Translate complex relations into loci and combined regions

Read a complex relation as a set

A locus consists of every point satisfying the stated relation. Complex modulus measures Euclidean distance in the Argand plane

1. 1Coordinates Separate the variable and fixed point into real and imaginary coordinates \(\displaystyle z=x+iy,\quad\alpha=a+bi\)
2. 2Displacement Subtract the coordinates \(\displaystyle z-\alpha=(x-a)+i(y-b)\)
3. 3Distance Apply the distance formula \(\displaystyle |z-\alpha|^2=(x-a)^2+(y-b)^2\)

For |z−(1−i)|=2, enter the centre’s real coordinate

Circle

\(\displaystyle |z-(1-i)|=2\)

Enter the imaginary coordinate of the centre

Which points satisfy |z−(1−i)|=2?

Check a distance level before squaring

A negative modulus level gives the empty set. Level zero gives one point. Positive level gives a circle. Squaring |z−a|=−2 would produce a spurious circle because the original right side is negative

What is the locus |z−(2+i)|=−1?

Equal distances give a perpendicular bisector

For distinct foci α and β, |z−α|=|z−β| is their perpendicular bisector. Square the nonnegative distances and cancel x²+y². The line passes through the midpoint and is perpendicular to the focus segment

The foci are −1+2i and 3+2i. Their equal-distance locus is x=k. Enter k.

Coincident foci change the classification

If α=β, the equality |z−α|=|z−β| is true for every z. The result is the whole plane, not a unique line. A viewport never supplies an artificial outer boundary

An argument fixes a direction from a starting point

Use −π<Arg≤π. Arg(z−α)=θ is a ray from α in direction θ, with α excluded because Arg(0) is undefined. For θ=π/4 and α=2−i, write (x,y)=(2,−1)+t(1,1), t>0. Negative t follows the opposite ray

Ray

\(\displaystyle \begin{gathered}(x,y)=(2,-1)+t(1,1)\\t>0\end{gathered}\)

Parameter

\(\displaystyle t=2\)

At the given parameter value, enter the imaginary coordinate

What does the hollow point A mean in this ray sketch?

State principal arguments accurately

A left-pointing horizontal ray has principal argument π under our convention. Writing Arg(z−α)=−π would give no solutions under this exact convention; an arbitrary argument modulo 2π is a different statement. Our generated rays use values in (−π,π]

A sum of distances produces an ellipse

For two distinct foci, a sufficiently large fixed distance sum gives a nondegenerate ellipse

1. 1Focal data The focal separation is twice the focal half-distance \(\displaystyle \begin{gathered}|\alpha-\beta|=2c>0\\|z-\alpha|+|z-\beta|=2A,\quad A>c\end{gathered}\)
2. 2Semiaxes The centre is the midpoint and the major axis follows the foci \(\displaystyle B^2=A^2-c^2>0\)
3. 3Eccentricity Divide the focal half-distance by the major semiaxis \(\displaystyle e=\frac cA,\quad e^2=\frac{c^2}{A^2},\quad0<e<1\)
4. 4Horizontal foci Use the larger denominator along the focal axis; interchange denominators for vertical foci \(\displaystyle \frac{(x-h)^2}{A^2}+\frac{(y-k)^2}{B^2}=1\)

The foci are −2 and 2, and the distance sum is 2A=6. Find A.

Foci

\(\displaystyle \alpha=-2,\quad\beta=2\)

Distance sum

\(\displaystyle |z+2|+|z-2|=6\)

Find the squared minor semiaxis of the ellipse

Separate empty, segment, point and circle cases

The distance sum L cannot be less than the focal separation 2c. If L=2c>0, the locus is the closed focus segment. If both foci coincide, L>0 gives a circle of radius L/2, L=0 gives the common point and L<0 gives no points

For distinct foci −2 and 2, what does |z+2|+|z−2|=4 describe?

An intersection requires every condition

For |z−(1−i)|≤2 and Re(z)<2, first draw the closed disk, then retain the portion left of x=2. The line is excluded. Some points on the circle survive, but only if they also have x<2.

