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Pre-Calculus Complex numbers (advanced)

Distance and midpoint in the complex plane

20 practice questions 0 video lessons Theory + worked examples

Distance and Midpoint in the Complex Plane

Common Core Pre-Calculus • Standard N-CN.6 • Complex Numbers

Distance and Midpoint in the Complex Plane is a topic in Complex Numbers in the Common Core State Standards. It is aligned to Standard N-CN.6, which requires students to calculate the distance and midpoint between numbers in the complex plane.

Distance between two complex numbers is \(|z_1-z_2|\) and the midpoint is \(\dfrac{z_1+z_2}{2}\), just as in the coordinate plane.

Common Core Pre-Calculus › Complex Numbers › Distance and Midpoint in the Complex Plane  —  Standard N-CN.6

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Theory

Because the complex plane is just the coordinate plane with axes relabeled, distance and midpoint work exactly as in coordinate geometry:

  • Distance between \(z_1\) and \(z_2\) is \(|z_1-z_2|\) — the modulus of their difference.
  • Midpoint is \(\dfrac{z_1+z_2}{2}\) — the average.
\(|z-w|\) is a distance. So \(|z-w|=r\) describes a circle of radius \(r\) centered at \(w\).
Distance between two complex numbers The distance between two complex numbers equals the modulus of their difference. Re Im z₁ z₂
The distance between two complex numbers is \(|z_1-z_2|\).
Distance & midpoint Distance & midpoint Distance & midpoint distance = |z₁ − z₂| midpoint = (z₁ + z₂)/2
Distance and midpoint formulas.

Distance and midpoint:

\[d(z_1,z_2)=|z_1-z_2|=\sqrt{(a_1-a_2)^2+(b_1-b_2)^2},\qquad M=\dfrac{z_1+z_2}{2}\]
distance is the modulus of the difference; midpoint is the average of the two numbers
\(|z-w|=r\) is a circle; \(|z-w|<r\) its interior.

How to find distance and midpoint

  1. Distance: subtract the numbers, then take the modulus.
  2. Midpoint: add the numbers and divide by 2.
  3. Loci: read \(|z-w|=r\) as a circle centered at \(w\).
Example 1 — Distance between two points
Find the distance between \(z_1=1+i\) and \(z_2=4+5i\).
Solution

Distance is the modulus of the difference.

\(z_2-z_1\)\(=\)\((4-1)+(5-1)i=3+4i\)
\(|z_2-z_1|\)\(=\)\(\sqrt{3^2+4^2}=5\)
distance is 5
Example 2 — Midpoint
Find the midpoint of \(z_1=2+3i\) and \(z_2=6-i\).
Solution

Average the two numbers.

\(\dfrac{z_1+z_2}{2}\)\(=\)\(\dfrac{(2+6)+(3-1)i}{2}\)
\(=\)\(\dfrac{8+2i}{2}=4+i\)
midpoint is 4 plus i
Example 3 — Distance with a negative part
Find the distance between \(z_1=-2+i\) and \(z_2=1-3i\).
Solution

Subtract, then take the modulus.

\(z_2-z_1\)\(=\)\((1+2)+(-3-1)i=3-4i\)
\(|z_2-z_1|\)\(=\)\(\sqrt{9+16}=5\)
distance is 5
Example 4 — A circle in the complex plane
Describe all \(z\) with \(|z-(2+i)|=3\).
Solution

This is the set of points a distance 3 from \(2+i\) — a circle.

\(|z-(2+i)|\)\(=\)\(3\)

A circle of radius 3 centered at \(2+i\).

a circle of radius 3 centered at 2 plus i

Common pitfalls

Distance is the modulus of the difference, not the difference of moduli.
Subtract carefully with negative parts. \((-2)-1=-3\), etc.
Midpoint stays complex. Keep the real and imaginary parts separate when averaging.

Frequently asked questions

How do you find the distance between two complex numbers?

Take the modulus of their difference: \(|z_1-z_2|\).

How do you find the midpoint of two complex numbers?

Average them: \(\dfrac{z_1+z_2}{2}\), keeping real and imaginary parts separate.

What does |z - w| = r represent?

A circle of radius \(r\) centered at \(w\) — all points a fixed distance from \(w\).

Is distance in the complex plane the same as in coordinate geometry?

Yes. The complex plane is the coordinate plane, so the distance and midpoint formulas are identical.

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