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Let $f(z) = 2z^{4}+5z^{2}$ and $g(z)=z^{4}+10z^{2}+1$. Prove that $f$ and $g$ have the same number of zeros inside the open unit disc as well as the same number of zeros outside the unit disc but inside the disc of radius $4$ centered at $0$.

Now, for the zeroes inside the unit disc we can apply Rouche's theorem we have $|g(z)-2f(z)|\le4<|2f(z)|\le14 $, when $|z|=1$ so $g$ and $f$ has the same number of zeroes inside the unit disc.

what about the other part of the question?

Thanks.

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If $|z|=1$, then $|2z^4|<|5z^2|$ and $|z^4+1|<|10z^2|$. So, both $f$ and $g$ have $2$ zeros when $|z|<1$.

Now, if $|z|=4$, then $|5z^2|<|2z^4|$ and $|10z^2+1|<|z^4|$. So, both $f$ and $g$ have $4$ zeros when $|z|<4$.

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  • $\begingroup$ Does the second part hold for radius less than $4$? $\endgroup$
    – Math1
    Apr 26, 2020 at 14:24
  • $\begingroup$ It holds when the radius is greater than $\sqrt{5+\sqrt{26}}\approx3.18$. $\endgroup$ Apr 26, 2020 at 14:28
  • $\begingroup$ Thanks for help. $\endgroup$
    – Math1
    Apr 26, 2020 at 14:41

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