Location of Complex Roots Here is a problem I think dealing with Rouche's theorem:
How many roots does the equation 
$$
\frac{1}{2}e^z+z^4+1=0
$$
have in the left half plane $Re(z)<0$
I see that in order to have a root in this region we must have
$$
|z^4+1|<\frac{1}{2}
$$
but how does this give us a region in the left half plane?
Thanks!
 A: Hint:
The set $|z^4+1| = 1/2$ consists of four simple closed loops, two of which lie in the left half-plane.  Use Rouche's theorem on these two loops.
More:
The graph of $|z^4+1|$ looks like floating sheet which has been pinned to the ground at $z=e^{i\pi/4}$, $e^{i3\pi/4}$, $e^{i5\pi/4}$, and $e^{i7\pi/4}$:

The sets $|z^2+1| = \text{const}.$ are the level curves of this graph---you can think of them as the intersection of a flat surface with the graph.  If we choose $\epsilon > 0$ very small then the surface at height $\epsilon$ only intersects the graph very near to these four points:

and so the set $|z^4+1| = \epsilon$ consists of four simple closed loops around the points $e^{ik\pi/4}$ listed above.  By the minimum modulus principle, any component of $|z^4+1| \leq \epsilon$ must contain a zero of $z^4+1$, so you will not run into any rogue components appearing elsewhere as you increase $\epsilon$.

Exercise.
For $0 \leq a < 1$, show that the set $|z^4+1| = a$ consists of four components, each properly contained in the four quadrants of the complex plane.

