How Do you Evaluate $z=(2+2i)^{4i}$

Question

I am trying to evaluate $z=(2+2i)^{4i}$ in the form $u+iv$ but I unsure how to. I am aware of the notion of complex logarithm and I initially thought it to be related to logs in some way but I could not get the correct answer. I would preferably like to just be pointed in the right direction so that I can try to get the correct answer for myself.

Thanks in advance!

$(2+2i)^{4i}=e^{4i(\log2+\log(1+i))}$, then use the fact that $1+i=\sqrt{2}e^{i\pi/4}$ to get (using the principal $\log$) that $(2+2i)^{4i}=e^{4i(\log(2)+\frac12 \log(2)+i\pi/4}=e^{6i\log(2)-\pi}=2^{6i}e^{-\pi}$ — Evariste, Nov 01 '21 at 14:04

score 2 · Accepted Answer · answered Nov 01 '21 at 14:26

You can have a look at my general presentation for $z^u$ when both $z$ and $u$ are complex numbers.

https://math.stackexchange.com/a/3729281/399263

So as I indicated express $z=2i+2$ in polar form and write the periodicity explicitly: $$2+2i=2\sqrt{2}\exp\left(i\frac{\pi}4+2ik\pi\right)\text{ for }k\in\mathbb Z$$

Then multiply the inners of the exponential by $u=4i$ in cartesian form:

$(2+2i)^{4i}=\exp\Big(4i\times\big(\underbrace{\ln(2\sqrt{2})}_{\frac32\ln(2)}+i\frac{\pi}4+2ik\pi\big)\Big)=\exp\Big(6i\ln(2)-\pi-8k\pi\Big)$

To make the principal value appear, separate what's depends on $k$ and what's not:

$$\begin{cases}z_{[0]}=\exp\big(6i\ln(2)-\pi)\big)=e^{-\pi}\Big(\cos(6\ln(2))+i\sin(6\ln(2))\Big)\\\\w^k=\exp\big(-8k\pi\big)=(e^{-8\pi})^k\end{cases}$$

Finally you get the multivalued expression $(2+2i)^{4i}=z_{[0]}\times w^k$

With $ k\in\mathbb Z$, where $z_{[0]}$ is called the principal value, and $w$ the multiplicative factor.

Better than I wanted to do it. – Ataulfo Nov 01 '21 at 14:31 — Ataulfo, Nov 01 '21 at 14:31

How Do you Evaluate $z=(2+2i)^{4i}$

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