Answer to Question #261407 in Complex Analysis for Bre

Question

Consider the polynomial of a complex variable z given by f(z)=z³-8z²+25z-26

(a) Show that z-3+2i is factor of f(z)

(b) Develop the quadratic factor of f(z)

(c) Hence, completely solve the equation f(z)=0

Answer 1

Consider the polynomial of a complex variable $z$ given by $f(z)=z^3-8z^2+25z-26.$

(a) Let us show that $z-3+2i$ is factor of $f(z).$ Since

$f(3-2i)=(3-2i)^3-8(3-2i)^2+25(3-2i)-26\\ =27-54i+36i^2-8i^3-8(9-12i+4i^2)+75-50i-26\\ =27-54i-36+8i-72+96i+32+75-50i-26=0,$

we conclude that $3-2i$ is a root of $f,$ and hence according to Bezout Theorem, $z-3+2i$ is factor of $f(z).$

(b) Let us develop the quadratic factor of $f(z).$ Since the coefficients of $f$ are the real numbers, $3+2i$ is also a root of $f,$ and hence $z−3-2i$ is factor of $f(z).$ It follows that $(z−3+2i)(z−3-2i ) =(z-3)^2-4i^2=z^2-6z+9+4=z^2-6z+13$

is the quadratic factor of $f(z).$

(c) Hence, let us completely solve the equation $f(z)=0.$

It follows that

$f(z)=z^3-8z^2+25z-26=(z^2-6z+13)(z-2),$

and thus $z=2$ is the third root of the equation.

We conclude that the roots of the equations are:

$3-2i,\ 3+2i$ and 2.