1. ## Complex roots

Having some difficulty with finding the four complex roots to the equation

Z^4 + 2i*Z^2 + 1

thanks

2. From what i understand you Factorize first, then use the difference of two square rule.

If in doubt just use the -b forumula though

3. Not that simple, but thanks.

I have broken it into two quadratics, but as its complex I need to represent the alternative roots in terms of complex exponentials.

I am having some trouble on this stage as I think i should square root something twice, whereas my answers tells me I don't need to

thanks again for trying

4. Originally Posted by Bryn
Having some difficulty with finding the four complex roots to the equation

Z^4 + 2i*Z^2 + 1

thanks

Let $u=z^2$

Then: $u^2+2iu+1=0$ or:

$u=\frac{-2i\pm i 2 \sqrt{2}}{2}$

then:

$z=\sqrt{-i(1\pm \sqrt{2})}$

So you got two cases in the square root, and each case has two square root.

5. The equation is a quadratic in $z^2$, with solutions $z^2 = -i\bigl(1\pm\sqrt2\bigr)$. The square roots of $-i$ are $\pm\tfrac1{\sqrt2}(1-i)$, so the solutions for z are $z = \pm\tfrac1{\sqrt2}(1-i)\sqrt{1+\sqrt2}$ and $z = \pm\tfrac1{\sqrt2}(1-i)i\sqrt{\sqrt2-1}$. I don't think there is any way of avoiding those repeated square roots.

6. Yes,

Thanks, I didn't explain exactly my problem I guess.

I need to write the solutions in terms of complex exponentials which is what I was finding tricky - anychance you could try that

thanks

7. Originally Posted by Bryn
I need to write the solutions in terms of complex exponentials which is what I was finding tricky - any chance you could try that
The solutions for $z^2$ are $z^2 = -i\bigl(1+\sqrt2\bigr)$ and $z^2 = i\bigl(\sqrt2-1\bigr)$. These have polar form $z^2 = re^{i\theta}$, where for the first root $r=1+\sqrt2$ and $\theta = -\pi/2$. For the second root $r = \sqrt2-1$ and $\theta = \pi/2$. To find the values of z, you need the square roots of these. That means taking the square root of r and halving the value of theta.