Multiplicative Inverse: Reciprocal of ℂ

Multiplicative Inverse in Real Numbers

For any nonzero real number

a

, the multiplicative inverse is

\frac{1}{a}

, also written

a^{-1}

. The defining property is:

Multiplicative Inverse in Real Numbers

a \cdot \frac{1}{a} = 1

The product of a number and its multiplicative inverse always equals one.

The inverse of

5

is

\frac{1}{5}

, since

5 \cdot \frac{1}{5} = 1

. The inverse of

-3

is

-\frac{1}{3}

, since

(-3) \cdot \left(-\frac{1}{3}\right) = 1

. The inverse of

\frac{2}{3}

is

\frac{3}{2}

, since

\frac{2}{3} \cdot \frac{3}{2} = 1

. Inversion swaps numerator and denominator.

On the number line, the multiplicative inverse relates to distance from the origin in a reciprocal fashion. Numbers with absolute value greater than one have inverses with absolute value less than one, and vice versa. The number

5

lies five units from the origin; its inverse

\frac{1}{5}

lies only one-fifth of a unit away. The number

\frac{1}{4}

lies close to the origin; its inverse

4

lies far away. Numbers with absolute value exactly one — just

1

and

-1

— are their own multiplicative inverses.

Zero has no multiplicative inverse. No real number

x

satisfies

0 \cdot x = 1

, because zero times anything is zero. This exception carries over to complex numbers.

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Reciprocal distance

For a nonzero real number

a

, the inverse

\frac{1}{a}

satisfies

a \cdot \frac{1}{a} = 1

. On the number line, its distance from the origin is the reciprocal of the distance of

a

.

Inverse of a real number

a \cdot \dfrac{1}{a} = 1

split by $|a|$ compared with $1$

↓

Case 1

|a| > 1

Inverse is closer to 0

e.g.

5 \to \frac{1}{5}

↑

Case 2

0 < |a| < 1

Inverse is farther from 0

e.g.

\frac{1}{4} \to 4

=

Case 3

|a| = 1

Its own inverse

just

1

and

-1

✗

Case 4

a = 0

No inverse

no real

x

satisfies

0 \cdot x = 1

Carries over

The exception for zero carries over to complex numbers.

The inverse of a real number·/complex-numbers/multiplicative-inverseLearn Math Class

Multiplicative Inverse of a Complex Number

For real numbers, the inverse is a flipped fraction — so it seems natural to flip each part of a complex number.

The inverse of

5

is

\frac{1}{5}

and the inverse of

4

is

\frac{1}{4}

. A complex number like

3 + 4i

is built from a

3

and a

4

, so a sensible first guess is

\frac{1}{3} + \frac{1}{4}i

.

Multiplying back shows the guess is wrong: the product is not 1.

An inverse must multiply with

3 + 4i

to give

1

. Expand the product piece by piece.

piece of (3 + 4i)(1/3 + i/4)	value
3 · 1/3	1
3 · i/4	3i/4
4i · 1/3	4i/3
4i · i/4	i² = −1
total	25i/12, not 1

Pitfall: Inverting each part separately

Writing

\frac{1}{3 + 4i}

as

\frac{1}{3} + \frac{1}{4}i

inverts the two parts on their own, and

(3 + 4i)\left(\frac{1}{3} + \frac{1}{4}i\right) = \frac{25}{12}i

, not

1

. The conjugate formula gives

\frac{3}{25} - \frac{4}{25}i

.

The failure was predictable: 3 + 4i is a sum, and a sum cannot be inverted piece by piece.

The same thing happens with plain numbers.

\frac{1}{2 + 3}

is

\frac{1}{5}

, not

\frac{1}{2} + \frac{1}{3}

. When two complex numbers are multiplied, every part meets every other part and

i \cdot i

turns into

-1

, so the real and imaginary parts cannot be handled one at a time. Any inverse has to be found for the number as a whole.

	invert the whole	invert each piece
real: 2 + 3	1/5 = 0.2	1/2 + 1/3 ≈ 0.83
complex: 3 + 4i	3/25 − 4i/25	1/3 + i/4

The definition

The multiplicative inverse of a nonzero complex number

z

, written

z^{-1}

or

\frac{1}{z}

, is the complex number

w

with

z \cdot w = 1

.

For a nonzero complex number

z = a + bi

, the multiplicative inverse is the unique complex number

w

satisfying

z \cdot w = 1

. The formula is:

Multiplicative Inverse

z^{-1} = \frac{\overline{z}}{|z|^2} = \frac{a - bi}{a^2 + b^2}

Learn more about this formula: Multiplicative Inverse →

Browse all complex numbers formulas →

This expression uses the conjugate

\overline{z} = a - bi

in the numerator and the square of the modulus

|z|^2 = a^2 + b^2

in the denominator.