Disk condition

\(\displaystyle |z-(1-i)|\le2\)

Half-plane condition

\(\displaystyle \operatorname{Re}(z)<2\)

Point

\(\displaystyle z=1\)

Does the given point satisfy both conditions? Enter 1 for yes or 0 for no

Disk condition

\(\displaystyle |z-(1-i)|\le2\)

Half-plane condition

\(\displaystyle \operatorname{Re}(z)<2\)

Point

\(\displaystyle z=3-i\)

Does the given point satisfy both conditions? Enter 1 for yes or 0 for no

For |z−(1−i)|≤2 and Re(z)<2, must every point of the disk's circular boundary be in the intersection?

Prove the bisector description, ungraded

For distinct α=a+bi and β=c+di, derive the exact line equation and prove that it passes through the midpoint and is perpendicular to the focus segment. Explain separately what changes if the foci coincide. This proof is ungraded

Model: reversible squaring and a normal vector

Both distances are nonnegative. Squaring and cancelling yields 2(c−a)x+2(d−b)y=c²+d²−a²−b². Substituting ((a+c)/2,(b+d)/2) verifies the midpoint. The normal vector (c−a,d−b) is parallel to the focus segment, so the line is perpendicular to it. Distinct foci make this normal nonzero. For coincident foci the original equality holds everywhere

Derive the ellipse equation, ungraded

With centre at the origin and foci (−c,0),(c,0), assume A>c>0. Prove that the distance sum 2A is equivalent to X²/A²+Y²/(A²−c²)=1. Check the converse rather than merely squaring. This proof is ungraded

Model: control the signs in both directions

Derive the equation without losing the signs of the distances, then verify the converse

1. 1Definitions Use the centred horizontal focal axis and a nondegenerate ellipse \(\displaystyle \begin{gathered}A>c>0\\d_-=\sqrt{(X+c)^2+Y^2}\\d_+=\sqrt{(X-c)^2+Y^2}\end{gathered}\)
2. 2Difference of squares Use the known distance sum to find the difference \(\displaystyle \begin{gathered}d_-+d_+=2A\\(d_--d_+)(2A)=4cX\\d_--d_+=\frac{2cX}{A}\\d_-=A+\frac{cX}{A}\end{gathered}\)
3. 3Square and collect The linear terms cancel and the remaining coefficient is positive \(\displaystyle \begin{gathered}(X+c)^2+Y^2=\left(A+\frac{cX}{A}\right)^2\\X^2\left(1-\frac{c^2}{A^2}\right)+Y^2=A^2-c^2\\\frac{X^2}{A^2}+\frac{Y^2}{A^2-c^2}=1\end{gathered}\)
4. 4Converse signs The ellipse equation bounds the horizontal coordinate \(\displaystyle \begin{gathered}|X|\le A\\A\pm\frac{cX}{A}\ge A-c>0\end{gathered}\)
5. 5Recover the distances Their squared values are the displayed positive squares, so take positive roots and add \(\displaystyle \begin{gathered}d_-=A+\frac{cX}{A},\quad d_+=A-\frac{cX}{A}\\d_-+d_+=2A\end{gathered}\)
6. 6Other positions Translate the centre; interchange the coordinates for a vertical focal axis \(\displaystyle X=x-h,\quad Y=y-k\)

Construct and check a combined region, ungraded

Sketch |z−(1−i)|≤2 with Re(z)<2, using equal unit scales. Mark which boundary pieces survive, and test 1, 3−i and 2−i. Use technology to check the distance inequalities and compare with your sketch. This construction and technology task is ungraded

Model: the inequalities decide each boundary point

Use a solid circle centred at (1,−1), radius 2; shade its interior to the left of the dashed line x=2. Retain only the circular arc with x<2. The line and its intersections with the circle are excluded. z=1 belongs; 3−i fails the half-plane; 2−i is inside the disk but fails the strict line condition. Graph the separate constraints and their logical intersection, checking test points numerically. A plotted pixel is not evidence that a strict boundary is included

Combine a fixed modulus and a direction

For r>0, |z−α|=r and Arg(z−α)=θ determine the single point α+r cis θ. If r=0, the modulus forces z=α but the argument is undefined, so the simultaneous system has no solution

For |z−(1−i)|=2 and Arg(z−(1−i))=π, enter the real coordinate of the simultaneous solution

Use drawings as communication, not hidden assumptions

Show the whole mathematical set: boundary style, excluded points, equal scales and directions beyond the viewport. For ellipses, label the centre, foci, major and minor vertices; do not include the interior in an equality locus. General sketches and proofs remain ungraded