Answer to Question #117190 in Complex Analysis for Popcy

Question

Given that 2−3i is a root of the equation z^3+pz^2+qz+13 = 0, find the other two roots and the values of the real constants p and q.

Answer 1

$(a+b)^3=a^3+3a^2b+3ab^2+b^3=>(2-3i)^3=8-36i-54+27i=-46-9i\newline (2-3i)^2=4-12i-9=-5-12i$

So we have:

$-46-9i+p(-5-12i)+q(2-3i)+13=0\newline (2q-5p)+(-12p-3q)i=33-9i=>\newline \begin{cases} 2q-5p=33\\ 12p+3q=-9 \end{cases}\newline \begin{cases} 2q-5p=33\\ q=-3-4p \end{cases}\newline \begin{cases} -6-8p-5p=33\\ q=-3-4p \end{cases}\newline \begin{cases} p=-3\\ q=9\\ \end{cases}\newline$

It means that:

$z^3-3z^2+9z+13=0$

$\dfrac{z^3-3z^2+9z+13}{z-2+3i}=z^2+(-1-3i)z+(-2-3i)$

$z^2+(-1-3i)z+(-2-3i)=0\newline D=(-1-3i)^2-4(-2-3i)=1+6i-9+8+12i=18i\newline \sqrt{D}=3\sqrt{2i}\newline if \sqrt{i}=\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}i, thenD=3+3i\newline if \sqrt{i}=-\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}i, thenD=-3-3i\newline D=3+3i\newline z_2=\dfrac{1+3i+3+3i}{-2-3i}=-2\newline z_3=\dfrac{1+3i-3-3i}{-2-3i}=\dfrac{2}{2+3i}\newline D=-3-3i\newline z_2=\dfrac{1+3i-3-3i}{-2-3i}=\dfrac{2}{2+3i}\newline z_3=\dfrac{1+3i+3+3i}{-2-3i}=-2\newline$