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

Question

I know it must be pretty easy but i just can't get it. I try to use the fact that $$e^{clogz}=z^c$$ $c \in C$ but still i can't. Can you write all of the steps in order to understand what i'm missing in solving equations of this type.

I need to solve $$z^{1+i}=4$$

Answer 1

\begin{align*} z^{1+i} &= 4 \\ (1+i) \log z &= \ln 4+2n\pi i \\ \log z &= \frac{\ln 4+2n\pi i}{1+i} \\ &= \frac{(\ln 4+2n\pi i)(1-i)}{2} \\ &= \left( \ln 2+n\pi \right)+i(n\pi-\ln 2) \\ z &= 2e^{n\pi} \operatorname{cis} (n\pi-\ln 2) \end{align*}

where $n\in \mathbb{Z}$

Answer 2

Hint: Note that $4=\exp( \log(4)+2\pi i k)$ with $k\in {\Bbb Z}$ and find the solutions to $$(1+i)\log z = \log(4)+2\pi i k$$

Answer 3

Generalize the problem, when $\text{q}\in\mathbb{C}$ and $\text{a}\in\mathbb{R}^+$, solve $\text{z}$:

$$\text{z}^{\text{q}}=\text{a}\Longleftrightarrow$$

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Take the natural logarithm of both sides and bring the power out in front:

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$$\text{q}\ln(\text{z})=\ln(\text{a})+2\pi ki\Longleftrightarrow$$

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Where $k\in\mathbb{Z}$.

Divide both sides by $\text{q}$:

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$$\ln(\text{z})=\frac{\ln(\text{a})+2\pi ki}{\text{q}}\Longleftrightarrow$$

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Cancel logarithms by taking exp of both sides:

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$$\text{z}=\exp\left[\frac{\ln(\text{a})+2\pi ki}{\text{q}}\right]$$