De Moivre's theorem: x^2 +(-1-3i)x+(-2+2i)=0

[btrfly24]

x^2 +(-1-3i)x+(-2+2i)=0

I'm using the quadratic equation first, then the De moivre's theorem but I just can't seem to get past the quadratic equation! hahaha! This is how far I've gotten

-1(-1-3i) +/- sqrt (-1-3i)^2-4(-2+2i) all over 2. Then:

(1+3i) +/- sqrt (1+6i-9)+(-8-8i) all over 2. From here I'm clueless. Am I even headed in the right direction? Please help! Catherine.

 

[arthur ohlsten]

x^2-[1+3i]x+[-2+2i]=0

by quadratic equation
x=[1+3i] +/-[1-9+6i+8-8i]^1/2 all over 2
2x=[1+3i] +/-[-2i]^1/2

we will use DeMoivres theorem to find the square roots of -2i
-2i=2[ cos270+i sin 270]
[-2i]^1/2= sqrt2[cos 135 +i sin 135]
[-2i]^1/2 =+/- [-1 +i]

2x=[1+3i]+/-[-1+i]
2x=4i OR 2x=2+2i
x=2i answer
x=1+i answer

Arthur

 

[Denis]

Re: De Moivre's theorem

x = (1+3i) +/- sqrt (1+6i-9)+(-8-8i) all over 2.

Simplify the sqrt contents:

x = [(1 + 3i) +- sqrt(-2i)] / 2

 

[btrfly24]

thanks so much for you guys replies, they helped out a ton!