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Complex Conjugate



Key Terms
Definition of the Conjugate
Conjugate Notation
Geometric Interpretation
Fundamental Properties
Conjugate and Modulus
Classification Theorems
Useful Identities
Applications to Division
Conjugate Pairs in Polynomials
Summary: What z̄ Tells You About z
Common Pitfalls
Complex Conjugate FAQ



The Mirror Operation


The Mirror Operation

Key Terms

Complex Conjugate— the reflection z‾=a−bi\overline{z} = a - bi
Modulus— satisfies ∣z∣2=z⋅z‾|z|^2 = z \cdot \overline{z}
Multiplicative Inverse— computed using the conjugate
Complex Plane— conjugation reflects across the real axis

See All Complex Numbers Definitions →


Definition of the Conjugate

Every point in the complex plane has a mirror image across the real axis.
Fold the plane along the horizontal axis. A point above the axis lands on a point below it, at the same distance from the fold, and a point on the axis stays where it is. Each complex number gets a partner this way.
The mirror image of −1 + 4i is −1 − 4i.
−1+4i-1 + 4i sits one unit left of the origin and four units up. Its mirror image is still one unit left, but four units down: the real part stays, the imaginary part changes sign.
real partimaginary partpoint
−1 + 4i−14(−1, 4)
its mirror image−1−4(−1, −4)
ReIm-10−10i-8−8i-6−6i-4−4i-2−2i22i44i66i88i1010i|z| = 4.1zz̅
z=−1+4iz = -1 + 4i and its mirror image −1−4i-1 - 4i, the same distance from the origin. The same state is on the conjugate and modulus tool.
This partner is the one that cancels the imaginary part.
Put the pair together. Adding them cancels the imaginary parts and leaves a real number. Multiplying them gives (−1)2−(4i)2=1+16(-1)^2 - (4i)^2 = 1 + 16, again a real number with no ii left. Only subtracting keeps the imaginary part, and then it doubles.
combine the pairworkresult
add(−1 + 4i) + (−1 − 4i)−2
multiply(−1)² − (4i)² = 1 + 1617
subtract(−1 + 4i) − (−1 − 4i)8i
That is why this partner comes up so often: it is how an ii is removed from a sum, a product or a denominator.
The definition
The complex conjugate of z=a+biz = a + bi is a−bia - bi: the same real part, the opposite imaginary part.
Complex Conjugate
z‾=a−bifor    z=a+bi\overline{z} = a - bi \quad \text{for} \;\; z = a + bi
Learn more about this formula: Complex Conjugate →
So the conjugate of −1+4i-1 + 4i is −1−4i-1 - 4i — its mirror image.
Conjugation is an operation: it takes one complex number and returns another.
The operation preserves the real part aa exactly while reversing the sign of the imaginary part. Where the original number has +bi+bi, the conjugate has −bi-bi. Where the original has −bi-bi, the conjugate has +bi+bi. Only the vertical component changes; the horizontal component remains untouched.
Concrete examples illustrate the pattern. The conjugate of 3+2i3 + 2i is 3−2i3 - 2i. The conjugate of −1−4i-1 - 4i is −1+4i-1 + 4i — the negative imaginary part becomes positive. The conjugate of 55 (a real number, written as 5+0i5 + 0i) is simply 55 — with no imaginary part to flip, nothing changes. The conjugate of 7i7i (pure imaginary, written as 0+7i0 + 7i) is −7i-7i — the entire number negates because the real part contributes nothing.
z Category z̄ Note
3 + 2i general complex (Q1) 3 − 2i imaginary part flips sign
−1 − 4i general complex (Q3) −1 + 4i negative imaginary becomes positive
5 real 5 unchanged — z̄ = z for every real z
7i pure imaginary −7i z̄ = −z for every pure imaginary z
0 zero (both real and pure imag.) 0 the unique fixed point of every reflection
(1⁄2) + (√3⁄2)i on the unit circle (|z| = 1) (1⁄2) − (√3⁄2)i here z̄ also equals z⁻¹
Two things to keep apart: the conjugate is not the negative, and it has two names.
The conjugate is not the same as negation. The negative of 3+2i3 + 2i is −3−2i-3 - 2i, changing both signs. The conjugate 3−2i3 - 2i changes only the imaginary sign. Confusing these operations leads to errors, particularly when manipulating equations involving both.
Two notations appear in mathematical literature. The overline zˉ\bar{z} dominates pure mathematics and most textbooks. The asterisk z∗z^* appears frequently in physics and engineering, particularly in quantum mechanics and signal processing. Both symbols denote the identical operation. This text uses the overline convention, but readers should recognize both forms as equivalent — both are listed among the complex number symbols.
The rest of the page covers the notation, the reflection picture, the properties of the conjugate, its link to the modulus, and its use in division and in polynomials.

