Find roots of $e^z=-3$ given that z=x+iy Can you give me some idea of how to do this?  I'm really stuck.  
 A: $$z=\ln(-3)=\ln 3+\ln(-1)= \ln 3+\pi i $$
In addition:
$$ \ln(z) = \ln(|z|) + i Arg(z) $$
A: Do you know this exponential form (polar form) of a complex number?
$$w = |w|e^{i\arg(w)}. $$
We have that $|-3| = 3$ and its argument (angle) is $\arg(-3) = \pi$. Adding or subtracting $2\pi$ of the angle does not change anything, so we can write
$$-3 = 3e^{i(\pi+2\pi p)} $$
for any $p\in\mathbb{Z}$. To finish up we have $3=e^{\ln 3}$, hence
$$-3 = e^{\ln 3}e^{i(\pi + 2\pi p)} = e^{i(\pi + 2\pi p) +\ln 3}, $$
so the roots are given by $\fbox{$z = \ln 3 + i(\pi+2\pi p)$}$, $p\in\mathbb{Z}$.
A: There are an infinite number of solutions in $\mathbb{C}$. Since
$$
e^x=-3=-1\cdot e^{\log(3)}
$$
we need to find a $z$ so that $e^z=-1$. Since $e^{iy}=\cos(y)+i\sin(y)$, we want $y$ to be an odd multiple of $\pi$. Thus, for any $n\in\mathbb{Z}$,
$$
z=\log(3)+(2n+1)\pi i
$$
A: Well, $e^z=e^xe^{iy}=e^x(\cos y +i\sin y)$.  So, $|e^z|=e^x=3$. What does that tell you about $x$?  For what value of $y$ is $\cos y+i\sin y=-1$?
A: Hint:
$$z=x+iy\implies e^z=e^x(\cos y+i\sin y)$$
If $e^z=-3$ then $\sin y=0$.
