Find all solutions in $\mathbb C$ for $z^4 = 1$

Question

To start, I write the equation in polar form: $$|z|^4(cos^4\theta + isin^4\theta) = 1(1 + 0i)$$

Next, I want to solve for $\theta$: $$cos4\theta = 1 \textrm{ and } sin4\theta = 0$$

$$4\theta = cos^{-1}(1) = 0$$ $$4\theta = sin^{-1}(0) = 0$$

Here is where I get stuck. I think that when solving for $\theta$, the $0$ also means $2\pi$ as well so I can get $4\theta = 2\pi + n2\pi$. This will turn into $\theta = \frac{\pi}{2} + \frac{n\pi}{2}$

My textbook says the values of $n$ yields values $\theta$ where $0 \leq \theta < 2\pi$ and it seems I feel like I am almost done with the problem, but I don't know how to deal with $n$ in the above equation.

There should $n$ solutions in $z^n$, so 4 solutions. I am not entirely sure why $n$ is added into the equation. I suspect it is link to the $i$ in the polar form. Am I missing a step between where I am with the $\theta$ value and the 4 solutions for $z^4$?

Answer 1

I am not entirely sure why n is added into the equation.

===

Ah!

Important.

The thing is that $\theta \equiv \theta + 2n\pi$ for any integer value of $n$ in that $e^{i\theta} = \cos(\theta) + i \sin(\theta) = \cos(\theta + 2n\pi) + i\sin(\theta + 2n\pi) = e^{i\theta}$.

So if you have $c = r*e^{i\gamma}$ and you want

$z^k = c=r*e^{i\gamma}$ so you want to find the $k$th roots of $c$. (Note: the root $k$ that you want to find has nothing to do with the $n$ in the expression $\theta + 2n\pi$)

you want to figure $z = \sqrt[k]{r} e^{i\frac \gamma k}$.

But we have an issue. That isn't the only root.

$c = re^{i\gamma}= re^{i(\gamma + 2\pi)}$ as well.

So $\sqrt[k]{r}e^{i(\frac \gamma k + \frac {2\pi}k)}$ will also be a root.

So will $c = re^{i\gamma}= re^{i(\gamma + 2\pi)}=re^{i(\gamma + 4\pi)}$ and $\sqrt[k]{r}e^{i(\frac \gamma k + \frac {4\pi}k)}$ will also be a root.

In fact we are going to have $k$ roots of $\sqrt[k]{r} e^{i\frac \gamma k}, \sqrt[k]{r}e^{i(\frac \gamma k + \frac {2\pi}k)}, \sqrt[k]{r}e^{i(\frac \gamma k + \frac {4\pi}k)}, .... \sqrt[k]{r}e^{i(\frac \gamma k + \frac {2(k-1)\pi}k)}$.

We write this as the $k$ roots $\sqrt[k]{r} e^{i(\frac \gamma k) + \frac {2n\pi}k}$ for all the $n$ where $n = 0,....., k-1$.

...

So in your case you have $4\theta = 0, 2\pi, 4\pi, 6\pi$ or $2n \pi$ for $n\in \mathbb Z$ and you what

$\theta = \frac {2n\pi}4$ for $n\in \mathbb Z$. If we assume $0 \le \frac {2n\pi}4 < 2\pi$, we have $n = 0, 1,2,3$ and $\theta = 0, \frac \pi 2, \pi, \frac {3\pi} 2$.

===== old answer====

$\theta = \frac \pi 2 + \frac {n\pi}2$ for $n\in \mathbb Z$ so $n = ...... ,-4, -3, -2, -1, 0 , 1 ,2, 3, 4,....$

So $\theta =....., \frac \pi 2 - \frac {4\pi}2, \frac \pi 2 - \frac {3\pi}2, \frac \pi 2 - \frac {2\pi}2, \frac \pi 2 - \frac {1\pi}2, \frac \pi 2 - \frac {0\pi}2, \frac \pi 2 + \frac {1\pi}2, \frac \pi 2 + \frac {2\pi}2, \frac \pi 2 + \frac {3\pi}2, \frac \pi 2 + \frac {4\pi}2,....$

So $\theta = ...... -\frac {3\pi}2, -\pi, -\frac \pi 2, 0, \frac \pi 2, \pi, \frac {3pi}2,2\pi, \frac {5\pi}2,....$

but if we only consider the values of $\theta$ that are unique up to equivalent values between $0$ and up to $2\pi$ we have

$\theta = 0, \frac \pi 2, \pi, \frac {3\pi}2$ and so

The roots are $e^0 = 1; e^{i\frac \pi 2} = i; e^{i\pi} =-1; e^{i\frac {3\pi}2}=-i$

.... which hopefully were the four roots you were expecting.

Geting a value of $\frac {n\pi}2$ or whatever just means to go through the possible integer values of $n$ that give values within the range $[0,2\pi)$. That's all.

Answer 2

The polar form $re^{i\theta}$ of complex numbers $z = x +iy$ is a wonderful thing, as are de Moivre's theorem

$(\cos \theta + i\sin \theta)^n = \cos (n\theta) + i\sin (n\theta), \tag 1$

and Euler's formula

$e^{i\theta} = \cos \theta + i \sin \theta; \tag 2$

but sometimes it is just easier to avoid all the hoopla with multiple values of $\theta$ separated by $2\pi$ and stick with plain old algebra, viz.

$z^4 - 1 = (z^2 - 1)(z^2 + 1); \tag 3$

$z^2 - 1 = (z - 1)(z + 1); \tag 4$

$z^2 + 1 = (z + i)(z - i); \tag 5$

each of the formulas $(3)$-$(5)$ is in fact a special case of the everyday polynomial identity

$a^2 - b^2 = (a + b)(a - b). \tag 6$

If we combine $(3)$-$(5)$ we obtain

$(z - 1)(z + 1)(z + i)(z - i) = (z^2 - 1)(z^2 + 1) = z^4 - 1; \tag 7$

since each of the distinct factors on the left is a monic polynomial we see we have completely factored $z^4 - 1$ and thus

$z = \pm 1, \; \pm i \tag 8$

are the roots of

$z^4 - 1 = 0. \tag 9$