Inverse tangent of a complex number

[bonfire09]

Homework Statement

I have to find $\tan^{-1}(2i)$.

Homework Equations

The Attempt at a Solution

So far I have $\tan^{-1}(2i)=z\iff tan z= 2i\iff \dfrac{sin z}{cos z}=2i$. From here I get that
$-3=e^{-2zi}$. I do no know how to take it further to get $z=i\dfrac{\ln 3}{2}+(\dfrac{\pi}{2}+\pi n)$. Should I use natural logarithms but the problem is that I have a $-3$ which won't allow me to take the natural log of both sides. Any help would be great thanks.

 

[haruspex]

I get that
$-3=e^{-2zi}$.

I get -(3+4i)/5 for that.

I have a $-3$ which won't allow me to take the natural log of both sides.

The log of a negative number is fine when complex answers are allowed. It's only 0 that has no log in the complex plane.

 

[Curious3141]

Homework Statement

I have to find $\tan^{-1}(2i)$.

Homework Equations

The Attempt at a Solution

So far I have $\tan^{-1}(2i)=z\iff tan z= 2i\iff \dfrac{sin z}{cos z}=2i$. From here I get that
$-3=e^{-2zi}$.

The solution is as simple as using $\tan{ix} = itanh{x}$.

But you can use your method and do the algebra, it's just a little more work. Comes to the same answer.