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Lesson 2 of 3

Complex Numbers and Quadratic Equations · Lesson 2 of 3

Complex Numbers on the Argand Plane

“See complex numbers as points and connect their distance and reflection properties to algebraic calculations.”

Learning Objectives

• Represent a + ib as the point (a, b) and read complex numbers from a diagram. • Identify the real and imaginary axes, including real, purely imaginary and zero values. • Explain modulus as distance from the origin using the Pythagorean theorem. • Interpret conjugation as reflection and distinguish it from taking a negative or a reciprocal. • Simplify a quotient of products before finding its conjugate. • Prove a modulus relationship by both component algebra and the product–quotient properties.

A pair of components gives a point

A complex number contains two real components, just as a coordinate point has two real coordinates. This lets us represent the entire complex-number system in a plane. The picture does not replace the algebra; it provides another way to understand the same number.

For z = a + ib, use the point P(a, b). The real part a is the horizontal coordinate and the imaginary part b is the vertical coordinate. Conversely, a point (a, b) determines exactly one complex number a + ib. This correspondence works in all four quadrants and on both axes.

Definition
Argand plane

The Argand plane, also called the complex plane, is the coordinate plane in which the point (a, b) represents the complex number a + ib.

Point–number correspondenceLaTeX
The horizontal coordinate is real part a; the vertical coordinate is imaginary part b. The point and the number are two representations of the same pair of real components.

The horizontal axis is called the real axis because all its points have imaginary part zero. The vertical axis is called the imaginary axis because all its points have real part zero. The origin represents 0 + 0i. An axis name does not change the fact that the coordinate values a and b themselves are real numbers.

Complex numberPointLocation
2 + 4i(2, 4)Quadrant I
−2 + 3i(−2, 3)Quadrant II
i(0, 1)Imaginary axis
2(2, 0)Real axis
−5 − 2i(−5, −2)Quadrant III
1 − 2i(1, −2)Quadrant IV
ReImO-6-5-4-3-2-1123456-3-2-112345A: 2 + 4iB: −2 + 3iC: iD: 2E: −5 − 2iF: 1 − 2iEach point keeps real part horizontal and imaginary part vertical.
Six complex numbers in the Argand plane— Compare each label with its horizontal and vertical coordinates. Real numbers and purely imaginary numbers lie on different axes.
Example — Plotting and reading a complex number

Problem
Locate −2 + 3i. Then write the complex number represented by (4, −1).

  1. 1.For −2 + 3i, the real part −2 sends you two units left from the origin; the imaginary part 3 sends you three units up.
  2. 2.Its point is (−2, 3), in quadrant II. The number is not 3 − 2i: exchanging the components gives a different point.
  3. 3.At (4, −1), the horizontal coordinate is 4 and vertical coordinate is −1. The corresponding number is 4 − i.
  4. 4.A negative vertical coordinate gives a negative imaginary component. The same sign rule works whether you begin with a number or with its point.
The imaginary coordinate is b, not ib

The point for 2 + 4i is (2, 4), not (2, 4i). Coordinate positions are measured by real numbers. The factor i belongs to the algebraic representation of the complex number.

Modulus is the distance from the origin

The two components locate a point, while modulus describes how far that point is from the origin. A right triangle connects these two descriptions. This geometry explains the square root in |a + ib| = √(a² + b²) and why modulus cannot be negative.

Drop a perpendicular from P(a, b) to the real axis. The triangle has horizontal length |a| and vertical length |b|, since geometric side lengths are nonnegative even when coordinates are negative. By the Pythagorean theorem, OP² = |a|² + |b|² = a² + b². Taking the nonnegative root gives OP = √(a² + b²) = |z|.

Distance represented by modulusLaTeX
O is the origin and P is the point (a, b). The distance is zero only at O.
ReImO-2-112345-1123456P(3, 4)3 units4 units|z| = 5z = 3 + 4i3² + 4² = 5²Distance OP = 5
Modulus from a right triangle— The horizontal and vertical lengths determine the sloping distance OP. The modulus is the distance, not either coordinate alone.
Example — Distance for a point in another quadrant

Problem
Find the modulus of −3 − 4i and interpret it geometrically.

  1. 1.The point is (−3, −4), three units left and four units down from the origin.
  2. 2.Its horizontal and vertical side lengths are 3 and 4. Thus the distance is √(3² + 4²) = √25 = 5.
  3. 3.The algebraic calculation is identical: |−3 − 4i| = √[(−3)² + (−4)²] = 5.
  4. 4.The point differs from (3, 4), but both have the same distance from the origin. Negative coordinates do not produce a negative distance.