Separating into real and imaginary parts:

Multiplicative Inverse of a Complex Number

z^{-1} = \frac{a}{a^2 + b^2} - \frac{b}{a^2 + b^2}i

So

(3 + 4i)^{-1} = \frac{3}{25} - \frac{4}{25}i

, not

\frac{1}{3} + \frac{1}{4}i

.

z⁻¹ is again a complex number, worked out from the conjugate and the modulus.

Consider

z = 3 + 4i

. The conjugate is

\overline{z} = 3 - 4i

and the modulus squared is

|z|^2 = 9 + 16 = 25

. The multiplicative inverse is:

Multiplicative Inverse of a Complex Number

z^{-1} = \frac{3 - 4i}{25} = \frac{3}{25} - \frac{4}{25}i

Verification:

(3 + 4i)\left(\frac{3}{25} - \frac{4}{25}i\right) = \frac{9}{25} - \frac{12}{25}i + \frac{12}{25}i - \frac{16}{25}i^2 = \frac{9}{25} + \frac{16}{25} = 1

.

For

z = 2i

, a pure imaginary number, the conjugate is

-2i

and the modulus squared is

4

. The inverse is

\frac{-2i}{4} = -\frac{1}{2}i

. For

z = 1 + i

, the conjugate is

1 - i

and the modulus squared is

2

. The inverse is

\frac{1 - i}{2} = \frac{1}{2} - \frac{1}{2}i

.

Read backwards, the definition is a test — and it shows why 0 is left out.

A number

w

is the inverse of

z

exactly when

z \cdot w = 1

, so any answer can be checked by multiplying it back; the guess

\frac{1}{3} + \frac{1}{4}i

fails that check,

\frac{3}{25} - \frac{4}{25}i

passes it. The same test explains the exclusion:

0 \cdot w = 0

for every

w

, so no number passes it for

z = 0

, as the section on why zero has no multiplicative inverse explains.

The next section shows where the formula comes from; after it come the notation, the picture of the inverse, its properties, the comparison with the conjugate, zero, division and uniqueness.

Why the Formula Works

The formula

z^{-1} = \frac{\overline{z}}{|z|^2}

is not arbitrary. It emerges from a natural strategy: eliminate the imaginary unit from the denominator.

Starting with

\frac{1}{z} = \frac{1}{a + bi}

, the goal is to rewrite this with a real denominator. Multiply numerator and denominator by the conjugate:

Why the Formula Works

\frac{1}{a + bi} = \frac{1}{a + bi} \cdot \frac{a - bi}{a - bi} = \frac{a - bi}{(a + bi)(a - bi)}

The denominator simplifies using the identity

z \cdot \overline{z} = |z|^2

:

Why the Formula Works

(a + bi)(a - bi) = a^2 - (bi)^2 = a^2 - b^2i^2 = a^2 + b^2

This is a real number — the square of the modulus. The result:

Why the Formula Works

\frac{1}{z} = \frac{a - bi}{a^2 + b^2} = \frac{\overline{z}}{|z|^2}

Pitfall: Forgetting that i squared is −1

Expanding

(2 + 5i)(2 - 5i)

to

4 - 25 = -21

takes

(5i)^2

to be

25

. Since

(5i)^2 = 25i^2 = -25

, the denominator is

4 + 25 = 29

, and

\frac{1}{2 + 5i} = \frac{2 - 5i}{29}

.

The key insight is that multiplying any complex number by its conjugate produces a real number. This property converts division by a complex number into division by a real number, which is straightforward.

Verification confirms the formula works:

Why the Formula Works

z \cdot z^{-1} = z \cdot \frac{\overline{z}}{|z|^2} = \frac{z \cdot \overline{z}}{|z|^2} = \frac{|z|^2}{|z|^2} = 1

The product of

z

and

\overline{z}

in the numerator equals

|z|^2

, canceling the denominator exactly.

The derivation above one step at a time, ending with the check that

z \cdot z^{-1} = 1

; use Next and Back to move through it.

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Derive

\displaystyle z^{-1} = \frac{\overline{z}}{|z|^2}

1 / 7

Multiply by the conjugate

\displaystyle\frac{1}{a + bi} = \frac{1}{a + bi} \cdot \frac{a - bi}{a - bi}

The goal is to rewrite

\frac{1}{a + bi}

with a real denominator.

\displaystyle z^{-1} = \frac{\overline{z}}{|z|^2}

Verification confirms the formula works.

Deriving the formula for the inverseLearn Math Class

Multiplying by the conjugate is the whole trick: it turns the denominator into the real number

|z|^2

.

