All possible values of $i^{\frac{1}{2} + i} = \exp((\frac{1}{2} + i) \log(i))$ in rectangular form

Question

Trying to find all possible values of $i^{\frac{1}{2} + i} = \exp((\frac{1}{2} + i) \log(i))$, so far I have the following:

$i^{\frac{1}{2} + i} = e^{\frac{1}{2} + i} e^{log(i)} \iff$

$i^{i} i^{-\frac{1}{2}} = e^{i} e^{\frac{1}{2}} \iff$

$\log(i^i i^{-\frac{1}{2}}) = i + 1/2 + 2 \pi k i, k \in \mathbb{Z} \iff$

$i \log(i) - \frac{1}{2} \log(i) - i = \frac{1}{2} + 2 \pi i k \iff$

$(\log(i) - \frac{1}{2} \log(i^{\frac{1}{i}}) - 1)i - \frac{1}{2} + 2 \pi i k = 0 $

I am not sure about the solution, particularly the last step where I do the following: $\frac{1}{2} \log(i) = i \frac{1}{i} \frac{1}{2} \log(i)= i \frac{1}{2} log(i^{\frac{1}{i}})$. Is this allowed?

The reason I do this is because I want a solution in rectangular form z = x + yi, so now I have $x = \frac{1}{2}$ and $y = \log(i) - \frac{1}{2} \log(i^{\frac{1}{i}}) - 1$.

Answer 1

Hint:

Rewrite $$i^{(\frac {1}{2} + i)}$$ as $$i^\frac {1}{2}(i^i)$$

For $i^\frac {1}{2}$, use DeMoivre's Theorem (with $(0+i)^{\frac {1}{2}}$). For $i^i$, rewrite as $e^{i \log i}$ and find the value of $\log i$.

(Note: "$\log$" in this case is the natural logarithm - I usually write "$\ln$".)

Answer 2

Evaluate as follows,

$$i^{\frac12 + i} = (e^{i\frac\pi2})^{\frac12 + i} =e^{i\frac\pi2(\frac12 + i)} =e^{i\frac\pi4-\frac\pi2}=e^{-\frac\pi2}(\cos\frac\pi4+i\sin\frac\pi4) =\frac1{\sqrt2 e^{\frac\pi2}}(1+i)$$

Answer 3

Well, this is a miserable problem. But I think the best way to look at your multivalued expression is as the product $i^{1/2}$ times the countably-valued but always-real quantity $i^i$.

The only two values for $i^{1/2}$ are $\pm\frac{1+i}{\sqrt2}$.

For the values of $i^i$ we write this as $\exp(i\log i)$, where what’s in the parentheses has the countably many values $i\bigl[i\frac\pi2+2ki\pi\bigr]$, for the various $k\in\Bbb Z$. Thus the infinitely many values of the second factor are $\exp(2k\pi-\frac\pi2)$.

Upshot? Your values are $\pm\frac{1+i}{\sqrt2}\exp(2k\pi-\pi/2)$, all $k\in\Bbb Z$, so all lying on the line $y=x$ in the Wessel-Argand-Gauss plane.