Complex Number, again... :(

[JasonRox]

This I'll give you the entire question and answer.

10. Solve:

z^4-2z^2+4=0

That's all I got.

Answer:

+-1/2(\sqrt{6}+-\sqrt{2i}) (four combinations of signs).

That is all.

I tried factoring, but I can't come up with anything. I also tried bisecting, but that is useless.

Knowing that the solution has complex numbers, I have no clue where they got this from.

[Dr Transport]

solve for z^{2} then take the square root of those........

[JasonRox]

Ok then.

z^2=(\frac{z^2}{\sqrt{2}}-2i)(\frac{z^2}{\sqrt{2}}+2i)

The square root of these?

[IamSorryOkay?]

my results are
z=+-\sqrt{1+i\sqrt{3}}
z=+-\sqrt{1-i\sqrt{3}}

I dont know whether they are correct or not..

[HallsofIvy]

let x= z2 then the equation z4- 2z2+ 4= 0 becomes x2- 2x+ 4= 0. That is the same as x2- 2x= -4 or
x2- 2x+ 1= (x- 1)2= -3. From that x= 1+- \sqrt{3}i.
Since x= z2, z^2= 1+- i\sqrt{3} so [itex]x= +- \sqrt{1+- i\sqrt{3}}.

[JasonRox]

I can see that, but where would their solution come from?

Using the quadratic formula yield the same result as yours, too.