Here's the question:
$1+3i$ is a root of the cubic $z^3+6z+20=0$. Identify the other two roots.
Obviously the conjugate complex number is another root so one is $1-3i$.
So I use the general formula $z^2-(\alpha+\beta)+\alpha\beta=0$, where $\alpha$ and $\beta$ are the complex conjugate pair, to work out that the cubic in question comes from, which I find to be:
$(z^2-z+10)(z+a)$, where $a$ is some number.
For the next part I set the cubic as $z^3+0z^2+6z+20=0$.
$\therefore$ $az^2-z^2=0z^2$
so $a=1$.
However, my problem arises when I try to find $a$ using the other powers:
With the same method,
$-za+10z=6z$
$\therefore a=4$
And with: $10a=20$, $a=2$.
Which value of $a$ is the correct one?
Surely I should get the same value for $a$ on each of these? Or am I making a silly error?
For varying questions of this nature, I've found that some work fine and some don't.
Best Answer
Using Complex conjugate root theorem, if $1+3i$ is a root. so will be $1-3i$
So, $z^3+6z+20=(z-a)\{z-(1+3i)\}\{z-(1-3i)\}=(z-a)(z^2-2z+10)$
as $\{z-(1+3i)\}\{z-(1-3i)\}=(z-1)^2-(-3i)^2=z^2-2z+1-(-9)=z^2-2z+10$
$z^3+6z+20=z^3-z^2(a+2)+z(2a+10)-10a$
Compare the coefficient any power (that involves $a$) of $z$ to find $a=-2$