How many roots does
$P(z)=2z^4+z^3-5z^2+z+2$ have in $\{ \operatorname{Re}(z)<0 \}$?
I was told that I should compute $P(it)$ for $t \in \Bbb R$ which is:
$P(it)=2t^4 +5t^2 +2 +it(1-t^2) $.
From that I somehow should get that the curve goes around $0$ twice, so from some theorem I should get that there are two roots.
But I cannot imagine what curve goes around $0$ and why it does so. :(
To follow the hint.
The hint is suggesting to use the argument principle.
The idea is to consider the change of the argument of $P(it)$ when $t$ moves along the circuit $\gamma$ composed of the real axis from $-R$ to $R$, for $R$ huge followed by the arc $Re^{i\theta}$ with $\theta\in[0,\pi]$. The roots of $P(z)$ with $\text{Re}(z)<0$ are now in the upper half-plane and they come in a finite number. So, for $R$ large this circuit contains the roots we want to count. By the argument principle the number of roots inside the circuit is equal to the number of times $P(i\gamma)$ winds around the origin.
Notice that the real part of $P(it)$ is never zero when $t$ is real, it is always greater or equal to $2$. So the image of $[-R,R]$ by $P(it)$ remains in the semi-plane $\text{Re}(z)\geq2$.
Now we follow the arc. We get
$$p(iRe^{i\theta})=2R^4e^{4i\theta}+5R^2e^{2i\theta}+2+iRe^{i\theta}(1-R^2e^{2i\theta})$$
For $R$ very large all terms are negligible compared to $2R^4e^{4i\theta}$. So, this will tell us how many times it winds around the origin. If it helps to see it better we can zoom-in by dividing by $R^4$. Zooming in doesn't change the winding number and the expression looks like $$2e^{4i\theta}+5R^{-2}e^{2i\theta}+2R^{-4}+iR^{-3}e^{i\theta}(1-R^2e^{2i\theta})=2e^{4i\theta}+{}\text{(itty-bitty-tiny numbers)}$$ Since $\theta\in[0,\pi]$ then $4\theta$ winds twice around the origin.
Therefore, by the argument principle the polynomial $P(it)$ has $2$ roots inside the circuit and therefore in the upper half-plane. This implies that the original polynomial $P(z)$ has $2$ roots in the left-half plane $\text{Re}(z)<0$.