2
$\begingroup$

We hadn't really covered much of Rouche's Theorem within class, so I'm kind of asking a trial question to see if I understand it.

I've been asked to find out how many zeros of the function $z^5 + \frac{1}{8}\cdot e^z + 1$, counting multiplicity, that lie inside the annulus $\{ z : \frac{1}{2} < |z| < 2\}$

So, I started off by considering the circle $|z| = \frac{1}{2}$, and let $f(z) = 1$, and $g(z) = z^5 + \frac{1}{8} \cdot e^z$. Then; $$|g(z)| = |z^5 + \frac{1}{8} \cdot e^z| = |z|^5 + \frac{1}{8} \cdot e^{|z|} = \frac{1}{2^5} + \frac{e^{\frac{1}{2}}}{8} < 1 = |f(z)|$$

Thus, there are no zeros of the function that exist within the circle $|z| \le \frac{1}{2}$.

Then, I considered the circle $|z| = 2$, and let $f(z) = z^5$, and $g(z) = \frac{1}{8} \cdot e^z + 1$. Then;

$$|g(z)| = |\frac{1}{8} \cdot e^z + 1| =\frac{1}{8} \cdot e^{|z|} + 1 =\frac{1}{8} \cdot e^{2} + 1 < 32 = 2^5 = |z|^5 = |f(z)|$$

Thus, we have five zeros within the circle $|z| \le 2$, and further, all of the zeros lie within the annulus.

Is this a proper application of Rouche's theorem, assuming that I placed "by Rouche's theorem...", etc?

$\endgroup$

1 Answer 1

1
$\begingroup$

Yes, this is a correct way to apply Rouché's theorem here.

However, the first of two applications could just as well be replaced by one-line argument with the reverse triangle inequality. Namely: when $|z|\le 1/2$, $$\left|z^5 + \frac{1}{8}\cdot e^z + 1\right|\ge |1|-\left|z^5\right| - \left|\frac{1}{8}\cdot e^z \right| \ge 1-\frac{1}{32} - \frac{1}{8}e^{1/2}>0$$ so there are no zeros there.

You do need the theorem to count the zeros inside of $|z|=2$, as you did.

$\endgroup$
1
  • $\begingroup$ Just a little late, but thanks for confirming my ideas!! $\endgroup$
    – Jack
    Commented Oct 18, 2014 at 11:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .