Finding all values in C of arcsech(i)

Pricipal argument:
Yes, your principal argument of $\pi / 2$ is fine as $(-1 + \sqrt{2} ) i$ is purely imaginary and is of form ai, where $a \in \mathbb{R}^{+}$.

As a check, try $\mathrm{sech}(z)$, where $z$ is the answer in your first post.

Though for part of your last equation, instead of $2ki$, you need $2 k \pi$.

Method for negative second case:
Yes, similar method for the second negative case.

For the real case, as $\mathrm{sech}(x)$ is an even function, the positive case is usually used of $\mathrm{arsech}(x)$ so to return positive values (depending on context of course).

Pricipal argument:
Yes, your principal argument of $\pi / 2$ is fine as $(-1 + \sqrt{2} ) i$ is purely imaginary and is of form ai, where $a \in \mathbb{R}^{+}$.

As a check, try $\mathrm{sech}(z)$, where $z$ is the answer in your first post.

Though for part of your last equation, instead of $2ki$, you need $2 k \pi$.

Method for negative second case:
Yes, similar method for the second negative case.

For the real case, as $\mathrm{sech}(x)$ is an even function, the positive case is usually used of $\mathrm{arsech}(x)$ so to return positive values (depending on context of course).

Hi thanks for your response I understand most of what you are saying I just don't see how there can be a real case

Also what would the Argand diagram look like for the positive case ?

Your solutions for the positive case are of form:

$\displaystyle \log ( -1 + \sqrt{2} ) + i( (\pi/2) + 2k \pi ),$ where $k \in \mathbb{Z}$.

This is just a complex value (of form a + bi) and using the pricipal argument, can you now see how we can create an argand diagram?

Yeah the a value is constant so it will just be a set of solutions on that vertical line repeating up and down the imaginary axis

Yes and of course the set of solutions for the negative case.

Brilliant, thank you for your help, i was just unsure about the diagrams, also a quick question if we had $i^{i\pi}$ and wanted to find all of it's values in $\Bbb C$ when we plot an argand diagram for these values just all lie on the negative real axis? using the fact that $z^{\alpha} =e^{\alpha \log(z)}$ using this i calculated $i^{i\pi} =e^{-\pi (\frac{\pi}{2}+2k\pi)},k \in \Bbb Z$.