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Friday, May 25, 2018

Using the Argument Principle for Counting Roots

Let $f(z)=z^{11}-2z^6+z^4+10i$. We want to find the number of roots of $f$ which are in the upper semiplane. By the Argument Principle, if $\gamma$ is a curve which has no roots of $f$ and $n_f$ is the number of roots of $f$ in the upper plane counting multiplicities, then
$$n_f=\frac{1}{2\pi i} \int_\gamma \frac{f''(z)}{f(z)}dz=\int_{f(\gamma)}\frac{dw}{w}=n(f\circ \gamma,0)$$
where $n(-,0)$ is the winding number of $-$ around $0$.

So, now take $R$ so big, that $f$ doesn't vanish when $|z|\geq R$, and such that $f(z)\sim z^{11}$.
Take $\gamma$ as the upper half circle of radius $R$ followed by the segment between $-R$ and $R$:


Also note that there are no real roots of $f$, since for $z$ real, $f(z)=0$ means $10i=0$. So, now we can use the Argument Principle, since there are no roots of $f$ on $\gamma$. Furthermore, $R$ is big enough to say that every root of $f$ is inside $\gamma$. Next we must see how $f\circ \gamma$ behaves.

When $\gamma$ does a half-twist around $0$, $f\circ \gamma$ does five twists and almost an extra half, ending at some point $x+10i$ where $x\in\mathbb{R}$. Then $f\circ\gamma$ goes somehow, around the line $y=10i$, from $x+10i$ to the starting point. Since that's done over the real axis, it doesn't make the extra twist around $0$. So $n(f\circ\gamma,0)=5$: That's the number of roots.


In the picture it can be seen the number of twists, and how the last one doesn't complete itself.
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