How to prove that the zeros of $z^4+iz^3+1$ are in the disk $D(0,\frac{3}{2})$, and determine how many of them are in the first quadrant? Prove that the zeros of $z^4+iz^3+1$ are in the disk $D(0,\frac{3}{2})$ and determine how many of them are in the first quadrant.
 A: On the circle of radius $R$, $|iz^3+1|\leq R^3+1$ and $|z^4|=R^4$, 
therefore, if $R=3/2$ then $|z^4|>|iz^3+1|$. By Rouché's theorem, the number of zeroes of $z^4$ inside the disc of radius $3/2$ equals the number of zeroes of $z^4+iz^3+1$ in the same disc.
This means that $z^4+iz^3+1$ has all the four roots in the given disc.
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 $f(t)=t^4+1 + it^3$, therefore $f(t)$ stays always in the first quadrant for $t\geq0$ (so the total change of argument along this part of the path is between $0$ and $\pi/2$); moreover $f(0)$ has argument equal to $0$ and for $M$ large $\arctan(M^3/M^4+1)=\textrm{arg}f(M)$ is near $0$ again, so the argument stays constant when $M\to\infty$.

*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 is real, hence its 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: Rouche's theorem on 
$$C_{3/2}:=\left\{z\in\Bbb C\;;\;|z|=\frac{3}{2}\right\}\;\;\;,\;\;f(z)=z^4\;\;,\;\;g(z)=iz^3+1$$
so
$$z\in C_{3/2}\Longrightarrow |g(z)|\le |z|^3+1=\frac{27}{8}+1<\frac{81}{16}=|f(z)| $$
So $\,f\,\,,\,\,f+g\,\,$ have the same number of zeros within $\,C_{3/2}\,$
