Geometrically describe the following subsets of $\CC$: $\abs{z-1} = 1$ $\abs{z-1} = 2\abs{z-2}$ $1/z = \bar z$ $\Re(z) = 3$ $\Im(z) = a$ with $a\in \RR$.

A circle of radius 1 about \(z=1\).

A circle, using that Apollonius circles are characterized as the locus of distances whose ratios to some fixed points \(A, B\) are constant. To actually compute this: \[ \abs{z-1}^2 &= 4\abs{z-2}^2 \\ \implies (x-1)^2 + y^2 - 4\qty{(x-2)^2 + y^2 } &=0 \\ -3x^2 + 14x - 3y^2 - 15 &= 0 && \star\\ \implies x^2 - {14\over 3}x + y^2 + 5 &= 0 \\ \implies (x- {14\over 2\cdot 3})^2 - {14\over 2\cdot 3}^2 + y^2 + 5 &= 0 \\ \implies (x-{14\over 6})^2 + y^2 = \qty{2\over 3}^2 ,\] which is a circle of radius \(2/3\) with center \(\qty{{14\over 6}, 0}\). To avoid the calculation, use \[ Ax^2 + Bxy + Cy + \cdots = 0,\quad A=1, B=0, C=1 \implies \Delta \da B^2 - 4AC < 0 ,\] which is an ellipse, and since \(A=C\) it is in fact a circle.

\(S^1\), using that \({1\over z} = {\bar{z} \over z\bar{z}} = {\bar z\over \abs{z}^2}\) and if this equals \(\bar{z}\), then \(\abs{z}^2=1\). Alternatively, \(1 = \bar{z}z = \abs{z}^2\).

Vertical line through \(z=3\).

Horizontal line through \(z=ia\).

Region to the right of the vertical line through \(z=a\).

Exterior of a circle: same calculation is (2), replacing \(=0\) with \(<0\). Note that the line marked \(\star\) involves dividing by a negative, so this flips the sign, and we get \(\cdots > \qty{2\over 3}^2\) at the end.