MHB Solving $z^4=-1$ with a Complex Number

[Dustinsfl]

For some odd reason, I can't think right now. How do I solve $z^4 = -1$ where z is a complex number?

 

[Prove It]

For some odd reason, I can't think right now. How do I solve $z^4 = -1$ where z is a complex number?

There's a few ways. The easiest is IMO this way...

\[ \displaystyle \begin{align*} z^4 &= -1 \\ \left(r\,e^{i\theta}\right)^4 &= e^{i\left(\pi + 2\pi n\right)}\textrm{ where }n \in \mathbf{Z} \\ r^4e^{4i\theta} &= 1e^{i\left(\pi + 2\pi n\right)} \\ r^4 = 1 \textrm{ and } 4\theta &= \pi + 2\pi n \\ r = 1 \textrm{ and } \theta &= \frac{\pi}{4} + \frac{\pi n}{2} \end{align*} \]

Now evaluate the values for $\displaystyle \theta $ which are in the region $\displaystyle \theta \in (-\pi, \pi]$.

 

[Sherlock1]

There's a few ways. The easiest is IMO this way...

\[ \displaystyle \begin{align*} z^4 &= -1 \\ z^4 - 1 &= 0 \end{align*} \]

Are you sure? :p

 

[Prove It]

Are you sure? :p

Oops, I meant +1 haha, editing :P

 

[Sherlock1]

If 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 ----------

Oops, I meant +1 haha, editing :P

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}

 

[Prove It]

If 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}

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 start :)