Conjugate Notation

Notation

Conjugate Notation

Four inherited strips across this section promised the bar would be explained somewhere — this is the somewhere. The bar itself, the bar as a scope mark, and the star the physicists use instead.
zz, a+bia + bi and C\mathbb{C} come from complex numbers basics, ii from the imaginary unit, Re/ImRe/Im from the algebraic form; the modulus ∣z∣|z| of Conjugate and Modulus below is owned by the absolute value page.
zˉ\bar{z}
z bar — the conjugate of z
A bar over a number flips the sign of its imaginary part: zˉ=a−bi\bar{z} = a - bi for z=a+biz = a + bi — Definition of the Conjugate above. One stroke, one reflection; nothing else about the number changes.
CasesReal numbers pass through unchanged — aˉ=a\bar{a} = a; pure imaginaries negate — zˉ=−z\bar{z} = -z; applied twice, the bar erases itself: zˉ‾=z\overline{\bar{z}} = z.
Do not confuseNegation. −z-z flips both signs, zˉ\bar{z} only the imaginary one: −(3+2i)=−3−2i-(3+2i) = -3-2i while 3+2i‾=3−2i\overline{3+2i} = 3-2i — equations mixing both marks punish the mix-up.
Same glyph elsewhereBars over letters are a crowded namespace: the sample mean xˉ\bar{x} of statistics and the set complement Aˉ\bar{A} of set theory — unrelated jobs sharing the stroke.
z1+z2‾\overline{z_1 + z_2}
the conjugate of z-one plus z-two
The bar stretches over whatever it conjugates — a scope mark as much as an operator. LaTeX splits the two jobs: \bar for the short accent on one letter, \overline for the stretching version; the LaTeX reference lists both.
CasesDistribution is a theorem, not a reading rule: z1+z2‾=z1ˉ+z2ˉ\overline{z_1 + z_2} = \bar{z_1} + \bar{z_2} and z1z2‾=z1ˉ z2ˉ\overline{z_1 z_2} = \bar{z_1}\,\bar{z_2} hold because Fundamental Properties below proves them — the notation alone promises nothing.
Do not confuseBar width read carelessly. z1ˉ z2\bar{z_1}\,z_2 and z1z2‾\overline{z_1 z_2} are different numbers — how far the bar reaches decides what gets conjugated, the same way a radical's vinculum decides what sits under the root.
z∗z^*
z star
The physics spelling of the same operation: z∗=zˉz^* = \bar{z}, dominant in quantum mechanics and signal processing — this site keeps the bar, but both run through the literature, as Definition of the Conjugate above notes.
CasesThe star scales where the bar cannot: in linear algebra it climbs to matrices as the conjugate transpose A∗A^*, joining the superscript-corner family of ATA^T and kin.
Do not confuseMultiplication. In programming and calculator syntax ∗* is the times sign, and z∗z^* reads dangerously like an unfinished product — the superscript position carries the meaning.
Same glyph elsewhereThe starred sample points xi∗x_i^* of Riemann sums — a "chosen element" decoration, no conjugation anywhere; that page's notation section points back here for exactly this reason.
DownloadPNG2×PDF
Notation · Cheat sheet
Complex Number Notation at a Glance
The marks from the notation cards across the complex-number pages: how each is read and what it means. Every group links to the full card on its page.
Basics
z,  wz, \; w
z, w: complex numbers
The usual letters for complex numbers, as xx and yy are for reals.
a+bia + bi
a plus b i
Real part aa, imaginary part bb; both are real numbers.
C\mathbb{C}
C, the complex numbers
The set of all a+bia + bi; it contains R\mathbb{R}.
Imaginary numbers
ii
i, the imaginary unit
The number with i2=−1i^2 = -1.
−1\sqrt{-1}
the square root of minus one
Another name for ii; root rules for reals fail with it.
bibi
b i
A pure imaginary number: real part 00.
Algebraic form
Re⁡(z),  Im⁡(z)\operatorname{Re}(z), \; \operatorname{Im}(z)
real part, imaginary part
For z=a+biz = a + bi: aa and bb. Im⁡(z)\operatorname{Im}(z) is bb, not bibi.
a+(−b)ia + (-b)i
a plus negative b, i
Written a−bia - bi; the imaginary part is −b-b.
z=x+yiz = x + yi
z equals x plus y i
Same form with plane letters: the point (x,y)(x, y).
Complex conjugateThis page
zˉ\bar{z}
z bar, the conjugate
For z=a+biz = a + bi, zˉ=a−bi\bar{z} = a - bi: the mirror image across the real axis.
z1+z2‾\overline{z_1 + z_2}
the conjugate of the sum
The bar covers the whole expression; it equals zˉ1+zˉ2\bar{z}_1 + \bar{z}_2.
z∗z^*
z star
The same conjugate, as written in physics and engineering.
Absolute value
∣z∣|z|
the modulus of z
Distance from 00: a2+b2\sqrt{a^2 + b^2}.
∣z∣2=zzˉ|z|^2 = z\bar{z}
mod z squared
Equals a2+b2a^2 + b^2, with no root to take.
∣z−c∣=R|z - c| = R
mod z minus c equals R
The circle of radius RR around cc.
Additive inverse
−z-z
negative z
Both parts change sign: −a−bi-a - bi.
−(a+bi)-(a + bi)
minus, quantity a plus b i
The brackets carry the minus to both parts.
z−wz - w
z minus w
Adding the inverse: z+(−w)z + (-w).
Multiplicative inverse
z−1,  1zz^{-1}, \; \frac{1}{z}
z inverse; one over z
The number with z⋅z−1=1z \cdot z^{-1} = 1; it equals zˉ∣z∣2\frac{\bar{z}}{|z|^2}.
(3+2i)−1(3 + 2i)^{-1}
the inverse of three plus two i
Found by multiplying top and bottom by the conjugate: 3−2i13\frac{3 - 2i}{13}.
Trigonometric form
r,  θr, \; \theta
r and theta, the polar pair
Distance from 00 and angle from the positive real axis.
arg⁡(z)\arg(z)
the argument of z
Any angle of zz; there are infinitely many, 2π2\pi apart.
Arg⁡(z)\operatorname{Arg}(z)
the principal argument
One chosen angle, usually in (−π,π](-\pi, \pi]; some texts use [0,2π)[0, 2\pi).
cis⁡θ\operatorname{cis}\theta
sis theta
Short for cos⁡θ+isin⁡θ\cos\theta + i\sin\theta.
Exponential form
eiθe^{i\theta}
e to the i theta
Equals cis⁡θ\operatorname{cis}\theta (Euler's formula): a point on the unit circle.
reiθre^{i\theta}
r e to the i theta
The whole number zz: length rr, angle θ\theta.
ei(θ+2πk)e^{i(\theta + 2\pi k)}
theta plus two pi k
The same number for every integer kk.
De Moivre's theorem
zkz_k
z sub k, the k-th root
The roots of zn=wz^n = w, numbered k=0,…,n−1k = 0, \ldots, n - 1.
ω\omega
omega
A primitive root of unity: its powers give all nn roots of 11.
R1/nR^{1/n}
R to the one over n
The real root of the length; the complex root w1/nw^{1/n} has nn values.
Same marks, different job
∣z∣|z|The modulus here; for a real number, the absolute value; for a matrix, the determinant.
zˉ\bar{z}The conjugate here; in statistics a bar means a mean, in sets a complement.
z−1z^{-1} vs z∗z^*The reciprocal, against the conjugate: they agree only on the unit circle.
w1/nw^{1/n}nn values for a complex ww; a single real root for a positive real number.
Complex number symbols tableComplex numbers
Complex Number Notation at a Glance·/complex-numbers/basicsLearn Math Class
ReIm-10−10i-8−8i-6−6i-4−4i-2−2i22i44i66i88i1010i|z| = 3.6zz̅
z and z̄ mirrored in the real axis, |z| = 3.6

