Having some difficulty with finding the four complex roots to the equation
Z^4 + 2i*Z^2 + 1
thanks
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
The equation is a quadratic in $\displaystyle z^2$, with solutions $\displaystyle z^2 = -i\bigl(1\pm\sqrt2\bigr)$. The square roots of $\displaystyle -i$ are $\displaystyle \pm\tfrac1{\sqrt2}(1-i)$, so the solutions for z are $\displaystyle z = \pm\tfrac1{\sqrt2}(1-i)\sqrt{1+\sqrt2}$ and $\displaystyle z = \pm\tfrac1{\sqrt2}(1-i)i\sqrt{\sqrt2-1}$. I don't think there is any way of avoiding those repeated square roots.
The solutions for $\displaystyle z^2$ are $\displaystyle z^2 = -i\bigl(1+\sqrt2\bigr)$ and $\displaystyle z^2 = i\bigl(\sqrt2-1\bigr)$. These have polar form $\displaystyle z^2 = re^{i\theta}$, where for the first root $\displaystyle r=1+\sqrt2$ and $\displaystyle \theta = -\pi/2$. For the second root $\displaystyle r = \sqrt2-1$ and $\displaystyle \theta = \pi/2$. 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.