Solving $z^{1+i}=4$.

Question

Solution:

Let $z=re^{i\theta}=r(\cos \theta + i\sin \theta)$. Then $z^{1+i}=e^{i\theta(1+i)}=e^{i\theta -\theta}=4$. So $e^{i\theta-\theta}=e^{-\theta}e^{i\theta}=e^{-\theta}(\cos \theta + i\sin \theta)=4(\cos 0 +i\sin 0) \Longleftrightarrow e^{-\theta}=4 \quad \text{and}\quad \theta =0$. But, $e^0 =1$. What am I doing wrong?

EDIT 10/26/18 Would like to provide a more systematic and traditional approach: \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}}\\ r&=4^{\frac{1-i}{2}}=[4^\frac{1}{2}]^{1-i}=2^{1-i} \end{align}

Answer 1

$$4 = 2^2 = z^{1 + i}$$

$$2 \log 2 = (1+i) \log z$$

$${2 \over 1+i} \log 2 = \log z$$

$$2^{2 \over 1 + i} = z$$

$$2^{2(1-i) \over (1+i)(1-i)}$$

$$2^{2(1-i) \over 2} = z$$

$$2^{1 - i} = z$$

Check:

$$\left( 2^{1-i}\right)^{1+i} = 2^{1+1} = 2^2 = 4$$

Answer 2

$4=2^2$

$2=(1-i)(1+i)$

$z^{1+i}=2^{(1-i)(1+i)}$

Taking $1+i$ th root of both sides we get:

$z=2^{1-i}$

Answer 3

A systematic way would go like this. Let the power function be defined as $z^a = e^{a \ln z}$ and the logarithm be defined as $\ln z = \ln |z| + i \arg z, \, -\pi < \arg z \leq \pi$. $e^z$ is periodic with period $2 \pi i$, therefore $$z^c = w \,\Leftrightarrow\, e^{c \ln z} = e^{\ln w} \,\Leftrightarrow\, c \ln z = \ln w + 2 \pi i k \,\Leftrightarrow \\ \ln |z| + i \arg z = \operatorname{Re} \zeta + i \operatorname{Im} \zeta \,\Leftrightarrow \\ z = e^\zeta \land -\pi < \operatorname{Im} \zeta \leq \pi, \\ \text{where } \zeta = \frac {\ln w + 2 \pi i k} c, \;k \in \mathbb Z.$$ The number of solutions will depend on $c$ and $w$. Substituting $c = 1 + i, \, w = 4$ gives $-\pi < \pi k - \ln 2 \leq \pi$, therefore there are two solutions $z = 2^{1 - i}$ and $z = -2^{1 - i} e^\pi$, corresponding to $k = 0$ and $k = 1$.