Fundamental theorem of algebra(FTA)   Every polynomial of degree $$n\geqslant1$$ with complex coeficients has a zero in $$\Bbb C$$.

Proof. Let $$p(z)=z^n+a_{n-1}z^{n-1}+\dotsb+a_1z+a_0$$ be a polynomial of degree $$n\geqslant1$$ and assume that $$p(z)\ne0$$ for all $$z\in\Bbb C$$. Then the function defined by $$\frac1{p(z)}$$ is entire. By Cauchy’s theorem,

$\int_{|z|=r}\dfrac{\mathrm dz}{zp(z)}=\dfrac{2\,\mathrm\pi i}{p(0)}\ne 0.$

Since

$|p(z)| \geqslant |z|^n\left(1-\frac{|a_{n-1}|}{|z|}-\dotsb-\frac{|a_0|}{|z|^n}\right) > \frac12|z|^n$

for $$z$$ sufficiently large, so

$\left|\int_{|z|=r}\dfrac{\mathrm dz}{zp(z)}\right| \leqslant 2\,\mathrm\pi r \cdot \max_{|z|=r}\dfrac1{|zp(z)|} = \dfrac{2\,\mathrm\pi }{\min\limits_{|z|=r}|p(z)|}\to0(r\to\infty),$

which is a contradiction, and therefore $$p(z)$$ has a zero.   $$\Box$$

$\int_0^{2\,\mathrm\pi}\dfrac{\mathrm d\theta}{p(2\cos \theta)}\ne 0.$

$\frac1i\int_{|z|=1}\frac{\mathrm dz}{zp(z+\frac1z)}=\frac1i\int_{|z|=1}\frac{z^{n-1}\,\mathrm dz}{q(z)},$

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