Why does the inversion of the unit circle centered at $-i$ give the horizontal line $y = \frac{1}{2}$?

Question

Inverting the circle centered at $-i$ with radius $1$, gives the horizontal line $y = \frac{1}{2}$, but why does the line have to be horizontal? Why not another straight line passing through the imaginary axis at $\frac{i}{2}$?

I understand that inverting the point on this circle that is furthest from the origin, namely $-2i$, gives the point on the image closest to the origin in the $w$-plane, namely $-\frac{1}{2i} = \frac{i}{2}$. But the notes that I am reading conclude at this point that the line must be horizontal.

Answer 1

Inversion preserves angles. The imaginary axis is preserved by inversion, and the imaginary axis intersects the circle perpendicularly at $-2i$. Thus, the inverse of the circle must intersect the imaginary axis perpendicularly at $\frac i2$. Thus, the line must be perpendicular to the imaginary axis; that is, horizontal.

Answer 2

You can also compute the inversion explicitly. Suppose $z$ is a point on the desired circle; then $|z+i|=1$, whence $z+i=e^{i\theta}$ for some $\theta$, and therefore $z=e^{i\theta}-i$.

Now the inversion of this is $z=\frac{1}{e^{i\theta}-i}=\frac{1}{\cos\theta + i(\sin\theta-1)}=\frac{\cos\theta - i(\sin\theta-1)}{\cos^2\theta+(\sin\theta-1)^2}=\frac{\cos\theta - i(\sin\theta-1)}{2-2\sin\theta}=\frac{\cos\theta}{2-2\sin\theta}-i\frac{\sin\theta-1}{2-2\sin\theta}=\frac{\cos\theta}{2-2\sin\theta}+\frac{i}{2}$

and you can see that as $\theta$ varies, the imaginary part is always $i/2$.