The mirror image z̄ turns awkward problems about z into clean real arithmetic.
Every complex number has a twin. Flip z = a + b i across the real axis — keep its real part, negate its imaginary part — and you get its conjugate z̄ = a − b i. This one move is the workhorse of the whole subject: multiply a number by its own conjugate and the imaginary parts cancel, leaving the real number a² + b² = |z|². That is exactly why the conjugate clears i out of a denominator (you met it in arithmetic of complex numbers), and it is the key that turns a tangled equation about z into ordinary real arithmetic. We will also see how the modulus — the length |z| = √(a² + b²) — travels through multiplication and division: lengths multiply, lengths divide.
Start with the definition. The conjugate of z = a + b i is
z̄ = a − b i — same real part a, imaginary part negated −b.
Geometrically that is the cleanest reflection there is: flip the point across the real axis. The real coordinate a does not move; the imaginary coordinate b swaps to −b. So 3 + 2i becomes 3 − 2i, and −1 − 4i becomes −1 + 4i.
Conjugation is reflection across the real axis. It obeys two facts you can read straight off the picture:
Conjugating twice undoes itself: z̄̄ = z (flip, then flip back). And z is real exactly when z = z̄ — a point equal to its own mirror image must sit on the real axis, where b = 0.
The conjugate flips only the imaginary part. The real part is untouched. A common slip is to negate everything — but the conjugate of 3 + 2i is 3 − 2i, not −3 − 2i. Negating both parts gives −z, a different number (a half-turn about the origin, not a mirror flip).
Dial the real part a and imaginary part b. Watch z̄ mirror z across the real axis — and watch z·z̄ land on the real axis as the real number |z|².
Here is the fact that makes the conjugate worth its weight. Multiply z = a + b i by its conjugate z̄ = a − b i and the cross terms cancel:
(a + b i)(a − b i) = a² − a b i + a b i − b² i² = a² − b²(−1) = a² + b².
The two middle terms (−a b i and +a b i) are equal and opposite, so they vanish; and −b² i² becomes +b² because i² = −1. What is left, a² + b², is a plain real number — and it is exactly the square of the length:
z·z̄ = a² + b² = |z|².
Worked: for z = 3 + 2i, (3 + 2i)(3 − 2i) = 9 + 4 = 13, and indeed |z|² = 3² + 2² = 13. This is the engine behind division: to clear i from a denominator you multiply by the conjugate precisely because the bottom turns into the real number |w|².
Two more identities fall out of the same flip. Adding z to its mirror cancels the imaginary part; subtracting cancels the real part:
z + z̄ = 2a = 2 Re z (a real number), z − z̄ = 2b i = 2i·Im z (purely imaginary).
So the conjugate lets you extract the real and imaginary parts with arithmetic alone: Re z = (z + z̄)/2, Im z = (z − z̄)/(2i).
Don't confuse z·z̄ with z². The product with the conjugate, z·z̄ = a² + b², is always real and never negative. But z² = (a + b i)² = a² − b² + 2ab i is generally complex. The mirror partner is what makes the imaginary part die.
The modulus is the arrow's length, |z| = √(a² + b²) ≥ 0. The headline rule is that length passes cleanly through multiplication: multiply two numbers and their lengths multiply.
|z₁ z₂| = |z₁| · |z₂| and |z₁ / z₂| = |z₁| / |z₂| (z₂ ≠ 0).
Worked: take z₁ = 3 + 4i (length 5, since 3² + 4² = 25) and z₂ = 1 + 2i (length √5). Their product is (3 + 4i)(1 + 2i) = 3 + 6i + 4i + 8i² = −5 + 10i, whose length is √(25 + 100) = √125 = 5√5. And on the other side, 5 · √5 = 5√5. The two routes agree — exactly as the rule promises.
Why is it true? Because |z|² = z·z̄ and conjugation passes through a product: (z₁z₂)‾ = z̄₁·z̄₂. So |z₁z₂|² = (z₁z₂)(z₁z₂)‾ = (z₁z₂)(z̄₁z̄₂) = (z₁z̄₁)(z₂z̄₂) = |z₁|²|z₂|². Take square roots and you have the rule. Conjugation in fact distributes over every operation:
(z + w)‾ = z̄ + w̄, (z w)‾ = z̄ · w̄, (z / w)‾ = z̄ / w̄.
Addition is the one place lengths do not simply combine. Since z₁ + z₂ closes the triangle of arrows (the tip-to-tail rule from vector operations), the third side can never beat the other two together: |z₁ + z₂| ≤ |z₁| + |z₂|, with equality only when the two arrows point the same way. Lengths multiply under × but only add an upper bound under +.
Two traps. First, |z| is never negative — it is a length, so |z| ≥ 0, and |z| = 0 only for z = 0. Second, conjugation hits both factors: (z w)‾ = z̄ · w̄, not z̄ · w. And since ℂ has no order, you can compare sizes |z₁| and |z₂| — but never write an inequality between the numbers z₁ and z₂ themselves.
