Here is my question – it is an example sheet question, completely non-examinable:
[I have managed this first part, but am including it to help give a sense of where the question is going.]
$(i)$ Show that $ z^4 + 12z + 1 $ has exactly three zeros in the annulus $ \{ 1 < |z| < 4 \} $.
$(ii)$ Prove that $z^5 + 2 + e^z $ has exactly three zeros in the half-plane $ \{ z \ | \ Re(z) < 0 \} $.
$(iii)$ Show that the equation $ z^4 + z + 1 = 0 $ has one solution in each quadrant. Prove that all solutions lie in the circle $ \{ z \ | \ |z| < 3/2 \} $.
The first part is a fairly straightforward consequence of Rouche's Theorem. It is straightforward because of the easy of counting the number of roots of polynomials, and we can just set $ |z| < 4 $ then $|z|<1$. This straightforward method cannot be applied to $(ii)$ and $(iii)$ since we involve exponentials and also looking at quadrants, not circles.
Any assistance (by that I don't mean and answer, but some (reasonably strong) hints) would be most appreciated! Thanks very much! 🙂
Best Answer
For problem $(ii)$, note that
That makes it a straighforward application of Rouché's theorem too.
For problem $(iii)$, note that $f$ is real (that is, $f(\mathbb{R}) \subset \mathbb{R}$, hence $f(\overline{z}) = \overline{f(z)}$ for all $z$), and has no real zeros. Then count the zeros in the left resp. right half-plane (two in each) with Rouché's theorem (that $z^4$ is real and non-negative on the imaginary axis may be helpful there), and use the same to note that all four zeros lie in the disk $\lvert z\rvert < \frac{3}{2}$.