Image of |z-1| ≤ 1 under w = 1/z in complex plane

[simick1712]

Hi everyone - I'm sure there's somebody here who can help with such a trivial question.

It's not a homework question before that's assumed - it's from a past exam paper, which I'm using for revision, sans answers.

It asks to describe, in the complex plane, the image of:

$$|z - 1| \leq 1$$

under the transformation

$$w = \frac{1}{z}$$

Now, I found the answer to be:

$$Re(w) \leq \frac{1}{2}$$

Is this correct?

Thanks for any help,

Simon.

 

[HallsofIvy]

Are you sure about the direction of that inequality? z= 1 itself satisfies |z-1|=0< 1 and z= 1 is mapped into w= 1 which has real part 1> 1/2.

 

[simick1712]

Oops - no, I'm not sure of that at all - what I actually meant (of course!) was

$$Re(w) \geq \frac{1}{2}$$

 

[HallsofIvy]

Yep. If z= x+ iy, then $w= \frac{1}{z}= \frac{1}{x+iy}\frac{x-iy}{x-iy}= \frac{x-iy}{x^2+y^2}$ which has real part $\frac{x}{x^2+ y^2}$.

If z satisfies $|z-1|\le 1$ then $|z-1|= \sqrt{(x-1)^2+ y^2}\le 1$ so $(x-1)^2+ y^2= x^2+ y^2- 2x+ 1\le 1$.
That is, $x^2+ y^2\le 2x$ and therefore $\frac{x}{x^2+ y^2}\ge \frac{1}{2}$.

 

[simick1712]

Excellent - thanks very much for your help.

Simon.