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

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

• Did you sketch your circle? It’s closest to the origin where? And you know that the image of a circle under $z\mapsto2/z$ is a circle or straight line, don’t you? May 16 '14 at 13:03
• Thanks Lubin would you like to explain me the concept of image of circle which you have used above. Thanks May 20 '14 at 14:47
• Sorry for the delay. Your question is well answered by the response below. More generally, you can show for yourself the image of any circle under $z\mapsto (ax + b)/(cz + d )$ where $ad -bc\ne0$ will be a circle or straight line. Same method. May 23 '14 at 21:00

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 $2$.

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

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

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

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

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.