Multiplicative Inverse Notation

Notation

Multiplicative Inverse Notation

The corner exponent, its fraction twin, and the parentheses that keep both honest. The

-1

superscript here joins a mark that crosses four topics — numbers, functions, trigonometry, matrices — each with its own meaning.

z

,

a + bi

and

\mathbb{C}

come from complex numbers basics,

i

from the imaginary unit; the conjugate bar and the modulus in the formula

z^{-1} = \bar{z}/|z|^2

belong to their pages.

z^{-1}

z to the minus one; z inverse

The

-1

superscript as an exponent in earnest:

z^{-1}

is the number with

z \cdot z^{-1} = 1

, computed by

\bar{z}/|z|^2

— Multiplicative Inverse of a Complex Number above. The same corner mark negative powers put on real numbers, now paid for by a conjugate.

CasesExtends along the exponent ladder —

z^{-n} = (z^{-1})^n = (z^n)^{-1}

; in polar dress the corner distributes:

z^{-1} = r^{-1}e^{-i\theta}

, modulus inverted, argument negated.

Do not confuseThe conjugate.

z^{-1} = \bar{z}/|z|^2

makes them parallel, and on the unit circle equal —

|z| = 1

forces

z^{-1} = \bar{z}

; off the circle they differ in length, and Multiplicative Inverse vs. Complex Conjugate below separates them.

Same glyph elsewhereThe corner changes job by base: undo-the-map on functions (

f^{-1}

), matrix inversion on

A^{-1}

— on numbers alone it is honest arithmetic.

\frac{1}{z}

one over z

The division spelling of the same number, wearing the fraction bar — the notation of choice when the next step is division, since

\frac{w}{z} = w \cdot \frac{1}{z}

.

CasesDisplayed fractions scope themselves —

\frac{1}{3+2i}

is unambiguous; inline, the slash binds tight:

1/3 + 2i

means

\frac{1}{3} + 2i

, so inline work writes

1/(3 + 2i)

.

Do not confuseA finished answer.

\frac{1}{z}

is a division still owed — evaluating it requires the conjugate trick of Why the Formula Works above; the bar promises the operation, not the result.

(3 + 2i)^{-1}

the inverse of three plus two i

Parentheses decide what the corner grips.

(3 + 2i)^{-1}

inverts the whole number; unparenthesized, the exponent binds only to the nearest symbol —

3 + 2i^{-1} = 3 + 2(-i) = 3 - 2i

, a different number entirely.

CasesNamed variables carry no ambiguity —

z^{-1}

grips all of

z

; the hazard lives where literals are inverted inline, the same place the slash trap above strikes.

Do not confuseA decoration. The

-1

is an operative exponent, not a label — dropping the parentheses does not roughly invert the number, it silently computes something else.

All complex number symbols →↑ Complex Numbers

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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, \; w

z, w: complex numbers

The usual letters for complex numbers, as

x

and

y

are for reals.

a + bi

a plus b i

Real part

a

, imaginary part

b

; both are real numbers.

\mathbb{C}

C, the complex numbers

The set of all

a + bi

; it contains

\mathbb{R}

.

Imaginary numbers

i

i, the imaginary unit

The number with

i^2 = -1

.

\sqrt{-1}

the square root of minus one

Another name for

i

; root rules for reals fail with it.

bi

b i

A pure imaginary number: real part

0

.

Algebraic form

\operatorname{Re}(z), \; \operatorname{Im}(z)

real part, imaginary part

For

z = a + bi

:

a

and

b

.

\operatorname{Im}(z)

is

b

, not

bi

.

a + (-b)i

a plus negative b, i

Written

a - bi

; the imaginary part is

-b

.

z = x + yi

z equals x plus y i

Same form with plane letters: the point

(x, y)

.

Complex conjugate

\bar{z}

z bar, the conjugate

For

z = a + bi

,

\bar{z} = a - bi

: the mirror image across the real axis.

\overline{z_1 + z_2}

the conjugate of the sum

The bar covers the whole expression; it equals

\bar{z}_1 + \bar{z}_2

.

z^*

z star

The same conjugate, as written in physics and engineering.

Absolute value

|z|

the modulus of z

Distance from

0

:

\sqrt{a^2 + b^2}

.

|z|^2 = z\bar{z}

mod z squared

Equals

a^2 + b^2

, with no root to take.

|z - c| = R

mod z minus c equals R

The circle of radius

R

around

c

.

Additive inverse

-z

negative z

Both parts change sign:

-a - bi

.

-(a + bi)

minus, quantity a plus b i

The brackets carry the minus to both parts.

z - w

z minus w

Adding the inverse:

z + (-w)

.

