Math Help - Revision-complex number

1. Revision-complex number

Find all roots of the equations
(a) z^4 + 2z^2 + 4 = 0

My step:

Let u = z^2
u^2 + 2u + 4 = 0

u = -1 + (3)^1/2 and u = -1 - (3)^1/2

i'm stuck at there...and tomorrow having complex number exam.

2. Originally Posted by wkn0524
Find all roots of the equations
(a) z^4 + 2z^2 + 4 = 0

My step:

Let u = z^2
u^2 + 2u + 4 = 0

u = -1 + (3)^1/2 and u = -1 - (3)^1/2

i'm stuck at there...and tomorrow having complex number exam.
Well, first your values for u are wrong. $u= -1+ i 3^{1/2}$ and $u= -1- i 3^{1/2}$. You forgot the "i".

Now use the fact that $u= z^2$.
$z^2= -1+ i3^{1/2}$ so $z= \pm\sqrt{-1+ i 3^{1/2}}$
$z^2= -1- i3^{1/2}$ so $z= \pm\sqrt{-1- i3^{1/2}}$.

I presume you will want to use d'Moivre's formula to take those square roots. Since $1^2+ (3^{1/2})^2= 4$ and $\frac{3^{1/2}}{-1}= -\sqrt{3}= arctan(\pi/3}$, in polar form, $-1+ i3^{1/2}= 2(cos(\pi/3)+ i sin(\pi/3))= 2e^{i\pi/3}$ and $-1- i3^{1/2}= 2(cos(\pi/3)- i sin(\pi/3))= 2e^{-i\pi/3}$.

3. Originally Posted by HallsofIvy
Well, first your values for u are wrong. $u= -1+ i 3^{1/2}$ and $u= -1- i 3^{1/2}$. You forgot the "i".

Now use the fact that $u= z^2$.
$z^2= -1+ i3^{1/2}$ so $z= \pm\sqrt{-1+ i 3^{1/2}}$
$z^2= -1- i3^{1/2}$ so $z= \pm\sqrt{-1- i3^{1/2}}$.

I presume you will want to use d'Moivre's formula to take those square roots. Since $1^2+ (3^{1/2})^2= 4$ and $\frac{3^{1/2}}{-1}= -\sqrt{3}= arctan(\pi/3}$, in polar form, $-1+ i3^{1/2}= 2(cos(\pi/3)+ i sin(\pi/3))= 2e^{i\pi/3}$ and $-1- i3^{1/2}= 2(cos(\pi/3)- i sin(\pi/3))= 2e^{-i\pi/3}$.

I forgot put i, its mean -1. there are four solution.