The roots of $z^n=1$ are called root of the unity.  Prove if $w=\cos{\frac{2\pi}{n}}+i\sin{\frac{2\pi}{n}}$ then the $n$ roots can be written as $1,w,w^2,...,w^{n-1}$  My attempt:  Let $z,w\in\mathbb{C}$ such that $w=z^n$ then $w^{1/n}=z$, by hypothesis we know that: $w=\cos{\frac{2\pi}{n}}+i\sin{\frac{2\pi}{n}}$ then,$
120402
Let $\omega$ be a primitive $n-$th root of unity.  (i) Show that its powers $\omega^k$, for $k ∈ {1, \ldots, n}$, are all different;  (ii) Deduce that they are precisely all the $n-$th roots of unity.  I know that the powers have to be different as otherwise the order of $ω$ would be less than $n$ so $\omega$ wouldn't be a primitive $n-$th root of unity but I don't know how to prove this rigorously.
90904
I have to prove that the following holds. A hint to use complex numbers has been given. I have tried to make a start but not to any result.      $$\prod_{k=1}^{n}\tan\left(\frac{\pi k}{2n+1}\right)=\sqrt{2n+1}$$     My Attempt:  Let us consider $(2n+1)^{\text{th}}$ roots of unity, $z_k=\exp\left(\frac{2k\pi i}{2n+1}\right)$. We can rewrite the product in terms of $\arg(z_k)$ as $\prod_{k=1}^{n}\tan\left(\frac{1}{2}\arg(z_k)\right)$. Or equivalently so, if we consider $(4n+2)^{\text{th}}$ roots of unity, we get this product as $\prod_{k=1}^{n}\tan(\arg(\zeta _k))$, where $\zeta_k= \exp\left(\frac{2k\pi i}{4n+2}\right)$.     I know it can be proved by proving the expression for $\prod\sin(\frac{1}{2}\arg(z_k))$. But I was wondering, is there a way to use telescopic products or purely the geometry of the complex roots of unity to arrive at this
5770
I have a dilemma. I have a task where I'm supposed to show that $|\cos z|^2+|\sin z|^2=1$ if and only if $z\in \mathbb{R} $. In my argument I have that $$\cos z\cdot\overline{\cos z} + \sin z\cdot\overline{\sin z}=\cos z\cdot \cos\overline{z} + \sin z\cdot \sin\overline{z}$$ but my teacher said that $\overline {\cos z} \neq \cos \overline{z}$.  Is this true?  But $$\overline{\cos(z)} = \overline{\frac{e^{iz}+e^{-iz}}{2}}=\frac{e^{-i\overline{z}}+e^{i\overline{z}}}{2}=\cos(\overline{z})$$
74361
Ok, so I have the following proof. Let $z$ and $w$ be complex numbers. Prove $\lvert z+w \rvert ^2 + \lvert z-w \rvert^2 = 2[\lvert z \rvert^2 + \vert w \rvert^2]$. Using $\vert z \rvert^2=z\bar{z}$, I have said   $\lvert z+w \rvert ^2 + \lvert z-w \rvert^2 = 2(\lvert z \rvert^2) + \vert w \rvert^2)$  $(z+w)\bar{(z+w)}+(z-w)(\bar{(z-w)}$  $(z+w)(z-w)+(z-w)(z+w)$  $z^2 - w^2 + z^2 - w^2$  $z^2 + z^2 + -w^2 + -w^2$  $2(z^2) + 2(-w^2)$  $2[z^2 + (-w)^2]$  $2[\lvert z \rvert^2 + \vert w \rvert^2]$  I guess my question is, is my logic correct? I am not so skilled with proof writing and though I feel that this is correct, I have been known to be wrong in the past. Just need some insight. Thank you.
22236
$$\cos A \cdot \cos 2A  \cdot \cos 2^{2}A  \cdot \cos 2^{3}A  ... \cos 2^{n-1}A = \frac { \sin 2^n A}{ 2^n \sin  A } $$  I am very much inquisitive to see how this trigonometrical identity can be proved.  PS:I am not much of interested about an inductive proof.
