# An application of Rouché for $z^4 - z + 5$

This is an assignment question I submitted but didn't recieve full marks for, so I'm trying to correct it. My reference is Stein and Shakarchi's Complex Analysis.

The question is in three parts, though the first and third parts are fine. I've included the third part for proof-verification though. In the first part (a), it is proved that the function $f(z) = z^4 -z + 5$ has four roots inside the annulus $4/3 < |z| < 5/3$. I had no problem with that and I'll omit the details, but I'll mention that it can be shown with Rouché's theorem applied with $G(z) = z^4 + 4$, and $F(z) = 1-z$.

The second part (b) is where I've had trouble. We wish to prove to prove that there is only one zero in the first quadrant. To go about this, I let $F(z) = z^4 + 6$, and $G(z) = -z-1$. Then I imagine the best approach would be to apply Rouché's Theorem on some square in the first quadrant that has sides along the real and imaginary axis. Unfortunately, when I tried this approach, I couldn't prove the needed inequalities along this contour.

Lastly, in (c), we show that there is exactly one root in each quadrant. Since the coefficients of $f(z)$ are real, it follows that roots of the polynomial come in conjugate pairs. Directly it follows that in the (-,+) quadrant that there is a root, and it must be unique to avoid a contradiction with (b). Now the remaining two roots must fall in the second and third quadrants. Considering the root $-a+ib$, we must have $-a-bi$, so given a root in one of the quadrants, we must have one in the other. We have now accounted for all of the roots of the polynomial, so we're done.

Consider $f(z) = z^4 - z + 5$ and $g(z) = z^4 + 5$. I claim $|f(z)) - g(z)| = |z| < |g(z)|$ on the positive real axis, on the positive imaginary axis, and on the circle $|z|=R$ for large $R$. Note that for either $z = x$ or $z = ix$ with $0 \le x \le 1$ we have $|f(z) - g(z)| \le 1 < 5 \le |g(z)|$, while if $x \ge 1$ then $|f(z) - g(z)| \le x^4 < |g(z)|$.