# Complex Number, again

1. Oct 11, 2004

### JasonRox

Complex Number, again... :(

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

10. Solve:

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

That's all I got.

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

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.

2. Oct 11, 2004

### Dr Transport

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

3. Oct 11, 2004

### JasonRox

Ok then.

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

The square root of these?

4. Oct 11, 2004

### 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..

5. Oct 11, 2004

### HallsofIvy

Staff Emeritus
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}}.

6. Oct 11, 2004

### JasonRox

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

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