I'm supposed to compute the following integral using the residue theorem:
$$\int_{|z+1| = 4} \frac{z}{e^{z}+3}dz$$
To be honest, I don't understand how I should go about it since there are no values of $z$ for which $e^{z}+3=0$. How can I use the residue theorem to solve this integral or how can I solve it without using the theorem?
There are solutions to $e^z=-3$, in the complex plane, as is already mentioned above. The complex exponential $e^z$ takes on all the values in the complex plane, except for $0$. Hence, to solve $e^z=-3$, you can say $e^x(e^{iy})=-3$, where I have written $z=x+iy$, with $x$ and $y$ being real numbers.
Hence (using the fact $e^{iy}=\cos y+i\sin y$), we note $e^x\cos y$ is $-3$ and (comparing imaginary parts) we see $e^x \sin y=0$, so $y=\pi$ (or another odd multiple of $\pi$) and thus $-e^x=-3$ and (so $e^x=3$). As $x$ is real, note now it should be fairly easy to solve for $z=x+iy$. Note - for the contour integral, we only care about the roots of $e^z=-3$ (if any) that lie within the circle $|z+1|=4$. Then if $I=\frac{1}{2\pi i}\int g(z) \, dz$ where $g=\frac{z}{e^z+3}$, we can find $I$ by residue theorem.