If $w_1$ and $w_2$ have unit modulus and arguments $0<\alpha_1<\alpha_2<\pi/2$, then $\arg(w_1-w_2)=\frac12(\alpha_1+\alpha_2-\pi)$

Question

The complex numbers $w_1$ and $w_2$ have modulus 1, and arguments $\alpha_1$ and $\alpha_2$ respectively, where $0 < \alpha_1 < \alpha_2 < \frac{\pi}{2}$. Show that $\arg(w_1 - w_2) = \frac1{2}(\alpha_1 + \alpha_2 - \pi)$.

How would you go about answering the question? I attempted it graphically, but the question is too abstract for me to really understand to complete it this way.

I then tried solving it algebraically, but again I'm not sure if what I'm doing is right, and secondly if it is I'm not sure how to go about completing the question.

If someone could give me a little hint or help me work out how to begin this question it would be greatly appreciated.

Thanks.

Answer 1

Proof 1 - geometry based, map problem to one on Euclidean geometry.

Identify complex plane $\mathbb{C}$ with the Euclidean plane $\mathbb{R}^2$.
Let $O,A,B$ be three points corresponds to complex number $0, w_1$ and $w_2$.

• Since $|w_1| = |w_2| = 1$, $|OA| = |OB| = 1$, $OAB$ is forming an issoceles triangle.

• The complex number $\rho = w_1 - w_2$ (or more precisely, $\frac{p}{|\rho|}$ ) corresponds to the direction of a ray start at $B$ pointing towards $A$. Since this ray lies on the line holding $AB$ and $\triangle OAB$ is isoceles, it is perpendicular to the angular bisector of $\angle AOB$.

• Since $\arg w_1 = \alpha_1$ and $\arg w_2 = \alpha_2$, the line $OA$ and $OB$ is making an angle $\alpha_1$ and $\alpha_2$ with the $x$-axis. This means the angular bisector is making an angle $\frac12(\alpha_1+\alpha_2)$ with $x$-axis.
• This implies the ray is making an angle (up to a sign) $\frac12(\alpha_1+\alpha_2) \pm \frac{\pi}{2}$ with $x$-axis.

From this, we can deduce $\arg( w_1 - w_2) = \frac12\left(\alpha_1 + \alpha_2 \pm \pi \right)$. Since the ray start at $B$ is pointing downwards, we need to take the minus sign. As a result, $$\arg( w_1 - w_2 ) = \frac12\left(\alpha_1 + \alpha_2 - \pi\right)$$

Proof 2 - playing with complex numbers directly.

Recall for any $z \in \mathbb{C}\setminus \{0\}$, $\arg(z)$ is a number $\theta \in (-\pi,\pi]$ such that $z = |z| e^{i\theta}$.

Let $\alpha = \frac12(\alpha_1 + \alpha_2)$ and $\beta = \frac12(\alpha_2 - \alpha_1)$. In terms of them, we have $$\begin{align}w_1 - w_2 &= e^{i\alpha_1} - e^{i\alpha_2} = e^{i(\alpha - \beta)} - e^{i(\alpha + \beta)} = e^{i\alpha}(e^{-i\beta} - e^{i\beta})\\ &= -2i e^{i\alpha} \sin\beta = 2\sin\beta e^{i(\alpha - \frac{\pi}{2})}\end{align}$$

Notice when $0 < \alpha_1 < \alpha_2 < \frac{\pi}{2}$, $$\alpha - \frac{\pi}{2} \in \left(-\frac{\pi}{2},0\right) \subset (-\pi,\pi] \quad\text{ and }\quad \beta \in \left(0,\frac{\pi}{4}\right) \implies 2\sin\beta > 0 $$

This means expression $\alpha - \frac{\pi}{2}$ match the role of $\theta$ in definition of $\arg(w_1 - w_2)$. As a result,

$$\arg(w_1 - w_2) = \alpha - \frac{\pi}{2} = \frac12(\alpha_1 + \alpha_2 - \pi)$$