The point z and its conjugate z̄ sit at the same distance from the origin, one above the real axis and one below: conjugation keeps the real part and flips the sign of the imaginary part, so it is a reflection in the real axis. The modulus, the length of either arrow, is the same 3.6 for both. Drag z anywhere and watch its mirror image follow on the conjugate and modulus tool.

Every algebraic property of the conjugate below is a consequence of this reflection.

Geometric Interpretation

The complex plane transforms the conjugate from an algebraic rule into a visible action. If z=a+biz = a + bi corresponds to the point (a,b)(a, b), then zˉ=a−bi\bar{z} = a - bi corresponds to (a,−b)(a, -b). The horizontal coordinate stays fixed while the vertical coordinate negates. This is reflection across the real axis.

Fact: The mirror has no complex derivative
The map z↦zˉz \mapsto \bar{z} is differentiable at no point: its difference quotient z+h‾−zˉh=hˉh\frac{\overline{z + h} - \bar{z}}{h} = \frac{\bar{h}}{h} is 11 for real hh and −1-1 for imaginary hh. A complex derivative can only rotate and scale small shapes, and a mirror turns them over.

Picture the real axis as a horizontal mirror. Every point above the axis has a mirror image below; every point below has an image above. The conjugate operation sends each complex number to its reflection. The point 2+3i2 + 3i at coordinates (2,3)(2, 3) reflects to 2−3i2 - 3i at (2,−3)(2, -3). The point −1−4i-1 - 4i at (−1,−4)(-1, -4) reflects to −1+4i-1 + 4i at (−1,4)(-1, 4).

Points lying directly on the mirror remain stationary. These are precisely the real numbers — they have no vertical displacement to reverse, so conjugation leaves them unchanged. The number 55 sits at (5,0)(5, 0) on the real axis, and its reflection lands at the same spot. This geometric fact corresponds to the algebraic observation that aˉ=a\bar{a} = a for any real aa.

Points on the imaginary axis exhibit different behavior. They lie perpendicular to the mirror, equidistant above and below the real axis. Reflection sends each to the opposite side of the origin. The number 4i4i at (0,4)(0, 4) reflects to −4i-4i at (0,−4)(0, -4). For pure imaginary numbers, conjugation equals negation: zˉ=−z\bar{z} = -z.

The reflection interpretation explains why conjugating twice returns the original number. Reflect a point across a line, then reflect again across the same line — the point returns home. Two mirror operations cancel completely, regardless of where the point started.

Fundamental Properties

The conjugate operation interacts predictably with arithmetic. Five properties govern how conjugation passes through sums, products, and powers, enabling simplification of complicated expressions.

The involution property states that conjugating twice recovers the original:

Conjugate of a Conjugate
z‾‾=z\overline{\overline{z}} = z
Learn more about this formula: Conjugate of a Conjugate →


Apply the definition twice: if z=a+biz = a + bi, then zˉ=a−bi\bar{z} = a - bi, and zˉ‾=a−(−b)i=a+bi=z\overline{\bar{z}} = a - (-b)i = a + bi = z. The double sign flip restores the original imaginary part. Geometrically, reflecting twice across the same axis brings every point back to its starting position.

Additivity allows conjugation to distribute over sums:

Conjugate of a Sum
z1+z2‾=z1‾+z2‾\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}
Learn more about this formula: Conjugate of a Sum →


The conjugate of a sum equals the sum of the conjugates. Proof: let z1=a+biz_1 = a + bi and z2=c+diz_2 = c + di. Then z1+z2=(a+c)+(b+d)iz_1 + z_2 = (a + c) + (b + d)i, so z1+z2‾=(a+c)−(b+d)i=(a−bi)+(c−di)=z1ˉ+z2ˉ\overline{z_1 + z_2} = (a + c) - (b + d)i = (a - bi) + (c - di) = \bar{z_1} + \bar{z_2}. The same property extends to subtraction: z1−z2‾=z1ˉ−z2ˉ\overline{z_1 - z_2} = \bar{z_1} - \bar{z_2}.

Multiplicativity states that conjugation distributes over products:

Conjugate of a Product
z1⋅z2‾=z1‾⋅z2‾\overline{z_1 \cdot z_2} = \overline{z_1} \cdot \overline{z_2}
Learn more about this formula: Conjugate of a Product →


Pitfall: Forgetting to conjugate i
Conjugating i(2+3i)i(2 + 3i) as i⋅(2+3i)‾=i(2−3i)=3+2ii \cdot \overline{(2 + 3i)} = i(2 - 3i) = 3 + 2i leaves the factor ii unchanged. Both factors flip: i‾⋅(2+3i)‾=−i(2−3i)=−3−2i\overline{i} \cdot \overline{(2 + 3i)} = -i(2 - 3i) = -3 - 2i, the conjugate of i(2+3i)=−3+2ii(2 + 3i) = -3 + 2i.

