Originally Posted by

**lo2** Hi I have this equation

$\displaystyle (e^{\pi\cdot z})^{2}=i$

Where I have to find the roots, when modulus er less than 1.

Cannot really solve it...

I have tried to $\displaystyle t=e^{\pi\cdot z}$

And then tried to solve it is a quacdric equation.

Where I get that

$\displaystyle e^{\pi x}\cos(\pi y)=1$ and $\displaystyle e^{\pi x}\sin(\pi y)=-2$

And also for the other root

$\displaystyle e^{\pi x}\cos(\pi y)=-1$ and $\displaystyle e^{\pi x}\sin(\pi y)=2$

But I have that $\displaystyle e^{\pi x}< 1 \Leftrightarrow x<0$

So anyone who could help me out or give me some tips?