Let $\omega=\frac{\sqrt{3}i-1}{2}$. Prove $\frac{\omega a-\omega}{1+a+\omega}\in\Bbb{R}$ for all $a\in\Bbb C$ with $|a|=1$, $a\not=\omega^2$.

Question

$\newcommand{\i}{\mathrm i}$Let $\omega=\frac{\sqrt3\i-1}2$, and $a\in\Bbb C$ with $|a|=1$, $a\not=\omega^2$. Prove that $z=\dfrac{\omega a-\omega}{1+a+\omega}\in\Bbb R$.

There's an obvious bash: let $a=\cos x+\i\sin x$. So \[\frac{\omega a-\omega}{1+a+\omega}=\frac{2\cos x-2\sqrt3\sin x-2}{2\cos x+2\sqrt3\sin x+4}\in\mathbb R.\] But can we solve it by the properties of complex numbers?

For example, prove $z=\overline z$. But I got a strange result when computing this.

Answer 1

Note that $w=e^{\frac{2\pi }3i}$ is a cube root of unity. So we have $1+w+w^2=0$ and $z=\dfrac{\omega(a-1)}{a-\omega^2}$.

Show $z=\overline z$, as expected by OP

We have $\omega\overline\omega=1$ as well as $a\overline a = |a|^2=1$.
$$\overline z=\frac{\overline\omega(\overline a-1)}{\overline a-{\overline\omega}^2}=\frac{\omega^2a\,\overline\omega(\overline a-1)}{\omega^2a(\overline a-{\overline\omega}^2)}=\dfrac{\omega(1-a)}{\omega^2-a}=z$$

Another approach, compute the argument of $z$

$$\arg(p+q)=\frac{\arg(p)+\arg(q)}2\pmod{2\pi}$$ for all complex number $p$ and $q$ with $|p|=|q|$ and none of $p,q,p+q$ is $0$. This formula can be seen immediately by the geometric interpretation of complex numbers. Note that $|a|=|{-}1|=|{-}w^2|=1$. There are two cases.

• $a=1$.
  Then $z=0$ is real.
• $a\not=1$. $$\begin{aligned} \arg(z)&=\arg(w)+\arg(a+(-1))-\arg(a+(-w^2))\\ &=\frac{2\pi}3+\frac{\arg a +\pi}2 - \frac{\arg a+\frac\pi3}2\\ &=\pi \pmod{2\pi} \end{aligned}$$ Hence, $z$ is a negative number.

Answer 2

The conjugate of the denominator is $1+a^{-1}+\omega^{-1}$, since inverses and conjugates are the same thing for elements of the unit circle.

It suffices to show $(\omega a-\omega)(1+a^{-1}+\omega^{-1})$ is real. This evaluates to: $$(a-1)+\omega(a-1)+\omega(1-a^{-1})=a-1+\omega(a-a^{-1})$$Which is real if and only if the imaginary part is zero, if and only if: $$0=\Im a+\frac{-1}{2}(\Im(a)-(-\Im(a)))+\frac{\sqrt{3}}{2}(\Re(a)-\Re(a))$$

Which is true.

Answer 3

We know that a Mobius transformation $f(z)=\frac{az+b}{cz+d}$ sends circles/lines to circles/lines and it is enough to check $3$ values to determine the image of a circle/line.

Consider the Mobius transformation $f(z)=\frac{wz-w}{z+w+1}$ where $w=-\frac12+\frac{\sqrt3}{2}i$. Then:

1. $f(1)=0\in\Bbb R$
2. $f(-1)=-2\in\Bbb R$ and
3. $f(i)=-\frac{3+2\sqrt3}{4+2\sqrt3}\in\Bbb R.$

Therefore, if $a\in\{z\in\Bbb C\,\,|\,\,\,|z|=1\}$ then $f(a)\in\Bbb R.$

Answer 4

Being able to write $ a = cosx + isinx$ is an inherent property of complex numbers.

If this doesn't feel "complex" enough, remember that there is a bijection

$$F : [0, \infty[ \times [0, 2\pi[ \rightarrow \mathbb{C}, (r, \phi) \mapsto re^{i\phi}=rcos\phi + i r sin\phi$$

which you are basically using to evaluate your a. Because $\vert a \vert = 1$, $r=1$ therefore you can write a is you have done. So you're using indeed properties of complex numbers.

Answer 5

To show $\newcommand{\w}{\omega}\newcommand{\ww}{\bar{\omega}}\newcommand{\aa}{\bar{a}}$ $$ \frac{\w a - \w}{1+a+\w} = \frac{\ww\aa-\ww}{1+\aa+\ww}, $$ we can show $$ \w( a - 1)(1+\aa+\ww)-\ww(\aa-1)(1+a+\w)=0. \tag{1} $$ Using $\w\ww=1$, $a\aa=1$, after expanding, we have $$ \begin{align*} &\quad \w( a - 1)(1+\aa+\ww)-\ww(\aa-1)(1+a+\w) \\ &= \w(a-1)(1+\aa)+(a-1) - \ww(\aa-1)(1+a)-(\aa-1) \\ &= \w(a-\aa)-\ww(\aa-a)+(a-\aa) \\ &= (\w+\ww+1)(a-\aa). \end{align*} $$ Note that $\w+\ww=-1$, so $(1)$ is proved.