The conjugate of a product equals the product of the conjugates. Verification requires expanding both sides and comparing — the algebra confirms equality. This property proves invaluable when simplifying products or verifying identities.

Division follows the same pattern:

Conjugate of a Quotient
(z1z2)‾=z1‾z2‾\overline{\left(\frac{z_1}{z_2}\right)} = \frac{\overline{z_1}}{\overline{z_2}}
Learn more about this formula: Conjugate of a Quotient →


valid for z2≠0z_2 \neq 0. Conjugation passes through quotients just as it passes through products.

Powers inherit the multiplicative property:

Conjugate of a Power
zn‾=(z‾)n\overline{z^n} = (\overline{z})^n
Learn more about this formula: Conjugate of a Power →


for any integer nn. Repeated application of the product rule establishes this for positive integers, and the quotient rule extends it to negative integers. Conjugating a power equals powering the conjugate.
Property Formula In words
Involution z̄ = z conjugating twice returns the original
Additivity z₁ + z₂ = z̄₁ + z̄₂ conjugate of a sum = sum of conjugates
Subtraction z₁ − z₂ = z̄₁ − z̄₂ conjugate of a difference = difference of conjugates
Multiplicativity z₁ · z₂ = z̄₁ · z̄₂ conjugate of a product = product of conjugates
Quotient z₁ ⁄ z₂ = z̄₁ ⁄ z̄₂  (z₂ ≠ 0) conjugate of a quotient = quotient of conjugates
Powers zⁿ = (z̄)ⁿ  (integer n) conjugate of a power = power of the conjugate
Conjugation passes through every arithmetic operation, so conjugating an expression means conjugating each of its pieces.

Conjugate and Modulus

A fundamental identity connects the conjugate to the modulus:

Conjugate and Modulus
z⋅zˉ=∣z∣2z \cdot \bar{z} = |z|^2

The product of any complex number with its conjugate yields the square of its modulus — a real, non-negative value. This relationship lies at the heart of complex arithmetic.

The proof follows from direct expansion. Let z=a+biz = a + bi, so zˉ=a−bi\bar{z} = a - bi. The product becomes:

Conjugate and Modulus
z⋅zˉ=(a+bi)(a−bi)=a2−abi+abi−b2i2=a2−b2(−1)=a2+b2z \cdot \bar{z} = (a + bi)(a - bi) = a^2 - abi + abi - b^2i^2 = a^2 - b^2(-1) = a^2 + b^2

Pitfall: Forgetting that i squared is −1
Expanding (3+4i)(3−4i)(3 + 4i)(3 - 4i) to 9−16=−79 - 16 = -7 takes (4i)2(4i)^2 to be 1616. In fact (4i)2=16i2=−16(4i)^2 = 16i^2 = -16, so the product is 9−(−16)=25=∣3+4i∣29 - (-16) = 25 = |3 + 4i|^2.

The middle terms cancel, and i2=−1i^2 = -1 converts the final term to positive. The result a2+b2a^2 + b^2 matches the definition ∣z∣2=a2+b2|z|^2 = a^2 + b^2 exactly.

This identity explains why z⋅zˉz \cdot \bar{z} always produces a real number. The imaginary parts eliminate each other through the cancellation of −abi-abi and +abi+abi. No matter how complicated zz appears, multiplying by its conjugate guarantees a real outcome.

A related fact: conjugation preserves modulus. The numbers zz and zˉ\bar{z} lie at equal distances from the origin, as reflection across a line through the origin does not change radial distance. Algebraically, ∣zˉ∣=∣a−bi∣=a2+(−b)2=a2+b2=∣z∣|\bar{z}| = |a - bi| = \sqrt{a^2 + (-b)^2} = \sqrt{a^2 + b^2} = |z|. The identity ∣zˉ∣=∣z∣|\bar{z}| = |z| holds universally.

Taking square roots of z⋅zˉ=∣z∣2z \cdot \bar{z} = |z|^2 gives ∣z∣=z⋅zˉ|z| = \sqrt{z \cdot \bar{z}}, an alternative formula for modulus that sometimes proves more convenient than a2+b2\sqrt{a^2 + b^2}.
ReIm-10−10i-8−8i-6−6i-4−4i-2−2i22i44i66i88i1010iθ₁θ₂z₁z₂z₁z₂
z · z̄ lands on the positive real axis

The arguments of z and z̄ are opposite, so they cancel when the two are multiplied and the product has argument 0: it lies on the positive real axis at distance |z|². That is the identity z z̄ = |z|² seen as a rotation back to the axis. Multiply any number by its conjugate on the complex multiplication tool.

The identity gives the modulus without a square root, which is why it appears in every division.

Classification Theorems

The conjugate provides algebraic tests for determining whether a complex number belongs to special subcategories. Two classification theorems identify real numbers and pure imaginary numbers through their relationship with their conjugates.

The first theorem characterizes real numbers via the conjugate:

Real Number Test
z∈R  ⟺  z=z‾z \in \mathbb{R} \iff z = \overline{z}
Learn more about this formula: Real Number Test →


Fact: Half of a real signal's spectrum is spare
For real samples xnx_n, the Fourier transform satisfies XN−k=Xk‾X_{N-k} = \overline{X_k}: conjugation leaves each xnx_n alone and only flips the sign in e−2πikn/Ne^{-2\pi i kn/N}. So NN real samples need only about N2\frac{N}{2} of the outputs, and fast routines compute just those.

