Represent addition, subtraction, multiplication and conjugation geometrically, and calculate distance and midpoint in the complex plane

Learning goal: Represent addition, subtraction, multiplication and conjugation of complex numbers geometrically on the complex plane, use properties of that representation for computation, and calculate the distance between two complex numbers as the magnitude of their difference and the midpoint of a segment as the average of the numbers at its endpoints.

Scope note. N.CN.5 and N.CN.6 are marked (+) in Ohio's Learning Standards for mathematics, which Ohio uses to indicate standards "for students who are planning on taking advanced courses". They are taught here in full, as extension content for students on that path.

Complex numbers as arrows. Every complex number z = a + bi can be pictured as the point (a, b) or as the arrow from the origin to that point. The arrow view makes the operations visible. This lesson draws each operation, then uses the drawing to compute things that would be slow to compute by expanding.

Addition is tip-to-tail. To add z = 2 + i and w = 1 + 3i, place the arrow for w with its tail at the tip of z. The tip of w then lands at 3 + 4i, which is z + w. Equivalently, z and w are two sides of a parallelogram with a vertex at the origin, and z + w is the opposite vertex. This is exactly the coordinate rule "add real parts, add imaginary parts," drawn: 2 + 1 = 3 across and 1 + 3 = 4 up. The diagram for this lesson shows this parallelogram.

Subtraction is the arrow between the points. z − w = (2 + i) − (1 + 3i) = 1 − 2i. Geometrically, z − w is the arrow that starts at w and ends at z, moved so that its tail is at the origin. To get from the point 1 + 3i to the point 2 + i you move 1 right and 2 down, which is 1 − 2i. This picture is why the distance formula below works: the length of z − w is the length of the segment joining the two points.

Conjugation is reflection across the real axis. The conjugate of z = 2 + i is z̄ = 2 − i: same real part, opposite imaginary part. On the plane that is the mirror image of the point across the horizontal axis. A real number lies on the mirror line and is its own reflection, matching the earlier fact that a real number is its own conjugate. Because reflection preserves distance from the origin, |z̄| = |z|.

Multiplication: magnitudes multiply and arguments add. This is the fact that makes polar form powerful. If z has magnitude r₁ and argument θ₁, and w has magnitude r₂ and argument θ₂, then zw has magnitude r₁r₂ and argument θ₁ + θ₂. (The reason is the angle-sum identities for cosine and sine applied to r₁(cos θ₁ + i sin θ₁)·r₂(cos θ₂ + i sin θ₂); expanding and using i² = −1 produces r₁r₂[cos(θ₁ + θ₂) + i sin(θ₁ + θ₂)].) So multiplying by a positive real number stretches the arrow without turning it, multiplying by −1 turns it through 180°, and multiplying by i turns it through 90° counterclockwise without changing its length, because i has magnitude 1 and argument 90°. Check with numbers: i(2 + i) = 2i + i² = −1 + 2i, and the arrow from the origin to (2, 1) rotated a quarter turn counterclockwise does end at (−1, 2). Check the general rule with z = 2 + i and w = 1 + 3i: zw = 2 + 6i + i + 3i² = −1 + 7i. Magnitudes: |z| = √5, |w| = √10, and |zw| = √(1 + 49) = √50 = √5·√10, as the rule says. Arguments: z has argument about 26.6°, w about 71.6°, and −1 + 7i about 98.1°, which is the sum.

A worked model inside the lesson: (−1 + √3 i)³ = 8. Expanding by hand is possible: (−1 + √3 i)² = 1 − 2√3 i + 3i² = −2 − 2√3 i, and then (−2 − 2√3 i)(−1 + √3 i) = 2 − 2√3 i + 2√3 i − 6i² = 2 + 6 = 8. But the geometric representation gives the answer without expansion. The number −1 + √3 i has magnitude √(1 + 3) = 2 and argument 120°, as found in the previous lesson. Cubing multiplies the magnitude by itself three times, 2·2·2 = 8, and adds the argument three times, 120° + 120° + 120° = 360°. An argument of 360° is the same direction as 0°, the positive real axis. So the cube is the real number 8. Both methods agree, and the geometric one explains why the answer is real: three turns of 120° make one full turn. The same reasoning shows (1 + i)⁸ = 16: magnitude √2 to the eighth power is (√2)⁸ = 2⁴ = 16, and 8·45° = 360°, so the result is the positive real number 16. And (√3 + i)⁶ = −64: magnitude 2⁶ = 64 and argument 6·30° = 180°, the negative real axis.

