Find the principal value of a complex number

Question

I've been given the number \begin{equation} i^{2-3i} \end{equation} I've used the following formula \begin{equation} z^c=e^{(c\log{z})} \end{equation} Which has given me \begin{equation} e^{(2-3i)\log{i}} \end{equation} I'm unsure if this is the principal value? Or on what I have to do next

Answer 1

$$i^{2-3i}=e^{(2-3i)\text{Log}\,i}=e^{(2-3i)\left(\log|i|+i\arg i\right)}=e^{(2-3i)\cdot\pi i/2}=e^{\pi i+\frac32\pi}=-e^{\frac{3\pi}2}$$

The principal value of a complex number is usually accepted as having argument in $\;[0,2\pi)\;$

Answer 2

For a more general overview https://math.stackexchange.com/a/3729281/399263

Note that when we have $z^u$ we are interested in $\ln(z)$ principal value because we have to choose a branch of the logarithm, the exponent $u$ does not pose an issue.

In fact setting $\begin{cases} z=a+ib=re^{i\theta}\\u=c+id\end{cases}$

If we consider $\arg(z^u)=\big(d\ln(r)+c(\theta+2k\pi)\big)\bigg|_{k\in\mathbb Z}$ it may not have values inside $[0,2\pi)$ (if $c$ is large) or have many (if $c$ is small), you could eventually choose a $k_0$ that minimize the absolute value of the argument but would that be really relevant ?

It makes more sense to take $k=0$ which correspond to the principal value of $\ln(z)$, which is the operation that leads to multivaluation.

In our case $z=i$ and $u=2-3i$

So $\ i^{2-3i}=\exp((2-3i)(i\frac{\pi}2+2ik\pi))=\exp(\frac{3\pi}2+i\pi+6k\pi+4ik\pi)=z_0\times w^k$

Since $\exp(i\pi)=-1$ and $\exp(4ik\pi)=1$ we have $$i^{2-3i} = z_0\times w^k\quad\text{where } \begin{cases}z_0=-e^{\frac{3\pi}2}\\ w=e^{6\pi}\end{cases}\ \text{ and }\ k\in\mathbb Z$$

And this is this $z_0$ we call the principal value.