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Why do $f(z)=4z+2$ and $p(z)=z^6+4z+2=g(z)+f(z)$ have the same number of zeros in the open unit disc centered at zero?

It is justified by $|g(z)|=|z|^6 =1<2=4|z|-2\le |4z+2|=|f(z)|$ for all $x$ in the unit circle, which is the boundary of the unit disc.

Why does this inequality prove that the two functions have the same number of zeros in the interior of the unit disc?

(It's Rouche theorem)

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    $\begingroup$ Rouche's theorem? $\endgroup$ – Angina Seng Dec 29 '18 at 18:05
  • $\begingroup$ The question in the title does not match that in the body of the question. $\endgroup$ – Angina Seng Dec 29 '18 at 18:06
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Consider the family of equations $tz^6+4z+2$ for $t\in[0,1]$. From the implicit function theorem we know that the roots (resp. the root set as a whole) follow continuous paths for $t>0$. The inequality you computed also applies generalized here and proves that for no $t$ there can be roots on or near the unit circle. Thus the root $-1/2$ for $t=0$ spawns the only such path inside the unit circle, for any $t\in[0,1]$ there is only a single root inside the unit circle. As $1/2^6$ is small, the perturbation to the root from $t=0$ to $t=1$ is small as well.

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