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I have the following function:$$z^4+3iz^2+z-2+i$$
I need to find the number of zeros in the upper half plane. I wonder how one should go about solving such a problem. Help would be greatly appreciated!
P.S. This question has been asked before, but the answer in that question is sloppy, the links provided do not work.
Thanks!
As suggested by another answer, we can deduce the desired number of roots by consideration of argument. I'll like to present a slightly more geometric view of this in the language of winding number.
Let $f(z) = z^4+3iz^2+z-2+i$ be the polynomial at hand.
For any $R > 0$, let $D_R$ be the semicircle of radius $R$ anchored at $z = 0$. More precisely,
$$D_R = \bigg\{ z \in \mathbb{C} : |z| < R \land \Re z > 0 \bigg\}$$ Let $C_R = \partial D_R$ be the boundary of $D_R$ (oriented in the counterclockwise direction). If $R$ is chosen so that $f(z) \ne 0$ on $C_R$, then the number of roots of $f(z)$ inside $D_R$ is given by a contour integral
$$N_R = \frac{1}{2\pi i} \int_{C_R} \frac{f'(z)}{f(z)} dz\tag{*1}$$
If $R$ is large enough so that $D_R$ contains all roots of $f(z)$ in the upper-half plane, then $N_R$ is the number of root we seek.
We don't really need to evaluate $(*1)$ directly.
The function $\frac{f'(z)}{f(z)}$ has a local antiderivative $\log f(z)$. If we start from some point on $C_R$, say $-R$, walk along $C_R$ counterclockwisely and analytic continue $\log f(z)$ along the way. By the time $z$ reaches the starting point $-R$ again, the final value of $\log f(z)$ will differ from the initial value by an amount $2\pi i N_R$.
Taking exponential of the analytic continuation of $\log f(z)$ along $C_R$. The number $N_R$ becomes the number of times $f(z)$ wraps around the origin as $z$ walk along the $C_R$. This is the winding number of the "curve" $f(C_R)$ with respect to the origin.
Back to the original problem and assume we have picked a $R$ large enough. $C_R$ consists of two pieces, a line segment and a semicircle
$$C_R = [ -R, R ] \cup \big\{ R e^{i\theta} : \theta \in [0,\pi ] \big\}$$
For $z \in [ -R, R ] \subset \mathbb{R}$, we have $$\begin{cases} \Re f(z) &= z^4 + z - 2,\\ \Im f(z) &= 3z^2+1 \end{cases} \quad\implies\quad \Im f(z) > 0 $$ This means $f([-R,R])$ lies completely inside upper half-plane. Notice when $z \to \pm \infty$, $\Re f(z) \to \infty$, $\Im f(z) \to \infty$ while $\frac{\Im f(z)}{\Re f(z)} \to 0$. The endpoints of $f([-R,R])$ lies in the $1^{st}$ quadrant near the $+ve$ $x$-axis. The contribution from $[-R,R]$ to the winding number is smaller than $\frac14$ and vanishes as $R \to \infty$.
On the semicircle $z = Re^{i\theta}$. When $R$ is large, $f(z)$ is dominated by the $z^4$ term. As $\theta$ varies from $0$ to $\pi$, $z^4 = R^4 e^{i4\theta}$ wraps around the origin twice. So the contribution from the semicircle to the winding number is $2$ plus another small number.
Combine these two pieces, we find the desired winding number $$N_R = \verb/small number/ + ( 2 + \verb/another small number/ )$$
Since $N_R$ is always an integer, $N_R = 2$ for large $R$. As a result, the function $f(z)$ has two roots on the upper half-plane.
At the end is a picture showing what happens to $f(C_R)$ when $R = 2$. The red section is $f([-R,0])$, the orange section is $f([0,R])$ and the green and blue sections are those for $f(R e^{i\theta})$ where $\theta$ belongs to $[0,\frac{\pi}{2}]$ and $[\frac{\pi}{2},\pi]$ respectively. As one can see, $R = 2$ is large enough and the image $f(C_2)$ wraps around the origin twice. This means the two roots in the upper half-plane satisfy $|z| < 2$.
$\hspace0.75in$ 
The usual method to attack such problems is Rouché's theorem. For that, we need to reduce the problem to a bounded region. But it's not hard to find a bound for the modulus of the zeros of a polynomial. Here we have
$$\lvert z^4 + 3iz^2 + z - 2 + i\rvert \geqslant \lvert z\rvert^4\biggl( 1 - \frac{3}{\lvert z\rvert^2} - \frac{1}{\lvert z\rvert^3} - \frac{3}{\lvert z\rvert^4}\biggr) \geqslant \lvert z\rvert^4 \biggl(1 - \frac{3}{9} - \frac{1}{27} - \frac{3}{81}\biggr) = \frac{19}{27}\lvert z\rvert^4 > 0$$
for $\lvert z\rvert \geqslant 3$, so we need to find the number of zeros of $f(z) = z^4 + 3iz^2 + z - 2 + i$ in the half-disk $D = \{z\in \mathbb{C} : \lvert z\rvert < 3,\, \operatorname{Im} z > 0\}$.
It remains to find a suitable function $g$ such that $\lvert f-g\rvert < \lvert f\rvert + \lvert g\rvert$ on $\partial D$ and the number of zeros of $g$ in $D$ is easy to determine.
On the circle $\lvert z\rvert = 3$, the dominating component of $f$ is the monomial $z^4$, so that needs to be included in $g$. One might be tempted to try $g(z) = z^4$, but that alone doesn't work, since it has a zero (of multiplicity four) on $\partial D$. We need to include some other term to avoid that.
Since $z^4$ is real and nonnegative on the real line, we can make sure that $g$ has no zeros on $\partial D$ by adding a (small enough) positive constant to $z^4$. The constant mustn't be too large in order to retain the inequality $\lvert f-g\rvert < \lvert f\rvert + \lvert g\rvert$ on the semicircular part of $\partial D$. In our case, we have a lot of wiggle room on the semicircle, so we can use $g(z) = z^4 + 1$. Verifying the premises of Rouché's theorem for that choice and determining the number of zeros of $g$ in $D$ isn't hard.
HINT: You can do an argument variation. Integrate the function along a contour which surrounds the upper half plane. A popular choice is the half circle with straight line along real axis from -R to R and with radius R and then letting $R \to \infty$.
