AMC 10 · 2004 · #16

Grade 11 algebra
Archetype Coordinate Plane Figure Computation └ Sub-archetype Line Intersection Vertices then Area
complex-numberscircle-equationquadratic-equations convert-to-algebraidentify-subproblems ↑ Prerequisites: complex-numbers
📏 Medium solution 💡 3 insights
Problem
A rule conjugates a complex number and then turns it a quarter turn. Count the numbers this rule leaves in place that also sit at distance 5 from the origin.

Pick an answer.

(A)
0
(B)
1
(C)
2
(D)
4
(E)
8
How to solve
Strategy Introduce a Variable

The usual reflex for counting solutions is to read off the degree of a polynomial, but z is not a polynomial in z, so degree tells nothing here. The way past that is to stop treating z as one object: name its real and imaginary parts, and the single complex equation becomes two real equations that can be handled with ordinary algebra. Once both conditions are written in the coordinates a and b, one of them describes a line and the other a circle, and the count can be checked twice - once by picture, once by solving.

1STEP 1

Split z into two real parts

Splitting into two real parts shows the rule simply swaps their roles.

f(a+bi)=i(a-bi)=ai-bi²=b+ai
2STEP 2

Match real and imaginary parts

Matching parts gives one condition, so the fixed numbers form a line.

b+ai=a+bi⇔ a=b
3STEP 3

Put the fixed points on the circle

That line runs through the centre, so it cuts the circle twice.

a²+b²=25 (circle of radius 5), b=a (line through the centre)
4STEP 4

Solve for the actual numbers

Solving confirms exactly two numbers, one on each side.

2a²=25 → a=±5√(2)/2, z=±5√(2)/2(1+i)
5STEP 5

Check both, then count

Both pass the check and nothing else can, so the count is 2, choice (C).

|5√(2)/2(1+i)|=5√(2)/2·√(2)=5, i·5√(2)/2(1-i)=5√(2)/2(1+i)
Answer
2
The shape of the two conditions caps the count immediately: one condition pins z to a line, the other to a circle, and a line meets a circle in at most two points, so 4 and 8 are impossible before any computation. Zero is impossible because this particular line runs through the centre of the circle and therefore has to cut it. One would require tangency, which a line through the centre can never have. That leaves 2, and the explicit numbers ±5√(2)/2(1+i) confirm it. One subtlety is worth naming: showing that solutions must lie on the line only bounds the count above by 2. The count is exactly 2 because the implication also runs backwards - every point of that line really is fixed by f.
💡Key takeaway

Split a complex number into its two real coordinates and a strange equation turns into ordinary algebra - here into a line, and a line through the centre of a circle always cuts it exactly twice.

  • Split z into two real parts
  • Match real and imaginary parts
  • Put the fixed points on the circle
  • Solve for the actual numbers
  • Check both, then count

More like this

Same archetype — closest grade level first.