All numbers with a given modulus r > 0 lie on the circle of radius r centred at the origin. Indeed, their coordinates satisfy a² + b² = r², the usual circle equation. This observation gives a visual interpretation of an algebraic modulus condition. Modulus alone does not specify a unique complex number because many points can have the same distance.

Example — Several numbers with the same modulus

Problem
Give four different complex numbers of modulus 2 and locate them.

  1. 1.Choose 2, −2, 2i and −2i. Their coordinate points are (2, 0), (−2, 0), (0, 2) and (0, −2).
  2. 2.For each number, a² + b² = 4, so its modulus is √4 = 2.
  3. 3.The points are the four axis intersections of the radius-2 circle. Other points on that circle also give modulus 2.
  4. 4.Thus an equality of moduli does not imply equality of complex numbers: it specifies distance rather than both coordinates.

Conjugation reflects a point across the real axis

Conjugation changes a + ib into a − ib. In coordinate form, it preserves the horizontal coordinate and reverses the vertical one. This is exactly reflection across the real axis, so the algebraic sign change has a precise geometric meaning.

If P(a, b) represents z, then Q(a, −b) represents z̄. Both have the same distance from the real axis and the same distance from the origin. Reflection twice returns the starting point, matching the algebraic fact that conjugating z̄ gives z.

ReImO-5-4-3-2-112345-4-3-2-11234P: 3 + 2iQ: 3 − 2iSame real part; opposite imaginary parts; equal modulus.
Conjugates as mirror images— P and Q lie equally far above and below the real axis. Their horizontal position stays fixed.
Example — Reading a conjugate from coordinates

Problem
For z = −2 + 3i, find z̄ and −z, and compare their points.

  1. 1.The point for z is (−2, 3). Reflecting across the real axis gives (−2, −3), so z̄ = −2 − 3i.
  2. 2.Taking the negative changes both components: −z = 2 − 3i, whose point is (2, −3).
  3. 3.These are different points because conjugation preserved the real part −2, while negation reversed it to 2.
  4. 4.All three numbers have modulus √(4 + 9) = √13. Equal distance does not erase the distinction between the operations.

A real number lies on the reflection axis, so its conjugate equals itself. A nonzero purely imaginary number lies on the vertical axis, so its conjugate is also its negative. These special cases explain why the operations can occasionally give the same result even though their general definitions are different.

A conjugate, negative and reciprocal are different

For z = 3 + 2i, z̄ = 3 − 2i, −z = −3 − 2i, and 1/z = (3 − 2i)/13. A reciprocal uses the conjugate divided by |z|². Conjugation alone gives a reciprocal only when |z| = 1.

A conjugate of a compound expression

When a number is given as a quotient of products, first identify the complete value or use conjugate-arithmetic properties consistently. The task may require multiplication, rationalisation and one final sign change. Keeping the stages separate avoids conjugating only part of the expression.

Example — Products, a quotient, then a conjugate

Problem
Find the conjugate of [(3 − 2i)(2 + 3i)]/[(1 + 2i)(2 − i)].

  1. 1.Expand the numerator: 6 + 9i − 4i − 6i² = 12 + 5i.
  2. 2.Expand the denominator: 2 − i + 4i − 2i² = 4 + 3i. This is nonzero.
  3. 3.Rationalise (12 + 5i)/(4 + 3i) by multiplying by (4 − 3i)/(4 − 3i). The denominator becomes 4² + 3² = 25.
  4. 4.The numerator is 48 − 36i + 20i − 15i² = 63 − 16i. Thus the original value is 63/25 − (16/25)i.
  5. 5.Its conjugate is 63/25 + (16/25)i. In the Argand plane, the final step reflects the point across the real axis.

You could instead conjugate every factor before performing the arithmetic: the result is [(3 + 2i)(2 − 3i)]/[(1 − 2i)(2 + i)]. The quotient-conjugate property guarantees agreement with the method above. Whichever route you choose, the bar applies to the complete expression and the denominator must remain nonzero.

A quotient whose modulus is one

A number and its conjugate have equal modulus. Dividing one by the other, when they are nonzero, therefore gives a quotient of modulus one. In coordinates this means that its point lies on the circle of radius one centred at the origin.

Example — A component proof of a unit-modulus relationship

Problem
If x + iy = (a + ib)/(a − ib), with real a and b not both zero, prove x² + y² = 1.

