Question

If the imaginary part of the expression $\frac{z-1}{e^{i\theta }}+\frac{e^{i\theta }}{z-1}$
be zero, then determine the locus of the point $z$.

Answer 1

Let $z-1=r(\cos \alpha +i\sin \alpha )=r{e}^{i\alpha }$
Or $(x-1)+iy=r\cos \alpha +ir\sin \alpha$
$\therefore r^2=(x-1)y^2$ and $\tan \alpha =\frac{y}{x-1}$
$\therefore$ Given expression, $r{e}^{i(\alpha -\theta )}+\frac{1}{r}{e}^{-i(\alpha -\theta )}$
Its imaginary part is $r$ $\sin (\alpha -\theta )-\frac{1}{r}\sin (\alpha -\theta )=0$
or $(r-\frac{1}{r})\sin (\alpha -\theta )=0$
Either $(r-\frac{1}{r})=0\Rightarrow r^2=1\Rightarrow (x-1{)}^2+y^2=1$
Above represents a circle centred at $(1,0)$ and radius $1$.
or $\sin (\alpha -\theta )=0$ $\therefore \alpha -\theta =0$
or $\tan \theta =\tan \alpha$
or $\frac{y}{(x-1)}\tan \alpha \therefore y(x-1)\tan \alpha$
Above represents a straight line passing through the point $(0,1)$.