Dtyjlui

Bit stuck on this question: The aim of this exercise is to determine all elements of finite order in $mathbb{C}^{∗}$, the multiplicative group of non-zero complex numbers.

What is the order of $z=cos(frac{2pi}{3})+isin(frac{2pi}{3})$?

I know $z$ can be of the form $z=re^{itheta}$ but don't know what to do to determine the 'order' or what I can do to the formula to solve this.

Any help would be great, thanks.

Hint: $z=cos(frac{2pi}{3})+isin(frac{2pi}{3})=e^{i2pi/3}$. You need to find the smallest $nin Bbb N $ such that $z^n=1$,i.e, $e^{in2pi/3}=1$. Note that $e^{i2pi}=1$. Can you find such an $n $?

Let us define $e^{iA}= cos A +isin A$ when $Ain Bbb R$ without any thought to the motivation for it. By DeMoivre's Theorem we have $e^{inA}=(e^{iA})^n$ for $nin Bbb Z^+ $ and $Ain Bbb R.$

Let $0ne zin Bbb C.$ Then $z=re^{it}$ where $|z|=rin Bbb R^+$ and $t$ is some member of $Bbb R.$

(1). Suppose $z^n=1$ for some $nin Bbb Z^+.$ Then $1=|z^n|=|r^n(e^{it})^n|=|r^ne^{nit}|=r^n,$ so $r=1.$ So $1=z^n=e^{nit},$ which requires $ntin {2pi m:min Bbb Z}.$ So $t/2pi in Bbb Q.$ So $zin {e^{2pi iq}:qin Bbb Q}.$

(2). Suppose $zin {e^{2pi iq}:qin Bbb Q}.$ Then $z=e^{2pi i m/n}$ for some $nin Bbb Z^+$ and some $min Bbb Z,$ so $z^n=e^{2pi i m}=1.$

(3). From (1) and (2), $z$ has a finite order iff $zin {e^{2pi i q}: qin Bbb Q}=$ $={cos 2pi q +isin 2pi q: qin Bbb Qcap [0,1)}.$

Note. For any real $u,v$ with $u^2+v^2=1$ there exists $tinBbb R$ such that $(cos t, sin t)=(u,v).$ So if $0ne zin Bbb C,$ let $z=x+iy$ with $x,y in Bbb R.$ Since $x,y$ are not both $0,$ we have $|z|=sqrt {x^2+y^2},>0.$ Now let $(u,v)=(x/|z|, y/|z|).$ Since $u^2+v^2=1,$ there exists $tin Bbb R$ with $(u,v)=(cos t,sin t). $ So $z=|z|(u+iv)=|z|(cos t +isin t)=|z|e^{it}.$

If the order of $z$ is the smallest positive integer $n$ such that $z^{n}=1$, then this is fairly simple.

$z$ has a finite order iff $vert z vert = 1$, and $arg (z)=pi cdot q$ where $q$ is a rational number. To find the order, just find the smallest positive integer n such that $q cdot n = 2k$ for some integer $k$. This is because multiplying the complex numbers, their arguments add together. Hopefully this makes sense and I interpreted the question right.

Find the smallest natural $n$ so that $z^n = 1$.

$z=cos(frac{2pi}{3})+isin(frac{2pi}{3})=e^{ifrac {2pi}{3}}$

So $z^n = e^{ifrac {2npi} 3}$

Now $e^{i 2kpi} = 1$ for all integer $k$ and $e^{itheta} = 1$ only if $theta = 2kpi$ for some integer $k$.

So $z^n = e^{ifrac {2npi} 3} = 1 iff frac {2npi}3 = 2pi*k$ for some integer $k$.

So we need to find then smallest natural $n$ so and $frac {2npi}3 = 2pi *k$ for some integer value of $k$.

That's.... not a hard question.