+i}e^{i\theta(1+i)}&=r^{1+i}e^{-\theta}(\cos\theta + i\sin\theta)=4(\cos 0 +i\sin 0)=4\
r^{1+i}e^{-\theta}&=4 \quad \text{and}\quad \theta=0\\
\log_e(r^{1+i}e^{-\theta})&=\log_{e}(4)\\
\log_(r^{1+i})+\log(e^{-\theta})&=\log_{e}(4)\\
(1+i)\log_{e}(r)-\theta&=\log_{e}(4)\\
\log_{e}(r)&=\frac{\log_e(4)}{1+i}\\
\log_{e}(r)&=\log_{e}(4^{\frac{1}{1+i}})\\
r&=4^{\frac{1}{1+i}}
\end{align}

 

[mjc123]

In step 1 you appear to be making the false assumption that r¹⁺ⁱ is real. (Express it as e^lnr(1+i) and see.)
I think it would be easier if you start by taking logs
ln4 = (1+i)lnz
and see where that gets you.