Exercise of complex variable, polynomials. Calculate the number of zeros in the right half-plane of the following polynomial:
$$z^4+2z^3-2z+10$$
Please, it's the last exercise that I have to do. Help TT. PD: I don't know how do it.
 A: Proceed like the previous problem for first quadrant.
You will find one root.
And note that roots will be conjugates.
So 2 roots in the right half-plane..
For zero in the first quadrant, consider the argument principle: if $Z$ is the number of zeroes of $f$ inside the plane region delimited by the contour $\gamma$, then $\Delta_\gamma(\textrm{arg}f)=2\pi Z$, i.e. the variation of the argument of $f$ along $\gamma$ equals $Z$ times $2\pi$.
Take a path from the origin, following the real axis to the point $M>0$, then make a quarter of circle or radius $M$, reaching the point $iM$ and then go back to the origin along the imaginary axis. Now try to determine the variation of the argument of $f(z)$ along this path for $M\to\infty$:


*

*along the real axis, the function is $f(t)=t^4-2t+2t^3+10$, therefore $f(t)$ is real for $t\geq0$ so the total change of argument along this part of the path is  $0$.

*along the path $Me^{i\theta}$ for $0\leq\theta\leq \pi/2$, if $M$ is very large, the function is near to $g(\theta)=M^4e^{i4\theta}$; therefore the argument goes from $0$ to $2\pi$.

*along the imaginary axis, the function's argument doesn't change.
So, the total change of the argument is $2\pi$, implying that the function has only one zero in that quadrant.

A: Consider the polynomial
$$
p(z) = z^4 + 2z^3 - 2z + \epsilon.
$$
When $\epsilon = 0$ we can solve for the zeros explicitly using the cubic formula.  They are $z=0$ and
$$
z \in \left\{\begin{array}{c}
-\frac{2}{3}+\frac{1}{3} \sqrt[3]{19-3 \sqrt{33}}+\frac{1}{3} \sqrt[3]{19+3 \sqrt{33}}, \\
-\frac{2}{3}-\frac{1}{6} \left(1+i \sqrt{3}\right) \sqrt[3]{19-3 \sqrt{33}}-\frac{1}{6} \left(1-i \sqrt{3}\right) \sqrt[3]{19+3 \sqrt{33}}, \\
-\frac{2}{3}-\frac{1}{6} \left(1-i \sqrt{3}\right) \sqrt[3]{19-3 \sqrt{33}}-\frac{1}{6} \left(1+i \sqrt{3}\right) \sqrt[3]{19+3 \sqrt{33}}
\end{array}\right\}.
$$
Now,
$$
\sqrt[3]{19-3 \sqrt{33}} > \sqrt[3]{19-3 \sqrt{36}} = 1,
$$
so that the first zero in this set satisfies
$$
\begin{align*}
& -\frac{2}{3}+\frac{1}{3} \sqrt[3]{19-3 \sqrt{33}}+\frac{1}{3} \sqrt[3]{19+3 \sqrt{33}} \\
&\qquad > -\frac{2}{3}+\frac{1}{3} \sqrt[3]{19-3 \sqrt{33}}+\frac{1}{3} \sqrt[3]{19-3 \sqrt{33}} \\
&\qquad = -\frac{2}{3}+\frac{2}{3} \sqrt[3]{19-3 \sqrt{33}} \\
&\qquad > 0.
\end{align*}
$$
The real part of the next two zeros is clearly negative:
$$
-\frac{2}{3}-\frac{1}{6} \sqrt[3]{19-3 \sqrt{33}}-\frac{1}{6} \sqrt[3]{19+3 \sqrt{33}} < 0.
$$
So when $\epsilon = 0$ we have one zero at $z=0$, one zero with $\Re(z) > 0$, and two zeros with $\Re(z) < 0$.
Next,
$$
p(iy) = y^4 + \epsilon - i2(y+y^3),
$$
so that $p(z)$ has no zeros on the imaginary axis when $\epsilon > 0$.
We now consider the zero of $p(z)$ which is located at $z=0$ when $\epsilon = 0$ as an analytic function $z_0 = z_0(\epsilon)$ with $z_0(0) = 0$.  Expand $z_0$ as a Taylor series
$$
\begin{align*}
z_0(\epsilon) &= z_0(0) + z_0'(0)\epsilon + O(\epsilon^2) \\
&= z_0'(0)\epsilon + O(\epsilon^2)
\end{align*}
$$
(valid for small $\epsilon$) and substitute this into the equation $p(z_0) = 0$ to get
$$
[1-2z_0'(0)]\epsilon = O(\epsilon^2).
$$
Divide both sides by $\epsilon$ and let $\epsilon \to 0$ to find that
$$
1-2z_0'(0) = 0
$$
or $z_0'(0) = 1/2$.  Thus for small $\epsilon > 0$ there are two zeros satisfying $\Re(z) > 0$.  Since $p(z)$ has no zeros with $\Re(z) = 0$ when $\epsilon > 0$ and the zeros of polynomials are continuous functions of the coefficients we conclude that this is true for all $\epsilon > 0$.
In particular, when $\epsilon = 10$ there are exactly two zeros with $\Re(z) > 0$ and two zeros with $\Re(z) < 0$.
