Why is 1 = $e^{2k{\pi}i}$ when solving nth root's of unity?

Question

Every time I look at an explanation regarding this concept this is automatically assumed.

Also, $n\phi =\theta+2k\pi$ is automatically thrown in without explanation?

I don't understand these, if anybody could run be through an example such as $z^6 = -1$ in as much detail as possible I'd greatly appreciate it since I have an exam tomorrow.

Thanks in advance.

Answer 1

Hint: Use Euler's formula. We thus have $$e^{2k\pi i} =\cos (2k\pi)+ i\sin (2k\pi) =1$$

---

For an example, take this MSE post on solutions to $z^6=-1$. Hope it helps.

Answer 2

Hint

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

Now use $\theta = 2k\pi$

P.S:

$\theta$ and $θ+2kπ$ are congruent angles. It means that all trigonometric functions take same value for both angles. For example, if you know that $\sin θ=x$ then $θ+2kπ$ is also a solution because $\sin(θ+2kπ)=x$ for any choice of $k$ integer

Answer 3

This has to do with Euler's famous result

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

and the natural extension:

$$z=re^{i\theta} = r(\cos\theta + i\sin\theta),\quad r\ge 0 ,\; $$

Since you can write any complex number in this form. Then from here we see the real values happen when $\theta \in \{\pi \cdot k: k\in\Bbb Z\}$ and it's just a hop a skip using periodicity of the trig functions to see you need only consider $\theta = {2\pi k\over n}, 0\le k < n$.

Answer 4

Euler's identity:

$e^{\pi i}=-1\implies$

$(e^{\pi i})^2=(-1)^2\implies$

$(e^{\pi i})^2=1\implies$

$e^{2\pi i}=1\implies$

$\forall{k}\in\mathbb{N}:(e^{2\pi i})^k=1^k\implies$

$\forall{k}\in\mathbb{N}:(e^{2\pi i})^k=1\implies$

$\forall{k}\in\mathbb{N}:e^{2k\pi i}=1$

Answer 5

This proeperty need not be automatically assumedn. It is a consequence of a particularly user-friendly definition of the involved functions:

The exponential function $\exp\colon \Bbb C\to\Bbb C$ (often suggestively written $z\mapsto e^z$) is defined as the unique(!) function $\Bbb C\to\Bbb C$ with the properties $$\tag1\exp(0)=1$$ and $$\tag2\exp'(z)=\exp(z).$$ From this, we derive important further properties. First of all, $$\tag3\exp(z+w)=\exp(z)\exp(w). $$ Indeed, if $\exp(w)\ne0$ then $z\mapsto \frac1{\exp(w)}\exp(z+w)$ clearly has properties $(1)$ and $(2)$, hence by uniqueness equals $\exp(z)$. This shows $(3)$ for the case $\exp(w)\ne 0$. But then if $\exp(z)=0$ for some $z$, we either have $\exp(z/2)=0$ or $0=\exp(z)=\exp(z/2)\exp(z/2)$ and again $\exp(z/2)=0$; by induction $\exp(z/2^n)=0$, and then by continuity, $\exp(0)=0$, contradicting $(1)$. Hence $\exp$ has no zeroes and $(3)$ holds for all $z,w$.

Next, we verify that $z\mapsto \overline{\exp(\overline z)}$ has properties $(1)$ and $(2)$, hence $$\tag4\exp(\overline z)=\overline{\exp(z)} $$ and in particular $\exp(t)\in\Bbb R$ for all real $t$. As there are no roots, $(1)$ and the IVT give us $$\tag5\exp(t)>0 \qquad\text{for all }t\in\Bbb R.$$ By $(2)$, this makes $\exp|_{\Bbb R}$ strictly increasing. By the MWT, for $t>0$ we have $\frac{\exp(t)-\exp(0)}t=\exp'(\underbrace{\tilde t}_{\in(0,t)})=\exp(\tilde t)> \exp(0)=1$, i.e., $\exp(t)> 1+t$ for $t>0$. Similarly, for $t<0$ we have $\frac{\exp(0)-\exp(t)}{-t}=\exp'(\tilde t)=\exp(\tilde t)< \exp(0)=1$, i.e., again $\exp(t)>1+t$, so in summary $$\tag6 \exp(t)\ge 1+t\quad\text{for all }t\in\Bbb R\quad\text{with equality iff }t=0.$$

If we define $$ \cos(z)=\frac{\exp(iz)+\exp(-iz)}{2},\qquad \sin(z)=\frac{\exp(iz)-\exp(-iz)}{2i},$$ we see that $\sin(0)=0$, $\cos(0)=1$, $\sin'=\cos$, $\cos'=-\sin$, and using $(4)$ we see that both map reals to reals. And of course $$\tag7\exp(iz)=\cos(z)+i\sin(z). $$ The derivative of $\sin^2(z)+\cos^2(z)$ is quickly verified to be constantly $0$, hence the expression is constant, i.e., $$\tag8\sin^2(z)+\cos^2(z)=1.$$ Assume $\cos(t)>0$ for all $t\in\Bbb R$. Then $\sin(t)$ is strictly increasing. In particular, $\sin(t)>\sin(1)>0$ for all $t>1$. But then $\cos(t)-\cos(1)<(t-1)\sin(t)$ for all $t>1$ and so $\cos(t)$ does become negative after all. We conclude that $\cos$ has a root on $(0,\infty)$. The minimal(!) positive root is defined as $\frac \pi 2$. As $\cos(t)$ must reach this first root while decreasing, we conclude (also using $(8)$) than $\sin(\frac \pi2)=1$. So $\exp(i\frac\pi2)=i$, $\exp(i\pi)=-1$, $\exp(2\pi i)=1$.

By $(3)$, the set $P:=\{\,z\in\Bbb C\mid \exp(z)=1\,\}$ is a subgroup (under addition) of $\Bbb C$ and is a closed set. As $\exp'(0)\ne 0$, $0$ is an isolated point of $P$ and so $P$ is a discrete subgroup of $\Bbb C$. Then either $P=\{0\}$ or $P=\varpi \Bbb Z$ for some $\varpi\ne 0$ or $P=\varpi_1\Bbb Z+\varpi_2\Bbb Z$ for non-zero $\varpi_1,\varpi_2$ with non-real quotient. The first case is excluded already by $2\pi i\in P$. In the third case, $\exp$ bounded on the compact parallelogram spanned by $\varpi_1$ and $\varpi_2$, hence by periodicity (i.e., by $(3)$) is bounded throughout, contradicting $(6)$. We conclude that we have the second case and more precisely $$\tag 9 \exp z=0\iff z\in 2\pi i\Bbb Z.$$