If |w||z|=1, show w=1/z

Question

$w, z \in \mathbb C$

Also, $arg(w)=arg(\bar z)$

If $|w||z|=1$, show $w=1/z$

I think this is obvious but I am not sure how to properly format my answer...

Answer 1

Let $\theta =arg(w)=\arg(\overset {-}z)$. Then $w=|w|e^{i\theta}$ and $\overset {-}z =|\overset {-}z| e^{i\theta}$. Taking conjugate on both sides in the second equation we get $z=|z|e^{-i\theta}$. Hence $wz=|w||z|e^{i\theta} e^{-i\theta} =1$.

Answer 2

Suppose that $z = r_1 (\cos \theta_1+i\sin\theta_1) $ and $w = r_2 (\cos \theta_2+i\sin\theta_2)$. Then $$arg(w) = \theta_2=arg(\bar z)=-\arg(z)=-\theta_1$$ Hence $w = r_2(\cos\theta_1-i\sin\theta_1)$. Also $$\vert w\vert \vert z\vert=r_1r_2=1$$ So $$\begin{align} wz&=\left[r_1(\cos\theta_1+i\sin\theta_1)\right]\left[r_2(\cos\theta_1-i\sin\theta_1)\right]\\ &=r_1r_2(\cos^2\theta_1-i^2\sin^2\theta_1)\\ &=(1)(\cos^2\theta_1+\sin^2\theta_1)\\ &=1 \end{align}$$ Hence it is proved that $$w=\frac1z$$

Answer 3

We know that any complex number can be written in the form $re^{i\theta}$. Hence let $w=re^{i\theta}$ ; then by hypothesis $|z|=1/|w|=1/r$ and $arg(w)=arg(\overline{z})$ $\Rightarrow arg(w)=-arg(z)=\theta$ , i.e $arg(z)=-\theta$. Therefore $z=(1/r) e^{-i\theta} \Rightarrow z=1/w $