For the forward direction, suppose zz is real, meaning z=a+0i=az = a + 0i = a for some a∈Ra \in \mathbb{R}. Then zˉ=a−0i=a=z\bar{z} = a - 0i = a = z. The conjugate equals the original.

For the reverse direction, suppose z=zˉz = \bar{z}. Writing z=a+biz = a + bi, the equation becomes a+bi=a−bia + bi = a - bi. Comparing imaginary parts: b=−bb = -b, which forces 2b=02b = 0 and thus b=0b = 0. With zero imaginary part, z=az = a is real.

Geometrically, real numbers sit on the real axis — the mirror line for conjugation. Points on a mirror remain fixed under reflection, so z=zˉz = \bar{z} characterizes exactly those points.

The second theorem characterizes pure imaginary numbers:

Pure Imaginary Test
z  is pure imaginary  ⟺  z‾=−zz \;\text{is pure imaginary} \iff \overline{z} = -z
Learn more about this formula: Pure Imaginary Test →


Forward: if z=biz = bi for real bb, then zˉ=−bi=−(bi)=−z\bar{z} = -bi = -(bi) = -z.

Reverse: if zˉ=−z\bar{z} = -z, then a−bi=−a−bia - bi = -a - bi. Comparing real parts: a=−aa = -a, so 2a=02a = 0 and a=0a = 0. The number has no real part and is pure imaginary.

Geometrically, pure imaginaries sit on the imaginary axis, perpendicular to the mirror. Reflection through the real axis sends each such point to its opposite through the origin, making zˉ=−z\bar{z} = -z.
Category of z Algebraic test Geometric meaning Example
Real (b = 0) z̄ = z lies on the real axis — the mirror itself, fixed by reflection z = 5: z̄ = 5 = z  ✓
Pure imaginary (a = 0) z̄ = −z lies on the imaginary axis — reflection sends to opposite point through origin z = 4i: z̄ = −4i = −z  ✓

Useful Identities

Three identities involving zz and zˉ\bar{z} appear constantly in calculations. Each extracts specific information from a complex number or produces a value with guaranteed properties.

The sum of a number and its conjugate isolates the real part:

Useful Identities
z+zˉ=(a+bi)+(a−bi)=2az + \bar{z} = (a + bi) + (a - bi) = 2a

The imaginary terms cancel, leaving twice the real part. Rearranging provides a formula: Re(z)=z+zˉ2Re(z) = \frac{z + \bar{z}}{2}. This identity guarantees that z+zˉz + \bar{z} is always real, regardless of the original number. Encountering this sum in any calculation signals that the result lies on the real axis.

The difference between a number and its conjugate isolates the imaginary part:

Useful Identities
z−zˉ=(a+bi)−(a−bi)=2biz - \bar{z} = (a + bi) - (a - bi) = 2bi

Pitfall: Dividing by 2 instead of 2i
For z=5−3iz = 5 - 3i, halving z−zˉ=−6iz - \bar{z} = -6i gives −3i-3i, which is not Im(z)Im(z): the imaginary part is the real number −3-3. Dividing by 2i2i gives it: −6i2i=−3\frac{-6i}{2i} = -3.

The real terms cancel, leaving twice the imaginary term. Rearranging: Im(z)=z−zˉ2iIm(z) = \frac{z - \bar{z}}{2i}. This identity guarantees that z−zˉz - \bar{z} is always pure imaginary. The result necessarily sits on the vertical axis.

The product of a number and its conjugate yields the squared modulus:

Useful Identities
z⋅zˉ=a2+b2=∣z∣2z \cdot \bar{z} = a^2 + b^2 = |z|^2

This result is always real and always non-negative. It equals zero only when z=0z = 0. The identity underlies division, modulus computation, and countless proofs.

These three identities — sum, difference, and product with the conjugate — form a toolkit for manipulating complex expressions. Recognizing when they apply often transforms an intimidating calculation into straightforward algebra.
ReIm-5−5i-4−4i-3−3i-2−2i-1−1i11i22i33i44i55iz₁+z₂ →z₁−z₂−z₂z₁z₂
z + z̄ on the real axis, z − z̄ on the imaginary axis

z = 3 + 2i and its conjugate 3 − 2i are mirror images across the real axis. Their sum, 6, cancels the imaginary parts and lands on the real axis; their difference, 4i, cancels the real parts and lands on the imaginary axis — the first two identities of this section, drawn. Move z and watch both results stay on their axes in the complex addition and subtraction tool.

Halving the sum and dividing the difference by 2i gives Re(z) = 3 and Im(z) = 2 back.

Applications to Division

Division of complex numbers requires expressing the quotient in standard algebraic form a+bia + bi. A complex denominator violates this requirement — the conjugate provides the remedy.

Consider the division wz\frac{w}{z} where both ww and zz are complex. The denominator z=c+diz = c + di contains an imaginary part, preventing direct interpretation as a number in standard form. The strategy: multiply both numerator and denominator by zˉ\bar{z}, the conjugate of the denominator.

Applications to Division
wz=wz⋅zˉzˉ=w⋅zˉz⋅zˉ=w⋅zˉ∣z∣2\frac{w}{z} = \frac{w}{z} \cdot \frac{\bar{z}}{\bar{z}} = \frac{w \cdot \bar{z}}{z \cdot \bar{z}} = \frac{w \cdot \bar{z}}{|z|^2}

Pitfall: Multiplying only the denominator
Multiplying just the denominator of 2+i1+i\frac{2 + i}{1 + i} by 1−i1 - i gives 2+i2\frac{2 + i}{2}, a different number. With the numerator multiplied too, (2+i)(1−i)(1+i)(1−i)=3−i2=32−12i\frac{(2 + i)(1 - i)}{(1 + i)(1 - i)} = \frac{3 - i}{2} = \frac{3}{2} - \frac{1}{2}i.

