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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.

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

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    $\begingroup$ Do you have a preferred method for accomplishing this task that does not involve taking of points? $\endgroup$ – abiessu Mar 30 at 23:05
  • $\begingroup$ 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 $-i/2$. $\endgroup$ – MPW Mar 30 at 23:10
  • $\begingroup$ @abiessu If I knew another method I wouldn't be asking here. I want someone to give me another method. $\endgroup$ – Mohamad Moustafa Mar 30 at 23:27
  • $\begingroup$ @MPW I know that, but how did you get it? I dont want to take points and see how they line up $\endgroup$ – Mohamad Moustafa Mar 30 at 23:27
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    $\begingroup$ 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) $\endgroup$ – MPW Mar 30 at 23:33
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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 $1/8$.

enter image description here

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