Question

Find the image of |z+1|=1 under the mapping w=1/z
​

Answer 1

Answer:

For the transformation w=1/z, find the image of

(a) the circle |z−2|=1

(b) the circle |z−1|=2

(c) the circle |z−1|=1

For (c), it's simple since z=1/w, so we get |1−w|=|w|, which is the straight line Rez=1/2.

However, for (a) and (b), I'm stuck. I tried to find it using the equation z=x+iy,w=u+iv, then x=uu2+v2,y=−vu2+v2. However, plugging this into the original equation, I get a very complicated equation that I cannot simplify. How may I approach (a) and (b)?

Rewrite w=1/z as z=1/w. I'll do the part (a).

It follows that z−2=1/w−2, and the condition |z−2|=1 implies that

∣∣∣1w−2∣∣∣=1⟹|1−2w|=|w|

(at this point one may refer to Apollonius' theorem to see that this is equation of circle)

Now you may write w=u+iv to see that above is equivalent to

(1−2u)2+4v2=u2+v2⟹3u2+3v2−4u+1=0⟹u2+v2−43u+13=0

and last equation can be written as (u−2/3)2+v2−19=0 which you should recognize as a circle.

You should be able to do part (b) now.