Assuming the principle of the argument, prove that any polynomial of degree $n$ has precisely $n$ zeros in $\mathbf{C}$, counted with multiplicity.

Consider a polynomial $p(z)=z^{4}+a z^{3}+b z^{2}+c z+d$, and let $R$ be a positive real number such that $|a| R^{3}+|b| R^{2}+|c| R+|d|<R^{4}$. Define a curve $\Gamma:[0,1] \rightarrow \mathbf{C}$ by

$$\Gamma(t)= \begin{cases}p\left(R e^{\pi i t}\right) & \text { for } 0 \leqslant t \leqslant \frac{1}{2} \\ (2-2 t) p(i R)+(2 t-1) p(R) & \text { for } \frac{1}{2} \leqslant t \leqslant 1\end{cases}$$

Show that the winding number $n(\Gamma, 0)=1$.

Suppose now that $p(z)$ has real coefficients, that $z^{4}-b z^{2}+d$ has no real zeros, and that the real zeros of $p(z)$ are all strictly negative. Show that precisely one of the zeros of $p(z)$ lies in the quadrant $\{x+i y: x>0, y>0\}$.

[Standard results about winding numbers may be quoted without proof; in particular, you may wish to use the fact that if $\gamma_{i}:[0,1] \rightarrow \mathbf{C}, i=1,2$, are two closed curves with $\left|\gamma_{2}(t)-\gamma_{1}(t)\right|<\left|\gamma_{1}(t)\right|$ for all $t$, then $n\left(\gamma_{1}, 0\right)=n\left(\gamma_{2}, 0\right)$.]