Multiplicative inverseThis page

z^{-1}, \; \frac{1}{z}

z inverse; one over z

The number with

z \cdot z^{-1} = 1

; it equals

\frac{\bar{z}}{|z|^2}

.

(3 + 2i)^{-1}

the inverse of three plus two i

Found by multiplying top and bottom by the conjugate:

\frac{3 - 2i}{13}

.

Trigonometric form

r, \; \theta

r and theta, the polar pair

Distance from

0

and angle from the positive real axis.

\arg(z)

the argument of z

Any angle of

z

; there are infinitely many,

2\pi

apart.

\operatorname{Arg}(z)

the principal argument

One chosen angle, usually in

(-\pi, \pi]

; some texts use

[0, 2\pi)

.

\operatorname{cis}\theta

sis theta

Short for

\cos\theta + i\sin\theta

.

Exponential form

e^{i\theta}

e to the i theta

Equals

\operatorname{cis}\theta

(Euler's formula): a point on the unit circle.

re^{i\theta}

r e to the i theta

The whole number

z

: length

r

, angle

\theta

.

e^{i(\theta + 2\pi k)}

theta plus two pi k

The same number for every integer

k

.

De Moivre's theorem

z_k

z sub k, the k-th root

The roots of

z^n = w

, numbered

k = 0, \ldots, n - 1

.

\omega

omega

A primitive root of unity: its powers give all

n

roots of

1

.

R^{1/n}

R to the one over n

The real root of the length; the complex root

w^{1/n}

has

n

values.

Same marks, different job

|z|

The modulus here; for a real number, the absolute value; for a matrix, the determinant.

\bar{z}

The conjugate here; in statistics a bar means a mean, in sets a complement.

z^{-1}

vs

z^*

The reciprocal, against the conjugate: they agree only on the unit circle.

w^{1/n}

n

values for a complex

w

; a single real root for a positive real number.

Complex number symbols table Complex numbers

Complex Number Notation at a Glance·/complex-numbers/basicsLearn Math Class

z = 3 + 4i and z⁻¹ = 0.12 − 0.16i

The inverse points in the mirrored direction, argument −θ, and its length is the reciprocal 1/5 of the original 5, which is why it needs a zoom to be seen: the modulus inverts and the argument negates. That is the formula z⁻¹ = z̄/|z|² as a picture. Watch the inverse shrink as z grows on the De Moivre visualizer.

Points inside the unit circle have inverses outside it, and the circle itself maps onto itself, although only 1 and -1 stay in place.

Geometric Interpretation

The multiplicative inverse transforms both the distance and direction of a complex number in the complex plane.

The modulus inverts:

Modulus of Inverse

|z^{-1}| = \frac{1}{|z|}

Learn more about this formula: Modulus of Inverse →

Browse all complex numbers formulas →

A number far from the origin has an inverse close to the origin, and vice versa. If

|z| = 5

, then

|z^{-1}| = \frac{1}{5}

. If

|z| = \frac{1}{3}

, then

|z^{-1}| = 3

. Numbers with modulus greater than one have inverses with modulus less than one. Numbers with modulus less than one have inverses with modulus greater than one.

Numbers on the unit circle — those with

|z| = 1

— have inverses also on the unit circle, since

|z^{-1}| = \frac{1}{1} = 1

. These numbers stay the same distance from the origin after inversion.

The argument negates:

Geometric Interpretation

\arg(z^{-1}) = -\arg(z)

Pitfall: Inverting only the modulus

For

|z| = 2

and

\arg(z) = \frac{\pi}{3}

, taking the number with modulus

\frac{1}{2}

and the same argument gives a product with

z

of argument

\frac{2\pi}{3}

, which is not

1

. The argument negates too:

\arg(z^{-1}) = -\frac{\pi}{3}

.

If

z

points in direction

\theta

, then

z^{-1}

points in direction

-\theta

. A number in the upper half-plane has its inverse in the lower half-plane, and vice versa. A number at angle

\frac{\pi}{4}

(northeast) has its inverse at angle

-\frac{\pi}{4}

(southeast).

Combining both effects: the multiplicative inverse reflects

z

across the real axis and scales its distance from the origin by

\frac{1}{|z|^2}

. Alternatively, in trigonometric form, if

z = r\,\text{cis}\,\theta

, then

z^{-1} = \frac{1}{r}\,\text{cis}(-\theta)

.

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The modulus inverts

The inverse of

z

has modulus

\dfrac{1}{|z|}

: a number far from the origin has an inverse close to it, and vice versa. Comparing

|z|

with

1

decides which.