Distance is the magnitude of the difference. The distance between two complex numbers z and w is |z − w|, the length of the arrow from w to z. For z = 2 + 5i and w = −4 − 3i: z − w = 6 + 8i and |6 + 8i| = √(36 + 64) = 10. This is the ordinary distance formula between the points (2, 5) and (−4, −3), written in complex notation; the horizontal change 6 and the vertical change 8 are the real and imaginary parts of the difference. Because |z − w| = |w − z|, the order does not matter, and the magnitude is never negative, as a distance must be.

Midpoint is the average of the endpoints. The midpoint of the segment joining z and w is (z + w)/2. For z = 2 + 5i and w = −4 − 3i: (2 + 5i + (−4 − 3i))/2 = (−2 + 2i)/2 = −1 + i. Check: the distance from z to the midpoint is |(2 + 5i) − (−1 + i)| = |3 + 4i| = 5, and from the midpoint to w is |(−1 + i) − (−4 − 3i)| = |3 + 4i| = 5. Each half is 5, which is half of the full distance 10, so −1 + i is genuinely the midpoint. Averaging the two numbers averages the real parts and the imaginary parts separately, which is the coordinate midpoint formula.

A misconception about distance. Students sometimes compute |z| − |w| for the distance between z and w. That is wrong: |z| and |w| are distances from the origin, and their difference is not the distance between the two points unless the points lie on the same ray from the origin, and even then only as the absolute value ||z| − |w||. For z = 3 and w = 3i, |z| − |w| = 0, but the distance |3 − 3i| = 3√2 is clearly not zero. Subtract first, then take the magnitude.

Worked geometric sum and difference: For z = 2 + i and w = 1 + 3i, compute z + w and z − w and describe each on the plane. z + w = 3 + 4i: the fourth vertex of the parallelogram with sides z and w. z − w = 1 − 2i: the arrow from the point w to the point z, since 1 + 3i moved 1 right and 2 down is 2 + i. Also w − z = −1 + 2i, the same arrow reversed. The lengths |z − w| and |w − z| are both √5.

Worked power by rotation: Compute (1 + i)⁸ using magnitude and argument, then confirm by repeated squaring. 1 + i has magnitude √2 and argument 45°. The eighth power has magnitude (√2)⁸ = 16 and argument 8·45° = 360°, which is the positive real direction, so (1 + i)⁸ = 16. Confirm: (1 + i)² = 2i; (2i)² = 4i² = −4; (−4)² = 16. Both routes give 16.

Worked distance and midpoint: For z = 1 + 7i and w = 4 + 3i, find the distance between them and the midpoint. Difference: z − w = −3 + 4i, so the distance is √(9 + 16) = 5. Midpoint: (z + w)/2 = (5 + 10i)/2 = 2.5 + 5i. Check: |z − midpoint| = |(1 + 7i) − (2.5 + 5i)| = |−1.5 + 2i| = √(2.25 + 4) = √6.25 = 2.5, which is half of 5.

Worked distance from a diameter: The points −2 + i and 6 − 5i are the ends of a diameter of a circle. Find the center and radius. Center is the midpoint: (4 − 4i)/2 = 2 − 2i. Diameter is the distance: |(−2 + i) − (6 − 5i)| = |−8 + 6i| = √(64 + 36) = 10, so the radius is 5. Check: |(−2 + i) − (2 − 2i)| = |−4 + 3i| = 5, as required for a point on the circle.

Four-operation diagram: On graph paper, plot z = 2 + i and w = 1 + 3i as arrows from the origin. Draw and label z + w as the parallelogram vertex, z − w as the arrow from w to z, z̄ as the reflection of z across the real axis, and iz as z rotated 90° counterclockwise. Beside each, write the algebraic computation (3 + 4i, 1 − 2i, 2 − i, −1 + 2i) and confirm that the drawn point matches. Then compute the magnitude and argument of z, w and zw, and show that the magnitudes multiply and the arguments add.

Independent performance: Part one: choose a complex number of magnitude 2 whose argument is a multiple of 30° but not a multiple of 90°, write it in rectangular form, and compute its sixth power two ways: by the magnitude-and-argument rule and by expanding (square, then cube the square). Both results must agree; explain what the argument of the answer tells you about whether it is real. Part two: choose two complex numbers in different quadrants, compute the distance between them as the magnitude of the difference and the midpoint as the average, and verify that the midpoint is the same distance from each endpoint. Success criteria: the two methods in part one agree exactly, every distance is computed by subtracting before taking the magnitude, and the midpoint check shows both half-distances equal.