# Thread: Equations with i

1. ## Equations with i

Hello i have some problems with solving two things connected with complex numbers:

1)
$\displaystyle |i^{\hat{z}}|=e$
I tried with logarythming both sides, so i get:
$\displaystyle \hat{z}=\frac{1}{ln i}=\frac{1}{i ( \frac{ \pi}{2}+ 2k \pi)}$
and here i stuck...

2)
$\displaystyle arg( i^{ \hat{z}})= \frac{ \pi}{2}$
Here i even do not have idea how to start

2. Originally Posted by Vermax
Hello i have some problems with solving two things connected with complex numbers:

1)
$\displaystyle |i^{\hat{z}}|=e$
I tried with logarythming both sides, so i get:
$\displaystyle \hat{z}=\frac{e}{ln i}=\frac{e}{i ( \frac{ \pi}{2}+ 2k \pi)}$
and here i stuck...

2)
$\displaystyle arg( i^{ \hat{z}})= \frac{ \pi}{2}$
Here i even do not have idea how to start
Please post the question exactly as it's worded in wherever you got it from.

3. Originally Posted by mr fantastic
Please post the question exactly as it's worded in wherever you got it from.
Yes sir. So the question is: "Solve the equation". In other words, "find z".
And sorry for posting this thread in Algebra.

4. There is something a bit off about this question.
I have settled on a solution if we restrict $\displaystyle i^{\overline{z}}$ its principle value.
Consider these: $\displaystyle \left| {\exp (z)} \right| = e^{\text{Re}(z)}$ and $\displaystyle i^{\overline{z}} = \exp (\overline{z}\log (i)) = \exp (\overline{z}(i(\pi /2)))$.

If $\displaystyle \left| {i^{\overline{z}}} \right| = e\; \Rightarrow \;\text{Im}(z) = \frac{2}{\pi}$ and $\displaystyle \text{Re}(z)$ is any real number

If $\displaystyle \arg (i^{\overline{z}} ) = \frac{\pi}{2}$ then $\displaystyle \text{Re}(z)=(4k+1)~\&~ \text{Im}(z) = \frac{2}{\pi}$

5. Plato used "\overline{z}" rather than "\hat{z}" to get $\displaystyle \overline{z}$, the standard notation for "complex conjugate".

6. Thank you Plato, however what is very weird for me that the answer for 2) is:
$\displaystyle z=\frac{1-4l}{1+4k}+iy$
,where y is a real number, and l,k are integers.
Does it have any sense?

According to the first equation, the solution is just simple:
$\displaystyle \overline{z}=\frac{1}{i ( \frac{ \pi}{2}+ 2k \pi)}$
so:
$\displaystyle z=x-\frac{1}{i ( \frac{ \pi}{2}+ 2k \pi)}$