Really not sure if this is the right place to post, but how can i write $\displaystyle (-e)^{i\pi}$ on $\displaystyle a+ib$ form?
This is not an easy problem. Recall that complex exponentiation does follow the usual rules that we use in the real numbers.
If each of $\displaystyle z~\&~w$ is a complex number then $\displaystyle w^z=\exp\(z\text{Log}(w))$
Here $\displaystyle \text{Log}(-e)=\ln|-e|+i\pi=1+i\pi$. That is the principal value.
So you want $\displaystyle [\exp[(i\pi)(1+i\pi)]=~?$
$\displaystyle (-e)^{i\pi} = (-1\cdot e)^{i\pi} = (-1)^{i\pi} \cdot e^{i\pi} = \left((e^{i\pi})^{i\pi}\right) \cdot e^{i\pi} = $ $\displaystyle e^{i^2\pi^2} \cdot e^{i\pi} = e^{-\pi^2} \cdot e^{i\pi} = -e^{-\pi^2} $
Note that I used this identity: $\displaystyle e^{i\pi}=-1$.
Does this help?
To to hopefully more directly answer the OP's question, recall that the real numbers are a subset of the complex numbers. So, the answer I detailed above is written in the complex notation $\displaystyle a=-e^{-\pi^2}$ and $\displaystyle b=0$.
This is the so called principal value.
More accurately you have $\displaystyle -1 = e^{i(\pi + 2k\pi)}$ where k is an integer (slight modification of your identity ;)).
If you substitute this, you'll find:$\displaystyle (-e)^{i\pi} = - e^{-\pi^2(1+2k)}$which is multi valued (all real) with the principal value you already found.