35974
I found a solution but when testing it, it does not work....?    $$\cos z + \sin z = i \\ \frac{e^{iz} + e^{-iz}}{2} + \frac{e^{iz} - e^{-iz}}{2i} = i \\ e^{2iz}(1+i) + 2e^{iz} + i - 1 = 0 \\ e^{iz} = \frac{-1 \pm \sqrt{2-i}}{1+i} \\ z = -i \mathrm{Log}\left(\frac{-1\pm\sqrt{2-i}}{1+i}\right) + k2\pi$$    Upon testing it in W|A with $k=0$, I don't get the required result $(=i)$. I'm not sure where I went wrong and I did this a few times already.
38332
If  $$\sin A + \cos A + \tan A + \cot A + \sec A + \csc A = 7$$ then prove that  $$\sin 2A \quad\text{ is a root of }\quad x^2 - 44x - 36 = 0$$  I have no idea how to solve it. Plz help.
26209
Next is what I have worked out to the moment.   $1$ and $-1$ are roots for all $n$.  $x \in \mathbb{Z}/n\mathbb{Z},\ $ $x^2\equiv1 \Leftrightarrow (x-1)(x+1)\equiv0 \Leftrightarrow \exists k \in \mathbb{Z}/n\mathbb{Z}: k(k+2)\equiv0 $. But how can it be applied to find other roots?
73185
Suppose $z$ is a nonzero complex number, so $z=re^{i\theta}$. Show that there are only $n$ distinct complex $n$-th roots, given by $r^{1/n}e^{i(2\pi k+\theta)/n}$ for $0\leq k\leq n-1$.   My proof: Let $z=re^{i\theta}\in\mathbb{C}$ and $a\in\mathbb{C}$ s.t. $a^{n}=z$. Let $a=\rho e^{i\varphi}$. Then, $a^{n}=(\rho e^{i\varphi})^{n}$. Setting $\rho^{n}e^{i\varphi n}=re^{i\theta}$ and rewriting in polar form, we get $$\rho^{n}(\cos(n\varphi)+i\sin(n\varphi))=r(\cos(θ)+i\sin(\theta)).$$ Thus $\rho^{n}=r$ $\rightarrow$ $\rho=r^{1/n}$ and $n\varphi=\theta\rightarrow\varphi=\theta/n$. Thus the roots of $a\in\mathbb{C}$ have the form $$r^{1/n}(\cos(\tfrac{\theta}{n})+i\sin(\tfrac{\theta}{n}))=r^{1/n}(\cos(\tfrac{\theta}{n}+2\pi j)+i\sin(\tfrac{\theta}{n}+2\pi j))$$ for some $j\in\mathbb{Z}$ due to the periodicity of trig functions. Then, $$r^{1/n}(\cos(\tfrac{\theta}{n}+2\pi j)+i\sin(\tfrac{\theta}{n}+2\pi j))=r^{1/n}e^{i\left(2\pi j+\tfrac{\theta}{n}\right)}=r^{1/n}e^{i(\theta+2\pi k)/n}$$ for $0\leq k\leq n-1 ,j=nk\in\mathbb{Z} $.  I feel like there could be a more concise way to put this.
23787
So I have this result  $1 + z + z^2 + ... + z^n = \frac{z^{n+1}-1}{z-1}$  which I proved already. Now I am supposed to use that result and De Moivre's formula to establish this identity     $1 + \cos\theta + \cos2\theta +... + \cos n\theta = \frac{1}{2} + \frac{\sin[(n+\frac{1}{2})\theta]}{2\sin(\frac{\theta}{2})}$   Can anyone help me?