Multiplicative inverse

|z^{-1}| = \dfrac{1}{|z|}

split by $|z|$ compared with $1$

↓

Case 1

|z| > 1

Moves closer

the inverse has modulus less than one

e.g.

|z| = 5 \Rightarrow |z^{-1}| = \frac{1}{5}

↑

Case 2

|z| < 1

Moves farther out

the inverse has modulus greater than one

e.g.

|z| = \frac{1}{3} \Rightarrow |z^{-1}| = 3

=

Case 3

|z| = 1

Stays on the unit circle

the same distance from the origin after inversion

The argument negates

In every case

\arg(z^{-1}) = -\arg(z)

: the inverse reflects

z

across the real axis and scales its distance by

\dfrac{1}{|z|^2}

.

The inverse of z: closer, farther or on the circle·/complex-numbers/multiplicative-inverseLearn Math Class

Properties of Multiplicative Inverse

The multiplicative inverse interacts with other operations in predictable ways. These properties parallel the corresponding rules for real numbers.

Double inversion returns the original number:

Properties of Multiplicative Inverse

(z^{-1})^{-1} = z

Taking the multiplicative inverse twice leaves

z

unchanged. If

w

is the inverse of

z

, then

z

is the inverse of

w

.

The inverse of a product equals the product of inverses:

Properties of Multiplicative Inverse

(z_1 \cdot z_2)^{-1} = z_1^{-1} \cdot z_2^{-1}

To invert a product, invert each factor. Order does not matter since multiplication is commutative.

Inversion and conjugation commute:

Properties of Multiplicative Inverse

(\overline{z})^{-1} = \overline{z^{-1}}

Conjugating first and then inverting produces the same result as inverting first and then conjugating.

The modulus inverts:

Modulus of Inverse

|z^{-1}| = \frac{1}{|z|}

Learn more about this formula: Modulus of Inverse →

Browse all complex numbers formulas →

The argument negates:

Properties of Multiplicative Inverse

\arg(z^{-1}) = -\arg(z)

Powers and inverses interact naturally:

Properties of Multiplicative Inverse

(z^n)^{-1} = (z^{-1})^n = z^{-n}

The inverse of

z^n

equals the

n

th power of

z^{-1}

, and both equal

z^{-n}

. This consistency allows negative exponents to work seamlessly with complex numbers, just as they do with real numbers.

Property	Formula	Meaning
Double inversion	(z⁻¹)⁻¹ = z	inverting twice returns the original number
Inverse of a product	(z₁ · z₂)⁻¹ = z₁⁻¹ · z₂⁻¹	to invert a product, invert each factor
Commute with conjugation	(z̄)⁻¹ = z⁻¹	conjugate-then-invert equals invert-then-conjugate
Modulus inverts	\|z⁻¹\| = 1 ⁄ \|z\|	distance from origin flips reciprocally
Argument negates	arg(z⁻¹) = −arg(z)	direction reflects across the real axis
Powers and negative exponents	(zⁿ)⁻¹ = (z⁻¹)ⁿ = z⁻ⁿ	negative exponents behave consistently for complex bases

Multiplicative Inverse vs. Complex Conjugate

Both the multiplicative inverse and the complex conjugate involve transforming a complex number, but they serve different purposes and produce different results.

The conjugate of

z = a + bi

is

\overline{z} = a - bi

. Only the imaginary part changes sign. The real part stays fixed.

The multiplicative inverse of

z = a + bi

is

z^{-1} = \frac{a - bi}{a^2 + b^2}

. Both parts are divided by

a^2 + b^2

, and the imaginary part also changes sign.

For

z = 3 + 4i

: the conjugate is

\overline{z} = 3 - 4i

, while the inverse is

z^{-1} = \frac{3 - 4i}{25} = \frac{3}{25} - \frac{4}{25}i

. These differ by a factor of

25 = |z|^2

.

Pitfall: Stopping at the conjugate

Taking

z^{-1} = \overline{z}

for

z = 1 + 2i

gives

1 - 2i

, but

(1 + 2i)(1 - 2i) = 5

, not

1

. Dividing by

|z|^2 = 5

gives

z^{-1} = \frac{1}{5} - \frac{2}{5}i

.

Geometrically, conjugation reflects across the real axis while preserving the modulus:

|\overline{z}| = |z|

. Inversion also reflects across the real axis but additionally scales:

|z^{-1}| = \frac{1}{|z|}

. The conjugate stays on the same circle centered at the origin; the inverse moves to a different circle.

The relationship between them is:

Multiplicative Inverse vs. Complex Conjugate

z^{-1} = \frac{\overline{z}}{|z|^2}

When are they equal? Only when

|z|^2 = 1

, meaning

|z| = 1

.

COROLLARY

On the unit circle,

z^{-1} = \overline{z}

.

