Find the locus of $2/z$ given that $|z-(1+i)| = 2$

Question

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 $\displaystyle \omega = \frac{2}{z}$

Answer 1

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 $2$.

Answer 2

Some ideas:

$$|z-(1+i)|=2\iff|z|\left|1-\frac{1+i}z\right|=2\iff|w|=\left|\frac2z\right|=\left|1-\frac{1+i}z\right|$$

If $\;z=x+iy\;$ , then

$$\frac{1+i}z=\frac{(1+i)\overline z}{|z|^2}=\frac{x+y}2+\frac{x-y}2i\implies 1-\frac{1+i}z=\frac{2-(x+y)}2-\frac{x-y}2i$$

But

$$2=|z-(1+i)|=|(x-1)+(y-1)i|\implies (x-1)^2+(y-1)^2=4$$

whereas

$$|w|^2=\frac14\left(4-4x-4y+x^2+2xy+y^2+x^2-2xy+y^2\right)=$$

$$=\frac12\left(x^2+y^2-2x-2y+2\right)=\frac12\left((x-1)^2+(y-1)^2\right)=2\implies$$

$$\color{red}{|w|=\sqrt2}$$

Answer 3

Locus of w is a circle centered at (-(4+2^1/2/14),(4+2^1/2/14)).Replace z with 2/w In the first equation and then square both sides of the equation.Rearrange the terms.