Vectors in Complex (1 Viewer)

asharnadeem · Oct 7, 2020

If alpha, beta, gamma and delta are complex numbers, then if alpha minus beta/gamma minus delta is imaginary, show that the join of alpha and beta is perpendicular to the join of gamma and delta.

CM_Tutor · Oct 8, 2020

$\begin{align*} \text{Given } \frac{\alpha - \beta}{\gamma - \delta} &= ki \text{, where $k\neq 0$ and $k\in\mathbb{R}$} \\ \arg{\frac{\alpha - \beta}{\gamma - \delta}} &= \arg{ki} \\ \arg{(\alpha - \beta)} - \arg{(\gamma - \delta)} &= \pm\frac{\pi}{2} \\ \arg{(\alpha - \beta)} &= \arg{(\gamma - \delta)} \pm \frac{\pi}{2} \end{align*}$

$\alpha - \beta$ is a vector from $\beta$ to $\alpha$ , which is presumably what is meant by the "the join of $\alpha$ and $\beta$ . The above shows that direction of $\alpha - \beta$ differs from that of $\gamma - \delta$ by 90 degrees in either the clockwise or anticlockwise direction. This means that the vectors must be perpendicular, as required.

This is an application of the rule that multiplication by $i$ corresponds to a anticlockwise rotation by 90 degrees. In other words, if A corresponds to the complex number $z$ and B corresponds to $iz$ and O is the origin then angle AOB is a right angle.

Vectors in Complex (1 Viewer)

asharnadeem

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