  1. 1.The denominator is nonzero because its squared modulus is a² + b² > 0. Multiply by (a + ib)/(a + ib).
  2. 2.The numerator becomes (a + ib)² = a² − b² + 2abi, while the denominator becomes a² + b².
  3. 3.Compare components: x = (a² − b²)/(a² + b²) and y = 2ab/(a² + b²).
  4. 4.Then x² + y² = [(a² − b²)² + 4a²b²]/(a² + b²)². Expand the numerator to a⁴ − 2a²b² + b⁴ + 4a²b² = a⁴ + 2a²b² + b⁴.
  5. 5.This is (a² + b²)², so the quotient equals 1. The nonzero condition justifies the division.

There is also a shorter proof once the modulus properties are understood. The moduli of a + ib and a − ib are equal and positive, so the modulus of their quotient is 1. Squaring |x + iy| = 1 gives x² + y² = 1. The component proof shows how the cancellation works; the modulus proof identifies the underlying reason.

ReImO-2-112-11P(−3/5, 4/5)|x + iy| = 1x² + y² = 1 describes the whole unit circle.
A quotient on the unit circle— For a = 1 and b = 2, the quotient is −3/5 + (4/5)i. Its squared coordinates add to 1, so it lies on the circle.
Example — Checking a particular unit-modulus quotient

Problem
Evaluate (1 + 2i)/(1 − 2i) and locate its point.

  1. 1.Multiply numerator and denominator by 1 + 2i. The denominator becomes 1² + 2² = 5.
  2. 2.The numerator (1 + 2i)² equals 1 + 4i + 4i² = −3 + 4i.
  3. 3.Thus the quotient is −3/5 + (4/5)i, represented by (−3/5, 4/5).
  4. 4.Its squared modulus is 9/25 + 16/25 = 1. The point lies in quadrant II on the unit circle, illustrating the general relationship proved above.

The picture tells you where the result can lie, while the arithmetic identifies its exact point. For a nonzero number of modulus one, the identity 1/z = z̄/|z|² simplifies to 1/z = z̄. That special conclusion follows from the modulus being one; it should not be applied to an arbitrary complex number.

Quiz

Quick check

Which point represents −5 + 2i?

Quick check

Which number lies on the real axis?

Quick check

What distance from the origin corresponds to −6 + 8i?

Quick check

What does conjugation do to the point (a, b)?

Quick check

Which statement is guaranteed if two numbers have equal modulus?

Quick check

If w = (a + ib)/(a − ib) with real a, b not both zero, where does its point lie?

Quick check

For a nonzero complex z, when does its conjugate equal its reciprocal?

Practice Problems

Practice Problems
  1. Plot the points for 2 + 4i, −2 + 3i, i, 2, −5 − 2i and 1 − 2i on one Argand diagram. Label each axis and point.
  2. Write the numbers represented by (−4, 0), (0, −3), (5, −2) and (−1, 4). State which lie on axes.
  3. For z = −6 + 8i, draw the right triangle connecting its point to the origin and calculate |z|.
  4. Plot z = −2 + 3i together with z̄ and −z. Describe the different coordinate changes.
  5. Give four distinct numbers with modulus 3, and explain why their points lie on one circle.
  6. Find the conjugate of [(3 − 2i)(2 + 3i)]/[(1 + 2i)(2 − i)] by conjugating the factors first, and compare with the worked method.
  7. For real a, b not both zero, prove that (a + ib)/(a − ib) has modulus 1, first by components and then by the quotient-modulus property.
  8. Evaluate (2 − i)/(2 + i), find its modulus, and locate its point on a labelled unit-circle diagram.
  9. Explain geometrically why |z̄| = |z| and why z̄ = z holds exactly when z is real.
  10. For z = 1 + 2i, calculate z̄ and 1/z. Show why reflection alone does not usually give a reciprocal.

Key Takeaways

Key Takeaways

• The number a + ib is represented by the point (a, b), with real part horizontal and imaginary part vertical. • The real and imaginary axes contain numbers whose other component is zero. • Modulus is distance from the origin, so its value is nonnegative and many different points can share it. • Conjugation reflects a point across the real axis; it changes only the imaginary component. • Conjugates have equal modulus, and their nonzero quotient lies on the unit circle. • Algebra identifies exact values, while the Argand plane explains their distance and reflection relationships.

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Complex Numbers and Their Algebra

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Chapter Summary and Practice