For

z = \frac{3}{5} + \frac{4}{5}i

, which has modulus

1

, the inverse and conjugate are both

\frac{3}{5} - \frac{4}{5}i

. Off the unit circle, they always differ.

Aspect	Conjugate z̄	Multiplicative inverse z⁻¹
Formula (for z = a + bi)	a − bi	(a − bi) ⁄ (a² + b²)
Real part	a (unchanged)	a ⁄ (a² + b²)
Imaginary part	−b	−b ⁄ (a² + b²)
Modulus	\|z̄\| = \|z\|	\|z⁻¹\| = 1 ⁄ \|z\|
Geometry	reflection across the real axis (same circle)	reflection + rescale to a different circle
Defined for which z	every z (including 0)	every z ≠ 0
When they coincide	z⁻¹ = z̄ if and only if \|z\| = 1 (on the unit circle)
Example (z = 3 + 4i)	3 − 4i	(3 − 4i) ⁄ 25 = 3⁄25 − (4⁄25)i

The inverse and the conjugate always point the same way; they differ only in length, and agree only on the unit circle.

Why Zero Has No Multiplicative Inverse

THEOREM

Zero is the sole complex number without a multiplicative inverse.

No complex number

w

satisfies

0 \cdot w = 1

.

The reason is fundamental: zero times any number equals zero, never one. Regardless of what

w

might be, the product

0 \cdot w

vanishes. The equation

0 \cdot w = 1

has no solution.

Fact: On a sphere, inverting is a half-turn

Complex analysis often adds one point

\infty

with

\frac{1}{0} = \infty

and draws

\mathbb{C}

as a sphere with

\infty

at the top. There

z \mapsto \frac{1}{z}

is a half-turn of the sphere about the axis through

1

and

-1

, swapping

0

and

\infty

. The extended system is no field:

0 \cdot \infty

stays undefined.

The formula

z^{-1} = \frac{\overline{z}}{|z|^2}

reveals the problem algebraically. For

z = 0

, the modulus is

|z| = 0

, so

|z|^2 = 0

appears in the denominator. Division by zero is undefined, and the formula produces no result.

This exception is unavoidable in any field. The field axioms guarantee multiplicative inverses for all nonzero elements, but explicitly exclude zero. The additive identity cannot have a multiplicative inverse — the two roles are incompatible.

The absence of

0^{-1}

is why division by zero remains undefined for complex numbers, just as for real numbers. The expression

\frac{z}{0}

has no meaning. Any calculation that leads to division by zero has gone astray.

Dividing by zero: nothing to draw

The division visualizer with z₁ = 4 + 2i and the denominator z₂ set to 0. Every other quotient has a length |z₁|/|z₂| and an angle θ₁ − θ₂; here the length would need a division by |z₂| = 0 and z₂ has no angle at all, so the tool draws no quotient. Move z₂ off the origin on the complex division visualizer.

The same is true of 0⁻¹ = 1/0: with z₁ = 1 the figure stays empty, because no w gives 0 · w = 1.

Connection to Division

Division of complex numbers is multiplication by the multiplicative inverse:

Connection to Division

\frac{z_1}{z_2} = z_1 \cdot z_2^{-1}

This definition reduces division to multiplication, which is already well understood. Every property of division follows from properties of multiplication combined with properties of the inverse.

Substituting the formula for

z_2^{-1}

:

Connection to Division

\frac{z_1}{z_2} = z_1 \cdot \frac{\overline{z_2}}{|z_2|^2} = \frac{z_1 \cdot \overline{z_2}}{|z_2|^2}

This is the standard method for dividing complex numbers: multiply the numerator by the conjugate of the denominator, then divide by the modulus squared.

Consider

\frac{5 + 3i}{1 + 2i}

. The conjugate of the denominator is

1 - 2i

, and

|1 + 2i|^2 = 1 + 4 = 5

. Computing:

Connection to Division

\frac{5 + 3i}{1 + 2i} = \frac{(5 + 3i)(1 - 2i)}{5} = \frac{5 - 10i + 3i - 6i^2}{5} = \frac{5 - 7i + 6}{5} = \frac{11 - 7i}{5} = \frac{11}{5} - \frac{7}{5}i

Pitfall: Dividing by the modulus, not its square

Dividing

(5 + 3i)(1 - 2i) = 11 - 7i

by

|1 + 2i| = \sqrt{5}

gives

\frac{11 - 7i}{\sqrt{5}}

, too long by a factor of

\sqrt{5}

. The divisor is

|1 + 2i|^2 = 5

, so the quotient is

\frac{11}{5} - \frac{7}{5}i

.

