Toronto Math Forum
MAT3342018F => MAT334Tests => Final Exam => Topic started by: Victor Ivrii on December 18, 2018, 06:18:35 AM

Consider $P(z)= z^3 +2z 3i$ and, using the argument theorem and Rouché's theorem calculate the number of its roots in each of the following domains:
(a) $\{z\colon z1<1\}$;
(b) $\{z\colon z1>1, z<2\}$,
(c) $\{z\colon z>2\}$.

P(z)=z^3 3 + 2z i = (z^3 – 3)+(2z i)
a, z1<1
0 < z < 2
Using Rouche Theorem, let z = 2, z^33= 5 > 2z  i, z^3 3 is the dominant,
z^33 =0 and z has 3 roots.
So it means when z<2, P(z) has 3 roots.
let z = 0, z^33 = 3 < 2z  i, 2z  i is the dominant,
2z  i =0 and z has 1 root.
So it means when z<0, f(z) has 1 root.
So when 0 < z < 2, f(z) has 31 =2 roots.
b, z1>1, z<2
z < 0, z > 2, 2 < z < 2 so 2 < z < 0
when 2 < z < 0, it has 1 root.
c, z > 2
when z < 2, it has 3 roots already, so when z > 2, it has 0 root.

Here's the answer

(a) P(z)=z^{3}+2z−3−i=(z1)^{3}+3(z1)^{2}+5(z1)i
at z−1=1, 5(z−1)≥(z1)^{3}+3(z1)^{2}i
so at z−1<1, 5(z−1) and P(z) have the same number of roots which is 1 (z=1).
(b) At z=2, z^{3}≥2z3i
so at z<2, z^{3} and P(z) have the same number of roots which is 3 (z=0 has the multiplicity of 3).
but we know z−1<1 is in z<2, z≠1, so at z−1>1,z<2, P(z) has the number of roots of 31=2.
(c) Because P(z) has the highest power of 3, so it has in all 3 roots, but we already find 3 roots at z<2,
so at z>2, the number of roots is 0.

(a) As $z=2$ consider $Q(z)=z^3$; then $Q(z)=8$,
$$
P(z)Q(z)=2iz 3i\le 4+3+i=4+\sqrt{10}<8;$$
therefore $P$ has as many roots in $\{z\colon z<2\}$ as $Q(z)$ has, which is $3$ (we count orders).
(c) Then there are no roots in $\{z\colon z>2\}$.
(b) Consider $z=w+1$ with $w=1$; then $P(w+1)=w^3+3w^2 +5w i$ and we set $Q(w)=5w$; then $Q(w)=5$,
$$
P(w)Q(w)=w^3+3w^2 i < 5.
$$
Indeed, $P(w)Q(w)\le w+3+1< 5$ except $w=1$, and for $w=1$ we have $w^3+3w^2 i=4i=\sqrt{17}<5$. Therefore $P(w+1)$ has as many roots in $\{w\colon w<1\}$ as $Q(w)$ has, which is $1$ (we count orders).
Finally, in the domain $\{z\colon z1>1, z<2\}$, there are $31=2$ roots.