Prove that a polynomial in $z$ without zeros is constant using Cauchy theorem.
Hint : If $P(z)$ is a polynomial that is not a constant, write $P(z)=P(0)+zQ(z)$, divide by $zP(z)$, and integrate around a large circle. This will lead to a contradiction if $P(z)$ has no zeros.
Consider $$\frac{P(z)}{zP(z)}=\frac{P(0)}{zP(z)}+\frac{zQ(z)}{zP(z)}$$
Integrating over circle of radius $R$
$$\int_R\frac{1}{z}dz=\int_R\frac{P(0)}{zP(z)}dz+\int_R\frac{Q(z)}{P(z)}dz$$
Assuming $P(z)$ has no zeros, we see that $\int_R\frac{Q(z)}{P(z)}dz=0$ using Cauchy theorem as $\frac{Q(z)}{P(z)}$ is analytic.
See that $\int_R\frac{1}{z}dz=2\pi i$ by considering parametrization $z(\theta)=Re^{i\theta}$.
I do not know how to compute $\int_R\frac{P(0)}{zP(z)}dz$ from basics but i can use Residue theorem here and assuming $P(z)$ has no zeros we see that $$\int_R\frac{P(0)}{zP(z)}dz=2\pi i\lim_{z\rightarrow 0}\frac{P(0)}{P(z)}=2\pi i$$
So, $$2\pi i=\int_R\frac{1}{z}dz=\int_R\frac{P(0)}{zP(z)}dz+\int_R\frac{Q(z)}{P(z)}dz=2\pi i$$
I do not see any contradiction.. What is wrong with my argument?