The work splits into two products. The numerator

5 + 3i

is multiplied by the conjugate

1 - 2i

, giving

11 - 7i

, and the denominator

1 + 2i

is multiplied by the same conjugate, giving the real number

5

. Each part of

11 - 7i

is then divided by that real number, and the quotient lands in standard form.

The connection explains why division by zero is undefined: zero has no multiplicative inverse, so

z_1 \cdot 0^{-1}

is meaningless.

For detailed coverage of division and other arithmetic, see operations.

z₁/z₂: arguments subtract, moduli divide

The quotient sits at the angle θ₁ − θ₂ and at distance |z₁|/|z₂|: division undoes the rotation and the scaling that multiplication performs. Multiplying numerator and denominator by the conjugate is the algebra that produces this picture. Move the divisor and watch the quotient counter-rotate on the complex division tool.

Every division is a multiplication by an inverse, which is why the two share one picture.

Uniqueness

THEOREM

Every nonzero complex number has exactly one multiplicative inverse.

Existence is established by the formula: for any

z = a + bi

with

z \neq 0

, the number

z^{-1} = \frac{\overline{z}}{|z|^2}

satisfies

z \cdot z^{-1} = 1

. Since

z \neq 0

, at least one of

a

or

b

is nonzero, so

|z|^2 = a^2 + b^2 > 0

and the formula is well-defined.

Uniqueness follows from a short argument. Suppose

w_1

and

w_2

both satisfy

z \cdot w_1 = 1

and

z \cdot w_2 = 1

. Then:

Uniqueness

w_1 = w_1 \cdot 1 = w_1 \cdot (z \cdot w_2) = (w_1 \cdot z) \cdot w_2 = 1 \cdot w_2 = w_2

Any two multiplicative inverses of the same number must be equal.

The uniqueness chain above, split at each equals sign, with the rule used at every step.

DownloadPNG2×PDF

ProveIf

z \cdot w_1 = 1

and

z \cdot w_2 = 1

, then

w_1 = w_2

1

w_1 = w_1 \cdot 1

identityOne is the multiplicative identity

2

= w_1 \cdot (z \cdot w_2)

substituteSince

z \cdot w_2 = 1

3

= (w_1 \cdot z) \cdot w_2

key stepAssociativity of multiplication

4

= 1 \cdot w_2

substituteSince

w_1

is an inverse of

z

5

= w_2

identityOne is the multiplicative identity

w_1 = w_2

Any two multiplicative inverses of the same number must be equal.

Proving the inverse is uniqueLearn Math Class

The gold line is associativity: regrouping

w_1 \cdot (z \cdot w_2)

as

(w_1 \cdot z) \cdot w_2

is what lets the two inverses meet.

This uniqueness is guaranteed by the field axioms. In any field, multiplicative inverses of nonzero elements are unique. The argument above uses only associativity of multiplication and the property of one as the multiplicative identity.

Uniqueness justifies the notation

z^{-1}

and

\frac{1}{z}

. If multiple inverses existed, these expressions would be ambiguous. Because exactly one inverse exists, the notation refers to a single, well-defined complex number.

Formula Summary

This summary collects the formulas for the multiplicative inverse in four families, shown one at a time with the buttons above the cards. Each card gives a formula with a line on what it says and a link to the section that explains it.

The first family finds the inverse of a nonzero number from its conjugate and its squared modulus, in two layouts, and notes when it equals the conjugate. The second gives its distance from the origin and its angle. The third collects the rules for a double inverse, products, conjugates and powers, and the last turns division into multiplication by the inverse.

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The inverse3 formulas

The reciprocal of a nonzero complex number.

\displaystyle z^{-1} = \frac{\overline{z}}{|z|^2} = \frac{a - bi}{a^2 + b^2}

The conjugate divided by the squared modulus, for

z \neq 0

, as the multiplicative inverse of a complex number shows.

\displaystyle z^{-1} = \frac{a}{a^2 + b^2} - \frac{b}{a^2 + b^2}i

The same inverse in standard form, with the real and the imaginary part apart, from the same section.

z^{-1} = \overline{z}

On the unit circle, where

|z| = 1

, the inverse equals the conjugate; off it they always differ (inverse vs. conjugate).

In the plane2 formulas

Where the inverse sits compared with the number.

\displaystyle |z^{-1}| = \frac{1}{|z|}

The inverse lies at the reciprocal distance from the origin, as the geometric interpretation shows.

\arg(z^{-1}) = -\arg(z)

Its angle is the negative of the original one, a reflection across the real axis (geometric interpretation).

Rules4 formulas

How the inverse combines with other operations.

(z^{-1})^{-1} = z

Inverting twice gives the number back, from properties of multiplicative inverse.

(z_1 \cdot z_2)^{-1} = z_1^{-1} \cdot z_2^{-1}

The inverse of a product is the product of the inverses, as the properties show.