The denominator becomes z⋅zˉ=∣z∣2z \cdot \bar{z} = |z|^2, a real number. The numerator w⋅zˉw \cdot \bar{z} is some complex number that can be computed by standard multiplication. Dividing a complex number by a real number simply scales both components, yielding standard form.

A complete example: compute 3+2i1−4i\frac{3 + 2i}{1 - 4i}.

The denominator is 1−4i1 - 4i, so its conjugate is 1+4i1 + 4i. Multiply numerator and denominator:

Applications to Division
3+2i1−4i⋅1+4i1+4i=(3+2i)(1+4i)(1−4i)(1+4i)\frac{3 + 2i}{1 - 4i} \cdot \frac{1 + 4i}{1 + 4i} = \frac{(3 + 2i)(1 + 4i)}{(1 - 4i)(1 + 4i)}

The denominator: (1−4i)(1+4i)=1−16i2=1+16=17(1 - 4i)(1 + 4i) = 1 - 16i^2 = 1 + 16 = 17.

The numerator: (3+2i)(1+4i)=3+12i+2i+8i2=3+14i−8=−5+14i(3 + 2i)(1 + 4i) = 3 + 12i + 2i + 8i^2 = 3 + 14i - 8 = -5 + 14i.

The quotient: −5+14i17=−517+1417i\frac{-5 + 14i}{17} = -\frac{5}{17} + \frac{14}{17}i.

Without the conjugate, no systematic method converts complex quotients to standard form. The identity z⋅zˉ=∣z∣2z \cdot \bar{z} = |z|^2 makes the technique work — it guarantees the denominator becomes real.
ReIm-10−10i-8−8i-6−6i-4−4i-2−2i22i44i66i88i1010iθ₁θ₂θ₁−θ₂z₁/z₂z₁z₂
Dividing by the conjugate clears the denominator

Multiplying numerator and denominator by z̄₂ turns the denominator into the real number |z₂|², and the picture shows the quotient at the angle θ₁ − θ₂ that results. Division is multiplication by the conjugate followed by a real scaling. Work any quotient through on the complex division tool.

Without the conjugate there would be no way to write a quotient in standard form.
The section's complete example, one move at a time. Watch the denominator: the conjugate turns it into the real number 1717.
DownloadPNG2×PDF
Applications to division

Dividing by multiplying with the conjugate

The identity z⋅zˉ=∣z∣2z \cdot \bar{z} = |z|^2 guarantees the denominator becomes real.
01
Complex denominatorBefore
3+2i1−4i\dfrac{3 + 2i}{1 - 4i}
↓Multiply by the conjugate— numerator and denominator both by 1 + 4i
02
Conjugate in
(3+2i)(1+4i)(1−4i)(1+4i)\dfrac{(3 + 2i)(1 + 4i)}{(1 - 4i)(1 + 4i)}
↓Multiply out— denominator 1 − 16i² = 17, numerator 3 + 14i + 8i² = −5 + 14i
03
Real denominator
−5+14i17\dfrac{-5 + 14i}{17}
↓Divide each part by 17— a real denominator only scales the two components
04
Standard formAfter
−517+1417 i-\dfrac{5}{17} + \dfrac{14}{17}\,i
Dividing (3 + 2i) by (1 − 4i)·/complex-numbers/complex-conjugateLearn Math Class
The same moves work for any quotient: the denominator always becomes ∣z∣2|z|^2, here 12+42=171^2 + 4^2 = 17, and the result lands in standard form a+bia + bi.

Conjugate Pairs in Polynomials

THEOREM
Polynomials with real coefficients exhibit remarkable structure in their complex roots: non-real roots always appear in conjugate pairs.

If z0z_0 solves the equation, so does z0ˉ\bar{z_0}.

The theorem states: let p(z)=anzn+an−1zn−1+⋯+a1z+a0p(z) = a_nz^n + a_{n-1}z^{n-1} + \cdots + a_1z + a_0 be a polynomial with all coefficients aka_k real. If p(z0)=0p(z_0) = 0 for some complex number z0z_0, then p(z0ˉ)=0p(\bar{z_0}) = 0 as well.

The proof exploits how conjugation interacts with polynomial evaluation. Since conjugation distributes over sums and products, and since conjugating a real number leaves it unchanged:

Conjugate Pairs in Polynomials
p(z0)‾=anz0n+⋯+a0‾=anz0n‾+⋯+a0=an(z0ˉ)n+⋯+a0=p(z0ˉ)\overline{p(z_0)} = \overline{a_nz_0^n + \cdots + a_0} = a_n\overline{z_0^n} + \cdots + a_0 = a_n(\bar{z_0})^n + \cdots + a_0 = p(\bar{z_0})

Pitfall: Pairing roots without real coefficients
Taking −i-i as a root of z2−2iz−1z^2 - 2iz - 1 because ii is one ignores the coefficient −2i-2i, which is not real. Substituting gives (−i)2−2i(−i)−1=−1−2−1=−4(-i)^2 - 2i(-i) - 1 = -1 - 2 - 1 = -4; the polynomial is (z−i)2(z - i)^2, and ii is its only root.

The key step uses ak‾=ak\overline{a_k} = a_k because each coefficient is real. If p(z0)=0p(z_0) = 0, then p(z0)‾=0ˉ=0\overline{p(z_0)} = \bar{0} = 0, so p(z0ˉ)=0p(\bar{z_0}) = 0.

Consequences flow immediately. A real quadratic with no real roots must have two complex conjugate roots — if 2+3i2 + 3i solves it, so does 2−3i2 - 3i. A real cubic always has at least one real root, since complex roots pair off and an odd number of roots cannot all be paired. A real polynomial of degree 4 might have four real roots, two real and two complex conjugates, or two pairs of complex conjugates — but never three real and one complex.

