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There's a few ways. The easiest is IMO this way...For some odd reason, I cant think right now. How do I solve $z^4 = -1$ where z is a complex number?
Are you sure?There's a few ways. The easiest is IMO this way...
\[ \displaystyle \begin{align*} z^4 &= -1 \\ z^4 - 1 &= 0 \end{align*} \]
Oops, I meant +1 haha, editingAre you sure?![]()
I think the idea of your earlier solution does work, though.Oops, I meant +1 haha, editing![]()
True, but to evaluate $\displaystyle \sqrt{i}$ and $\displaystyle i^{\frac{3}{2}}$ requires converting to polars, so it's easier to convert right at the startIf we use Euler's identity we can easily find that $\displaystyle z^n+1 = \prod_{k=0}^{n-1}\left(z-e^{i\left(\dfrac{\pi}{n}+\dfrac{2\pi}{n} k\right)}\right).$
---------- Post added at 00:31 ---------- Previous post was at 00:19 ----------
I think the idea of your earlier solution does work, though.
\begin{aligned}z^4+1 & = z^4-i^2 \\& = (z^2-i)(z^2+i) \\& = (z-\sqrt{i})(z+\sqrt{i})(z^2-i^3) \\& = (z-\sqrt{i})(z+\sqrt{i})(z-i^{3/2})(z+i^{3/2}). \end{aligned}