Describe geometrically the sets of points \(z\) in the complex plane defined by the following relations:
\(|z-z_1|=|z-z_2|\), where \(z_1,z_2\in\mathbb C\).
\(1/z=\overline z\).
\(\operatorname{Re}(z)=3\).
\(\operatorname{Re}(z)>c\) (respectively, \(\operatorname{Re}(z)\ge c\)), where \(c\in\mathbb R\).
\(\operatorname{Re}(az+b)>0\), where \(a,b\in\mathbb C\).
\(|z|=\operatorname{Re}(z)+1\).
\(\operatorname{Im}(z)=c\), where \(c\in\mathbb R\).
Solution
1. For (a), if \(z_1\ne z_2\), the locus is the perpendicular bisector of the line segment joining \(z_1\) and \(z_2\); if \(z_1=z_2\), the locus is all of \(\mathbb C\).
Squaring both sides gives \[ |z-z_1|^2=|z-z_2|^2 \iff 2\operatorname{Re}\!\bigl(z(\overline{z_2}-\overline{z_1})\bigr)=|z_2|^2-|z_1|^2. \] If \(z_1\ne z_2\), this is the equation of a real affine line. It consists exactly of the points equidistant from \(z_1\) and \(z_2\), hence is their perpendicular bisector. If \(z_1=z_2\), the original equality is tautological for every \(z\).
2. For (b), the locus is the unit circle \(\{z\in\mathbb C:|z|=1\}\).
The equation is defined only for \(z\ne0\). Multiplying by \(z\) gives \[ 1=z\overline z=|z|^2, \] which is equivalent to \(|z|=1\). Conversely every point of the unit circle satisfies \(1/z=\overline z\).
3. For (c), the locus is the vertical line \(\{x+iy:x=3\}\).
Writing \(z=x+iy\), one has \(\operatorname{Re}(z)=x\), so the equation is exactly \(x=3\).
4. For (d), the loci are respectively the open and closed right half-planes bounded by the vertical line \(x=c\).
Again writing \(z=x+iy\), the inequalities become \(x>c\) and \(x\ge c\).
5. For (e), if \(a\ne0\) the locus is an open half-plane; if \(a=0\), it is all of \(\mathbb C\) when \(\operatorname{Re}(b)>0\) and is empty when \(\operatorname{Re}(b)\le0\).
Write \(a=\alpha+i\beta\), \(b=u+iv\), and \(z=x+iy\). Then \[ \operatorname{Re}(az+b)=\alpha x-\beta y+u. \] If \(a\ne0\), then \((\alpha,-\beta)\ne(0,0)\), so \[ \alpha x-\beta y+u>0 \] defines one of the two open half-planes bounded by the line \(\alpha x-\beta y+u=0\). If \(a=0\), the inequality is the constant condition \(u>0\), giving the two stated possibilities.
6. For (f), the locus is the parabola\[ y^2=2x+1, \]which opens to the right and has vertex \((-1/2,0)\).
Write \(z=x+iy\). The equation \(|z|=x+1\) implies \[ x^2+y^2=(x+1)^2, \] hence \(y^2=2x+1\). Conversely, on this parabola, \[ x+1=\frac{y^2+1}{2}>0, \] so squaring introduced no extraneous points and \(|z|=x+1\) indeed holds.
7. For (g), the locus is the horizontal line \(\{x+iy:y=c\}\).
For \(z=x+iy\), one has \(\operatorname{Im}(z)=y\), so the equation is exactly \(y=c\).