Conjugate pairs multiply to give real quadratic factors. If z0=a+biz_0 = a + bi is a root, then (z−z0)(z−z0ˉ)=z2−2az+(a2+b2)(z - z_0)(z - \bar{z_0}) = z^2 - 2az + (a^2 + b^2), a quadratic with real coefficients. This factorization explains why every real polynomial factors completely into real linear and real quadratic terms — the Fundamental Theorem of Algebra guarantees complex roots exist, and conjugate pairing ensures they combine into real factors.
Real coefficients force non-real roots into mirror pairs, and each pair multiplies into a real quadratic factor.

Summary: What z̄ Tells You About z

The conjugate is most useful not for its own sake but for what it reveals when combined with z itself. The table below collects the identities and tests in which z̄ extracts a specific quantity from z, signals a real or non-negative result, or characterizes z as belonging to a special subset (the real axis, the imaginary axis, or the unit circle).
Expression involving z̄ Result Always… What it gives you
z + z̄ 2 · Re(z) = 2a real isolates the real part of z
z − z̄ 2i · Im(z) = 2bi pure imaginary isolates the imaginary part of z
z · z̄ |z|² = a² + b² real and non-negative the squared modulus; underpins division
|z̄| |z| equal to |z| confirms that reflection preserves distance from origin
z̄ = z condition for… true iff z is real classification test for membership in ℝ
z̄ = −z condition for… true iff z is pure imaginary classification test for membership in iℝ
z̄ = z⁻¹ condition for… true iff |z| = 1 classification test for points on the unit circle

Common Pitfalls

Six mistakes account for many of the wrong answers in work with the conjugate. Five are marked where they happen in the sections above; the confusion of the conjugate with the negative is the one the definition section already points out. Here each is worked through in full, with the wrong and the right working side by side.

Confusing the conjugate with the negative

Asked for the conjugate, both signs are flipped instead of only the sign of the imaginary part. The result is the negative of the number, its reflection through the origin rather than across the real axis.

Wrong working: 3+2i‾=−3−2i\overline{3 + 2i} = -3 - 2i.

Right working: the real part stays and only the imaginary sign changes, so 3+2i‾=3−2i\overline{3 + 2i} = 3 - 2i. The number −3−2i-3 - 2i is −(3+2i)-(3 + 2i).

Why it happens: both operations flip a sign and both are written with a single mark, a bar or a minus.

How to check: a number and its conjugate have the same real part, so their sum is real and equal to twice that part: (3+2i)+(3−2i)=6(3 + 2i) + (3 - 2i) = 6. A number and its negative add to 00.

Forgetting to conjugate i

A product that contains the factor ii is conjugated factor by factor, but the bar is applied to the other factor only, as if ii were a real constant.

Wrong working: i(2+3i)‾=i⋅(2−3i)=2i−3i2=3+2i\overline{i(2 + 3i)} = i \cdot (2 - 3i) = 2i - 3i^2 = 3 + 2i.

Right working: every factor is conjugated, and i‾=−i\overline{i} = -i. So i(2+3i)‾=−i(2−3i)=−2i+3i2=−3−2i\overline{i(2 + 3i)} = -i(2 - 3i) = -2i + 3i^2 = -3 - 2i. Directly, i(2+3i)=2i+3i2=−3+2ii(2 + 3i) = 2i + 3i^2 = -3 + 2i, whose conjugate is −3−2i-3 - 2i.

Why it happens: ii looks like a fixed coefficient, and coefficients usually pass through the bar unchanged. Only real coefficients do.

How to check: work out the product first, then flip the sign of its imaginary part. Both routes must give the same number.

Forgetting that i squared is −1

In the product z⋅zˉ=(a+bi)(a−bi)z \cdot \bar{z} = (a + bi)(a - bi), the difference-of-squares pattern gives a2−(bi)2a^2 - (bi)^2, and (bi)2(bi)^2 is then taken to be b2b^2.

Wrong working: (3+4i)(3−4i)=9−16=−7(3 + 4i)(3 - 4i) = 9 - 16 = -7.

Right working: (3+4i)(3−4i)=9−(4i)2=9−16i2=9+16=25(3 + 4i)(3 - 4i) = 9 - (4i)^2 = 9 - 16i^2 = 9 + 16 = 25. Since ∣3+4i∣=9+16=5|3 + 4i| = \sqrt{9 + 16} = 5, this is ∣3+4i∣2=25|3 + 4i|^2 = 25, as the identity requires.

Why it happens: the real pattern (x+y)(x−y)=x2−y2(x + y)(x - y) = x^2 - y^2 is applied with y=4y = 4 instead of y=4iy = 4i.

How to check: z⋅zˉz \cdot \bar{z} is never negative. A negative result such as −7-7 shows the slip at once; the answer must equal a2+b2a^2 + b^2.

Dividing by 2 instead of 2i

The imaginary part is extracted from z−zˉ=2biz - \bar{z} = 2bi by halving it, which leaves the term bibi instead of the real number bb.

Wrong working: for z=5−3iz = 5 - 3i, z−zˉ=(5−3i)−(5+3i)=−6iz - \bar{z} = (5 - 3i) - (5 + 3i) = -6i, so Im(z)=−6i2=−3iIm(z) = \frac{-6i}{2} = -3i.

Right working: Im(z)=z−zˉ2i=−6i2i=−3Im(z) = \frac{z - \bar{z}}{2i} = \frac{-6i}{2i} = -3.

Why it happens: the formula for the real part divides by 22, and the imaginary part is often pictured as the whole term bibi rather than its coefficient bb.