Set two numbers z₁ and z₂. The readout computes |z₁z₂| directly and |z₁|·|z₂| separately — they should always match (and the figure draws all three arrows so you can see the lengths multiply).
Now the payoff. Two standard moves crack almost any problem that mixes z and z̄:
Move 1 — write z = x + y i and match parts. A single complex equation is secretly two real equations (real = real, imaginary = imaginary). Solve those.
Move 2 — use z·z̄ = |z|². Whenever a product of conjugates appears, replace it by the real number |z|² and the imaginary clutter disappears.
Find z with z + 2z̄ = 9 + i. Write z = x + y i, so z̄ = x − y i. Then
(x + y i) + 2(x − y i) = (x + 2x) + (y − 2y) i = 3x + (−y) i.
Match this to 9 + i: the real parts give 3x = 9 ⇒ x = 3; the imaginary parts give −y = 1 ⇒ y = −1. So z = 3 − i. Check: (3 − i) + 2(3 + i) = 3 − i + 6 + 2i = 9 + i ✓.
The target is z + 2z̄ = 9 + i. Dial x and y in z = x + y i until the computed left side turns green — i.e. until both the real equation (3x = 9) and the imaginary equation (−y = 1) hold at once.
The same machinery answers questions phrased with the modulus. "Find all z with |z| = 5 and Re z = 3": from a² + b² = 25 and a = 3 we get b² = 16, so z = 3 + 4i or z = 3 − 4i — a conjugate pair, of course, since they share a real part and a modulus.
| Idea | The rule | Why it matters |
|---|---|---|
| Conjugate | z̄ = a − b i | mirror across the real axis; flips only Im z |
| Product with conjugate | z·z̄ = a² + b² = |z|² | always real; clears i from a denominator |
| Real / imaginary parts | z + z̄ = 2 Re z, z − z̄ = 2i Im z | extract a and b by arithmetic |
| Modulus & ×, ÷ | |z₁z₂| = |z₁||z₂|, |z₁/z₂| = |z₁|/|z₂| | lengths multiply and divide |
| Conjugation distributes | (z+w)‾ = z̄+w̄, (zw)‾ = z̄·w̄ | passes through every operation |
| Two solving moves | match parts or use z·z̄ | turn one complex equation into real arithmetic |
Next, in the trigonometric form, the modulus pairs up with the argument — and "lengths multiply" grows into the beautiful rule that multiplication adds angles.
Write the conjugate of each: (a) 5 − 3i, (b) −2 + 6i, (c) 7, (d) 4i.
Flip the sign of the imaginary part only. (a) 5 + 3i; (b) −2 − 6i; (c) 7 (a real number is its own conjugate); (d) −4i (a purely imaginary number negates entirely).
Compute z·z̄ for z = 5 + 12i, and state |z|.
(5 + 12i)(5 − 12i) = 25 + 144 = 169, a real number. Since z·z̄ = |z|², we get |z| = √169 = 13.
Verify |z₁z₂| = |z₁||z₂| for z₁ = 2 + i and z₂ = 1 − 3i.
Lengths: |z₁| = √5, |z₂| = √10, so |z₁||z₂| = √50 = 5√2. Product: (2 + i)(1 − 3i) = 2 − 6i + i − 3i² = 5 − 5i, length √(25 + 25) = √50 = 5√2. They match. ✓
Find z satisfying 2z − z̄ = 3 + 6i.
Let z = x + y i. Then 2(x + y i) − (x − y i) = (2x − x) + (2y + y) i = x + 3y i. Matching 3 + 6i: x = 3 and 3y = 6 ⇒ y = 2. So z = 3 + 2i. Check: 2(3 + 2i) − (3 − 2i) = 6 + 4i − 3 + 2i = 3 + 6i ✓.
Explain in one line why, if a real-coefficient equation has the root 2 + 5i, it must also have the root 2 − 5i.
Conjugation distributes over every operation, so conjugating the whole (real-coefficient) equation leaves it unchanged while sending the root z to z̄ — so 2 − 5i is a root too. (This is the conjugate-pair law, made precise in complex numbers and equations.)
Find every z with |z| = √13 and Im z = 2.
Write z = a + 2i. Then |z|² = a² + 4 = 13, so a² = 9 and a = 3 or a = −3. Thus z = 3 + 2i or z = −3 + 2i — two numbers of equal length on the line Im z = 2.
Six questions to lock it in. Tap the answer you think is right.
This lesson develops the conjugate and modulus of a complex number and the identity z·z̄ = |z|², aligning with the Common Core standards CCSS HSN-CN.A.3 (find the conjugate of a complex number and use it to compute moduli and quotients) and CCSS HSN-CN.A.1–2 (the form a + b i and its arithmetic). The geometric thread — conjugation as reflection across the real axis, and the multiplicative law |z₁z₂| = |z₁||z₂| — prepares the conjugate-pair theorem for polynomial roots (HSN-CN.C.7–9) taken up in 31.6. A good check for understanding: ask why multiplying by the conjugate is the standard trick for dividing complex numbers, and have the student state the answer "because the denominator becomes the real number |w|²."