72740
This is my attempt I know this is incomplete or may even be wrong.   Let $θ_1 \in \arg(z)$ and $θ_2 \in \arg(w)$. Then, $θ_1+θ_2 \in \arg(z)+\arg(w)$. Also, $θ_1+θ_2 \in \arg(zw)$. Is this sufficient for the proof or correct at all? Hope someone could help me out. Thanks.   EDIT: Just to avoid any confusions I will add how $\arg z$ is defined.   $\arg z=\{\text{Arg}\, z+2\pi k\mid k\text{ is an integer}\}$
83335
Prove that the equation $\tan(z)=z$ has only real roots. How to do it? The idea is that the increment of the argument need to look at the boundary of the square with a side of $\pi n$ and another that $\tan(z)-z$ has a pole at $0$. I do not know how to use it.
22060
This is Exercise EP $8$ from Fernandez and Bernardes's book Introdução às Funções de uma Variável Complexa (in Portuguese).     If $z,w\in \mathbb{C}^{\times}$, then $|z+w|=|z|+|w|$ iff $w=tz$ for   some $t\gt 0.$   I supposed that $|z+w|=|z|+|w|$ and by squaring it I got  $\operatorname{Re}(z\bar{w})=|z||w|$ and I am stuck.  I would appreciate a hint in this question.
25105
Let $n \ge 1$ and $u_{1},u_{2},...,u_{n}$ are all root of unity for $n+1$ and there are different from $1$.(a) Find polynomial $f \in \mathbb R[x]$ of degree $n$ for which: $f(u_{k})=0$ for $k=1,2,...,n$.(b) Calculate real coefficient and imaginary part for $(2-u_{1})(2-u_{2})\cdot...\cdot(2-u_{n}).$    I have a knowledge about root of unity but I completely don't know how to do it because I think that I need some extra trick to do it but my ideas finished. I know that $\varepsilon_n^{(k)} = \cos\left(\tfrac{2k\pi}{n}\right) + i\sin\left(\tfrac{2k\pi}{n}\right) = e^\frac{2\pi i k}{n}$ and for $n&gt;1$ $\sum _{{k=0}}^{{n-1}}e^{{\frac  {2\pi ik}{n}}}=0$ but I think it is too little knowledge to solve this task.Can you help me?
105769
$z_1$, $z_2$, and $z_3$ are 3 complex numbers. Prove that if they represent the vertices of an equilateral triangle then $z_1 + \omega z_2 + \omega^2 z_3 = 0$ where $\omega$ is a 3rd root of unity. Any help would be thoroughly appreciated.
41928
If w = $\frac {\sqrt2+\sqrt3}{\sqrt5-\sqrt3} $  Express the following in the simplest form (with a rational denominator)  i) w  ii) $\frac {1}{w} $  It must be like this right ?  w = $\frac {\sqrt2+\sqrt3}{\sqrt5-\sqrt3} $ x $\frac {\sqrt5+\sqrt3}{\sqrt5+\sqrt3} $  Please help me with this question coz I don't get it at all :/
60278
We know that for every polynomial of $n$.order Fundamental Theorem of Algebra guarantees $n$ complex roots.  Lets consider the complex exponential function $f(z)=\exp(z)$. As $f(z)$ is holomorphic, we are allowed to evaluate the power sum of $f(z)$.  $$f(z)=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots$$   We also know that   $$f(z)=\exp(z)=\exp(x+iy)=\exp(x)\exp(iy)=\exp(x)\cdot(\cos(y)+i\sin(y))$$  A zero of this expression would only be possible if $\exp(x)$ or $\cos(y)+i\sin(y)$ or both would vanish. But $\exp(x)&gt;0$ and $|\cos(y)+i\sin(y)|=1$, hence $\exp(z)$ doesn't vanish. As the exponential is fully defined by the powersum with infinite radius of convergence, we can conclude that the powersum, despite of beeing of "polynomial like type" does not have any zeros.  I would be glad if someone could give me an intuitive explanation why the fundamental theorem of algebra is only true for finite polynomials.