How to check: Im(z)Im(z) is always a real number. Read it off directly: in 5−3i5 - 3i the coefficient of ii is −3-3.

Multiplying only the denominator

To make the denominator real, it is multiplied by its conjugate, but the numerator is left as it is. The fraction has then been multiplied by zˉ\bar{z} instead of by zˉzˉ=1\frac{\bar{z}}{\bar{z}} = 1, so its value changes.

Wrong working: 2+i1+i→2+i(1+i)(1−i)=2+i2=1+12i\frac{2 + i}{1 + i} \to \frac{2 + i}{(1 + i)(1 - i)} = \frac{2 + i}{2} = 1 + \frac{1}{2}i.

Right working: 2+i1+i=(2+i)(1−i)(1+i)(1−i)=2−2i+i−i21−i2=3−i2=32−12i\frac{2 + i}{1 + i} = \frac{(2 + i)(1 - i)}{(1 + i)(1 - i)} = \frac{2 - 2i + i - i^2}{1 - i^2} = \frac{3 - i}{2} = \frac{3}{2} - \frac{1}{2}i.

Why it happens: the goal is a real denominator, and the step looks like a repair of the denominator alone.

How to check: multiply the answer by the original denominator; it must give the numerator back. (32−12i)(1+i)=32+32i−12i−12i2=2+i\left(\frac{3}{2} - \frac{1}{2}i\right)(1 + i) = \frac{3}{2} + \frac{3}{2}i - \frac{1}{2}i - \frac{1}{2}i^2 = 2 + i, while (1+12i)(1+i)=12+32i\left(1 + \frac{1}{2}i\right)(1 + i) = \frac{1}{2} + \frac{3}{2}i.

Pairing roots without real coefficients

The conjugate pair theorem is applied to a polynomial with a coefficient that is not real, and the conjugate of a known root is listed as a second root.

Wrong working: ii is a root of z2−2iz−1z^2 - 2iz - 1, since i2−2i⋅i−1=−1+2−1=0i^2 - 2i \cdot i - 1 = -1 + 2 - 1 = 0. So −i-i is a root too, and the roots are ±i\pm i.

Right working: substituting −i-i gives (−i)2−2i(−i)−1=−1+2i2−1=−1−2−1=−4(-i)^2 - 2i(-i) - 1 = -1 + 2i^2 - 1 = -1 - 2 - 1 = -4, not 00. The polynomial is (z−i)2(z - i)^2, so ii is a double root and −i-i is not a root. The proof of the theorem needs ak‾=ak\overline{a_k} = a_k, and here −2i‾=2i\overline{-2i} = 2i.

Why it happens: the theorem is remembered as "complex roots come in pairs", without its condition on the coefficients.

How to check: look at every coefficient before pairing roots. If one is not real, substitute the conjugate and see whether it gives 00.

Complex Conjugate FAQ

What do the notations z̄ and z* mean?

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Both denote the conjugate of z, obtained by flipping the sign of the imaginary part. The overbar dominates in mathematics while the asterisk is standard in physics and engineering, and some texts write conj(z). All three name the same operation, so the choice signals the field rather than any difference in meaning.Read more →

What is z multiplied by its conjugate equal to?

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The square of the modulus, which is always a real, non-negative number. For a + bi the cross terms cancel and what remains is a² + b². This single fact is why the conjugate clears denominators in division and why it links directly to distance in the complex plane.Read more →

Is the conjugate the same as negation?

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No. Negating flips both components, sending a + bi to −a − bi, which reflects the point through the origin. Conjugating flips only the imaginary part, reflecting across the horizontal axis instead. The two coincide only for pure imaginary numbers, where there is no real part to distinguish them.Read more →

How do you tell if a complex number is real or pure imaginary using its conjugate?

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A number equals its own conjugate exactly when it is real, since flipping a zero imaginary part changes nothing. It equals the negative of its conjugate exactly when it is pure imaginary. These two tests turn a question about components into a single equation, which is often easier to apply inside a proof.Read more →

Why do complex roots of real polynomials come in conjugate pairs?

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Because conjugation passes through addition and multiplication unchanged, so conjugating the whole polynomial equation leaves the real coefficients alone while turning a root into its conjugate. If one satisfies the equation the other must too. This is why real polynomials always have an even number of non-real roots.Read more →

Do complex roots still come in conjugate pairs if a coefficient is not real?

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No, the pairing needs every coefficient to be real. In z2−2iz−1z^2 - 2iz - 1 the number ii is a root, but −i-i gives (−i)2−2i(−i)−1=−1−2−1=−4(-i)^2 - 2i(-i) - 1 = -1 - 2 - 1 = -4, not 00. The polynomial is (z−i)2(z - i)^2, so ii is a double root with no partner. The proof breaks at one step: conjugating the coefficient −2i-2i gives 2i2i, and the conjugated polynomial is a different one.Read more →

Is complex conjugation differentiable?

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No, at no point at all, even though it looks as simple as z↦zz \mapsto z. Its difference quotient is z+h‾−zˉh=hˉh\frac{\overline{z + h} - \bar{z}}{h} = \frac{\bar{h}}{h}, which equals 11 when hh is real and −1-1 when hh is imaginary, so there is no single limit as h→0h \to 0. Geometrically, a complex derivative can only rotate and scale small shapes, and a mirror turns them over.Read more →

Why is half of the Fourier transform of a real signal redundant?

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Because of conjugate symmetry. For real samples xnx_n, conjugating the transform leaves each xnx_n alone and only flips the sign in the exponent of e−2πikn/Ne^{-2\pi i kn/N}, which gives XN−k=Xk‾X_{N-k} = \overline{X_k}. The second half of the outputs is then the mirror image of the first, so NN real samples need only about N2\frac{N}{2} of them, and fast routines for real data compute just those.Read more →