FP2 4.1 - Further loci and regions

You already know

For $z=x+\mathrm{i}y$, $|z-a|$ is the distance from the point $a$. Equal distances from two distinct fixed points give a perpendicular bisector. A modulus inequality selects a side of its boundary; simultaneous conditions mean intersection. Principal arguments lie in $(-\pi,\pi]$, and $\arg0$ is undefined.

A distance-ratio equation

For fixed distinct $a,b$ and $k\geq0$, write $z=x+\mathrm{i}y$. Then

is equivalent to

For $k>0,\ k\ne1$, this is a circle; for $k=1$, the perpendicular bisector. For $k=0$, it is the single point $a$. Squaring is valid because both sides are nonnegative.

Not in the formula booklet: learn it.

Worked example 1: unequal distances give a circle

Question. Find the locus $|z+1|=2|z-2|$, giving its centre and radius.

Plan. Square both distances, collect terms and complete the square.

Why this step? The completed-square form locates the centre directly and makes the radius unambiguous.

Answer. The circle has centre $3$, or coordinates $(3,0)$, and radius $2$.

Check. At $z=1$, the distances are $2$ and $1$, so the equality holds. This is the circle's leftmost point.

Worked example 2: the circle case becomes a line

Question. Find the locus $|z+1|=|z-2|$.

Plan. Use the same expansion and observe the cancellation of the quadratic terms.

Why this step? Equal coefficients make the quadratic terms cancel, giving the perpendicular bisector rather than a circle with a finite radius.

Answer. The complete vertical line $x=\tfrac12$.

Check. It passes through the midpoint of $(-1,0)$ and $(2,0)$, and is perpendicular to their horizontal joining line.

Watch out

The wrong squared right-hand side is $2((x-2)^2+y^2)$. The factor $2$ must also be squared, giving $4((x-2)^2+y^2)$. Otherwise the resulting circle is wrong.

Region and strip conditions

For nonnegative distance expressions and $k\geq0$, squaring preserves an inequality:

A real-part restriction

is the closed vertical strip $p\leq x\leq q$, when $p\leq q$. Strict comparisons exclude the corresponding boundaries. Combine conditions by intersection.

Not in the formula booklet: learn it.

Worked example 3: the exterior within a strip

Question. Describe the region

Find where its circular and vertical boundaries meet.

Plan. Simplify the ratio comparison, intersect with the strip, then substitute the active vertical edge into the circle.

Why this step? The exterior, rather than the interior, satisfies the original ratio comparison.

Answer. Keep the exterior of the circle centred at $(3,0)$, radius $2$, within $0\leq x\leq2$. All admissible edges are included. The boundary meetings are $2\pm\mathrm{i}\sqrt3$. The line $x=0$ does not meet the circle.

Check. At $z=0$, $1\leq4$, so the left edge belongs. At $x=0$, the circle would require $9+y^2=4$, which has no real solution.

Worked example 4: the circular cap within the same strip

Question. Describe

including all active boundaries.

Plan. Reverse the previous circle inequality, then identify which strip edges actually meet the surviving set.

Why this step? The circle starts at $x=1$, so $x=0$ is not an edge of this intersection.

Answer. The closed left cap of the disc cut by $x=2$, bounded by the relevant circular arc and the chord from $2-\mathrm{i}\sqrt3$ to $2+\mathrm{i}\sqrt3$. Its leftmost point is $1$.

Check. $z=2$ satisfies the original inequality $3\geq0$, so the chord's midpoint is included.

Watch out

The line $(x-3)^2+y^2\leq4$ is the wrong conclusion for $|z+1|\leq2|z-2|$. Dividing the simplified inequality by $-3$ reverses it, so the correct side is $(x-3)^2+y^2\geq4$.