# Thread: complex number equation

1. ## complex number equation

Hi,

I would like some help with this.

The equation $p(z)=z^4-2z^3+4z^2-2z+3$ is given.

Show that one of the roots are i.

So i verified that i is indeed a root. And since complex roots are "mirrored", another root must be -i.

The factor theorem.
$(z-i)(z+i) = z^2 +1$

Then i factored the equation using polynomial division and the result is
$(z^2+1)(z^2-2z+3)$

the second bracket is a simple quadratic.
$\frac{-2}{-2} \pm \sqrt{\bigg(\frac{-2}{2}\bigg)^2 -3}$

= $1 \pm \sqrt{1-3}$

This is where i get it wrong, isn't the square root expression supposed to evaluate to 2i ?

The correct answer is $1\pm i\sqrt{2}$ however, the result im getting is $1\pm 2i$

Any ideas?

2. Originally Posted by Jones
Hi,
= $1 \pm \sqrt{1-3}$

This is where i get it wrong, isn't the square root expression supposed to evaluate to 2i ?

The correct answer is $1\pm i\sqrt{2}$ however, the result im getting is $1\pm 2i$

Any ideas?
No, you have it right.

= $1 \pm \sqrt{1-3} = 1 \pm \sqrt{-2} = 1 \pm \sqrt{-1*2} = 1 \pm i\sqrt{2}$

Just remember that the square root doesn't go away just because you have a negative number under it. In the same way $\sqrt{8} = \sqrt{4*2} = 2\sqrt{2}$, right?