complex numbers $z^n = 1$ complex plane

Question

I would like to know if I answered the question below correctly or if I need to change something.

Question: Take $n ∈ \mathbb N$. Find all solutions for $z^n=1$. Where can we find the solutions in the complex plane? Draw them in the complex plane for some values of $n$.**

$z^n = 1$ and $ n ∈ \mathbb N$. Assumption: $ 1 = e^{i0} = e^{i2π} = e^{i4π} = e^{i2(2π)} $.

$z^n = 1$ and $ n ∈ \mathbb N$.

= $ \cos (0) + i \sin (0) $, (as $r = 1$)

= $ \cos (0+2kπ) + i \sin (0+2kπ)$

= $ \cos (2kπ) + i \sin (2kπ) $, this gives us $ e^{i(2kπ)} $ with $k ∈ \mathbb Z$.

$z^n = 1$, so $z=\sqrt[n]{1 }$ and $z = 1^{1/n}$.

Using deMoivre's theorem, we get $ z = [e^{i(2kπ)}]^{1/n} $ =

$ z = e^{i2π(k/n)} , k = 0,1,2,...,n-1 $ with $ n ∈ \mathbb N$ .

Using deMoivre's theorem again, we get $ z = [\cos(2kπ) + i \sin (2kπ)]^{1/n} = \cos (2kπ/n) + i \sin (2kπ/n) $.

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Now, we draw the solutions for some values of $n$ in the complex plane.

$ n = 1 \to z = 1 $ gives $ \cos (2kπ) + i \sin (2kπ) $, so $ z = 1 $.

$ n = 2 \to z = 1^{1/2} $ gives $ \cos (2kπ/2) + i \sin (2kπ/2) = \cos (kπ/2) + i \sin (kπ/2) $ , so $ z = 1, -1 $.

$ n = 3 \to z = 1^{1/3} $ gives $ \cos (2kπ/3) + i \sin (2kπ/3) $ , so $ z = -1/2 + 1/2√3 i, -1/2 - 1/2√3 i, 1 $ .

$ n = 4 \to z = 1^{1/4} $ gives $ \cos (2kπ/4) + i \sin (2kπ/4) = \cos (kπ/2) + i \sin (kπ/2) $, so $z = i, -1, -i, 1$.

Did I do it right? Should I change something in my notation?

Answer 1

A more straightforward way to solve this would be to use that two nonzero complex numbers are equal if and only if their modules are equal, and their arguments are equal modulo $2\pi$ in the following way.

Write

$$z=re^{i\theta},$$

and note that your equation then reads

$$r^ne^{in\theta}=1.$$

Since $r^n=\lvert z^n\rvert=1$ by matching modulo, we have that $r=1$, and so $z=e^{i\theta}\ne0$. Furthermore, by matching arguments,

$$\exists k\in\mathbb{Z},\quad n\theta=2\pi k,$$

i.e.

$$\exists k\in\mathbb{Z},\quad\theta=\frac{2\pi k}n.$$

This gives the possible solutions as

$$z=e^{i\frac{2\pi k}{n}}$$

for $k\in\mathbb{Z}$. It is also easy to check (by plugging into the original equation) that these are indeed solutions for any choice of $k\in\mathbb{Z}$. Finally, notice that

$$e^{i\frac{2\pi (k+n)}{n}}=e^{i\frac{2\pi k}{n}}e^{i2\pi}=e^{i\frac{2\pi k}{n}},$$

and so there are only $n$ solutions, which can be found by taking, for example, $k=0,1,\dots, n-1$.

Answer 2

An easier way to solve would be: $$ \begin{align*} z^{n} &= 1 \Rightarrow z^{n} = |z^{n}|\\ z &= |z| \cdot \mathrm{e}^{\arg(z) \cdot \mathrm{i}} (|z| \cdot \mathrm{e}^{\arg(z) \cdot \mathrm{i}})^{n} &= 1\\ |z|^{n} \cdot (\mathrm{e}^{\arg(z) \cdot \mathrm{i}})^{n} &= 1\\ |z|^{n} \cdot \mathrm{e}^{n \cdot \arg(z) \cdot \mathrm{i}} &= 1\\ \mathrm{e}^{n \cdot \arg(z) \cdot \mathrm{i}} &= 1 \quad\mid\quad (\text{ })^{\frac{1}{2}}\\ \mathrm{e}^{\arg(z) \cdot \mathrm{i}} &= 1^{\frac{1}{2}}\\ \mathrm{e}^{\arg(z) \cdot \mathrm{i}} &= \pm \sqrt{1}\\ \mathrm{e}^{\arg(z) \cdot \mathrm{i}} &= \pm 1\\ &\Rightarrow \mathrm{e}^{\arg(z) \cdot \mathrm{i}} = \Re(\mathrm{e}^{\arg(z) \cdot \mathrm{i}})\\ &=\Rightarrow \arg{z} = 2 \cdot \frac{k_{k \in \mathbb{Z}} + n}{n} \cdot \pi \end{align*} $$

That means that there are n solutions with $$ k = 1, ~2, ~..., ~n - 1, $$ wich means that there are mure solutios for z then $$ z = \left\{1, ~i, ~-1, ~-i\right\}, $$ aka you should change this. e.g. $$ \begin{align*} z &= \frac{\mathrm{1}}{\sqrt{2}} + \frac{\mathrm{i}}{\sqrt{2}}\\ z^{4} &= (\frac{\mathrm{1}}{\sqrt{2}} + \frac{\mathrm{i}}{\sqrt{2}})^{4}\\ z^{4} &= (1^{\frac{1}{4}})^{4}\\ z^{4} &= 1 \end{align*} $$