If complex numbers $z$ satisfy the equation $|z-(1+i)| = 2$ and $\displaystyle \omega = \frac{2}{z}$, then locus traced by $\omega$ in complex plane, is …
My try
I want to solve it geometrically. Here $|z-(1+i)| = 2$ Represent a Circle whose center is at $(1,1)$ and Radius $=2$.
So $z$ lies on a given circle.
But I did not understand how can we find locus of $\displaystyle \omega = \frac{2}{z}$
Best Answer
Because $|\frac{2}{w}-(1+i)|=2$, so $\bigg(\frac{2}{w}-(1+i)\bigg)\bigg(\frac{2}{\bar{w}}-(1-i)\bigg)=4$
$$\therefore \frac{4}{w\bar{w}}-(1-i)\frac{2}{w}-(1+i)\frac{2}{\bar{w}}+2=4\Rightarrow w\bar{w}-(1-i)w+(1+i)w=2$$ $$\Longrightarrow|w+(1-i)|^2=4$$
So we get $\frac{2}{w}$ is actually a circle with center $-1+i$ and radius $2$.