QUESTION: I have just come to know that, if it is given that $\frac{z-a}{z-b}$ is purely imaginary, where $z \in \Bbb{C}$, then it implies that the angle between the vectors $z-a$ and $z-b$ is $\frac{\pi}2$.
I have build up a reasoning for this one.. Obviously, we can write $\frac{z-a}{z-b}$ as $re^{i\theta}$ and therefore, setting $\theta$ to be $\frac{\pi}2$ makes the $cos(\theta)$ part of $e^{i\theta}$zero, which in turn makes $z$ a purely imaginary number.. Well and good.
So, from this very reasoning can we say that whenever such a condition is stated we can directly conclude that the locus of $z$ is a circle with the end points of the diameter as $a$ and $b$ ?
Now, extending on this concept, if we have a question stating that $\frac{z-a}{z-b}$ is purely real, then we can say that the angle between the vectors $z-a$ and $z-b$ is zero, or both of them lie on the same straight line.
I am not quite accustomed with the geometrical interpretations of complex division, so I need to know whether my above reasoning is correct..
I wish to know proper geometrical reasons of the above made observations that -
- Two vectors (in the argand plane) that are orthogonal, when divided, leave us with a purely complex number.
- Two vectors (in the argand plane) that are collinear, when divided, leave us with a purely real number.