2
$\begingroup$

An exercise from my Complex Analysis course (mostly focused on integration):

Show that all the roots of $f(z)=z^7-5z^3+12$ lie inside the annulus $1<|z|<2$.

Knowing that, being a polynomial function, $f(z)$ has 7 roots and no poles in $\mathbb{C}$, I thought I could try using the argument principle and show that $$\frac1{2\pi i}\oint\limits_{|z|=2} \frac{f'(z)}{f(z)}dz = 7$$ and $$\frac1{2\pi i}\oint\limits_{|z|=1} \frac{f'(z)}{f(z)}dz = 0$$ but it doesn't really seem to be a clever approach to me, since the integration would likely require finding the roots of $f(z)$ first... Any ideas?

$\endgroup$
  • 1
    $\begingroup$ Don't have time to explore the following possibility, but: en.wikipedia.org/wiki/Rouch%C3%A9's_theorem $\endgroup$ – avs Dec 15 '16 at 20:39
  • $\begingroup$ Yep. Rouché's theorem makes a short meal of it. $\endgroup$ – Daniel Fischer Dec 15 '16 at 20:44
  • $\begingroup$ Saw Rouché's Theorem already, but I still fail to see how to use it for the current situation... plus it's not part of the course $\endgroup$ – Ottavio Bartenor Dec 15 '16 at 20:46
  • 1
    $\begingroup$ On $\lvert z\rvert = 2$, you have $\lvert f(z) - z^7\rvert < \lvert z^7\rvert$. On $\lvert z\rvert = 1$, you have $\lvert f(z) - 12\rvert < \lvert 12\rvert$. $\endgroup$ – Daniel Fischer Dec 15 '16 at 20:49
  • $\begingroup$ Of course you can also show it without using anything fancy. For $\lvert z\rvert \geqslant 2$, we have $\lvert z\rvert^7 \geqslant 2\lvert z\rvert^6$, and so $\lvert f(z)\rvert \geqslant \lvert z\rvert^7 - 5\lvert z\rvert^3 - 12 \geqslant \bigl(\lvert z\rvert^6 - 5\lvert z\rvert^3\bigr) + \bigl(\lvert z\rvert^6 - 12\bigr) > 0$, and for $\lvert z\rvert \leqslant 1$, … $\endgroup$ – Daniel Fischer Dec 15 '16 at 20:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.