# Find the image under the Mobius transformation $f(z) = 1/z$ of $|z-3i|=1$ [closed]

I am given $$|z-3i|=1$$ which is a circle with radius $$1$$ and centre $$(0,3i)$$ on the complex plane. I want to find the image (to sketch it) under the transformation $$1/z$$ WITHOUT taking points and seeing how they change.

## closed as off-topic by abiessu, Saad, Cesareo, Lord Shark the Unknown, Paul FrostMar 31 at 9:02

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• Do you have a preferred method for accomplishing this task that does not involve taking of points? – abiessu Mar 30 at 23:05
• It will be some circle lying completely within the circle of radius $1/2$ about the origin. It will go through the points $-i/4$ and $-i/2$. – MPW Mar 30 at 23:10
• @abiessu If I knew another method I wouldn't be asking here. I want someone to give me another method. – Mohamad Moustafa Mar 30 at 23:27
• @MPW I know that, but how did you get it? I dont want to take points and see how they line up – Mohamad Moustafa Mar 30 at 23:27
• In fact the segment joining those two points will be a diameter of the circle, so it is now completely determined. How did I I get this? It’s just that the family of lines/circles is preserved by Moebius transforms, and there are certain symmetries that can be observed (original circle is symmetric wrt y-axis, and this map preserves that, etc) – MPW Mar 30 at 23:33

Write: $$z = \cos \theta + i (3 + \sin \theta)$$, and then take the inverse, extract the real and imaginary parts, which depend upon $$\theta$$.
The original solution is a function of $$\theta$$, and the transformed is centered on $$(0,-3/8)$$ or radius $$1/8$$.