(\overline{z})^{-1} = \overline{z^{-1}}

Inverting and taking the conjugate give the same result in either order (properties of multiplicative inverse).

(z^n)^{-1} = (z^{-1})^n = z^{-n}

The inverse of a power is the same power of the inverse, from the properties.

Division2 formulas

Dividing means multiplying by the inverse.

\displaystyle \frac{z_1}{z_2} = z_1 \cdot z_2^{-1}

Division by

z_2 \neq 0

is multiplication by its inverse, as connection to division shows.

\displaystyle \frac{z_1}{z_2} = z_1 \cdot \frac{\overline{z_2}}{|z_2|^2} = \frac{z_1 \cdot \overline{z_2}}{|z_2|^2}

Putting in the inverse formula gives the usual rule: multiply by the conjugate and divide by the squared modulus (connection to division).

Multiplicative inverse formulas·All complex number formulasLearn Math Class

A quick check on any inverse is to multiply it by the original number: the product must be exactly one. Forgetting to divide the conjugate by the squared modulus is the usual slip, and this check catches it at once unless the number lies on the unit circle.

Common Mistakes

Several errors recur when students compute multiplicative inverses of complex numbers.

Forgetting to use the conjugate is common. Students attempt

\frac{1}{3 + 2i} = \frac{1}{3} + \frac{1}{2}i

, treating real and imaginary parts separately. This is incorrect. The inverse of a sum is not the sum of inverses. The correct approach multiplies by

\frac{3 - 2i}{3 - 2i}

to obtain

\frac{3 - 2i}{13}

.

Errors in computing

|z|^2

are frequent. For

z = 3 - 4i

, the modulus squared is

3^2 + (-4)^2 = 9 + 16 = 25

, not

9 - 16 = -7

. The formula is

a^2 + b^2

, always a sum, never a difference. Signs in the original number do not affect the addition.

Confusing

z^{-1}

with

-z

conflates two unrelated operations. The multiplicative inverse

z^{-1}

satisfies

z \cdot z^{-1} = 1

. The additive inverse

-z

satisfies

z + (-z) = 0

. For

z = 2 + i

, the multiplicative inverse is

\frac{2 - i}{5}

; the additive inverse is

-2 - i

. These share nothing but a sign change in the imaginary part.

Confusing

z^{-1}

with

\overline{z}

is another pitfall. They are equal only on the unit circle. For

z = 3 + 4i

, the conjugate is

3 - 4i

while the inverse is

\frac{3 - 4i}{25}

. Forgetting to divide by

|z|^2

leaves the answer off by a factor of

25

.

Leaving a complex denominator unreduced is incomplete. A final answer like

\frac{2 + 3i}{1 - i}

requires rationalization. Multiply by

\frac{1 + i}{1 + i}

to obtain a real denominator and express the result in standard algebraic form.

Mistake	Wrong	Correct	Why
Inverting componentwise	1 ⁄ (3 + 2i) = 1⁄3 + (1⁄2)i	1 ⁄ (3 + 2i) = (3 − 2i) ⁄ 13	the inverse of a sum is not the sum of inverses
Sign error in \|z\|²	for 3 − 4i: 9 − 16 = −7	9 + 16 = 25	\|z\|² = a² + b² is always a sum; signs in z don't affect it
Confusing z⁻¹ with −z	inverse of 2 + i is −2 − i	inverse of 2 + i is (2 − i) ⁄ 5	−z is the additive inverse; z⁻¹ is the multiplicative inverse
Confusing z⁻¹ with z̄	for 3 + 4i: z⁻¹ = 3 − 4i	z⁻¹ = (3 − 4i) ⁄ 25	z⁻¹ = z̄ ⁄ \|z\|² — must divide by the modulus squared
Leaving a complex denominator	(2 + 3i) ⁄ (1 − i) as final answer	multiply by (1 + i) ⁄ (1 + i) to rationalize	standard form a + bi requires a real denominator

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1Inverting only the modulus

In polar form, the modulus of

z

is replaced by its reciprocal, but the argument is copied unchanged. The result has the right distance from the origin and points in the wrong direction.

Start

For

z = 2\,\text{cis}\,\frac{\pi}{3}

,

Wrong turn

Error here

z^{-1} = \frac{1}{2}\,\text{cis}\,\frac{\pi}{3}

.

Right path

The modulus inverts and the argument negates, so

z^{-1} = \frac{1}{2}\,\text{cis}\left(-\frac{\pi}{3}\right)

. Then

z \cdot z^{-1} = 2 \cdot \frac{1}{2}\,\text{cis}\left(\frac{\pi}{3} - \frac{\pi}{3}\right) = \text{cis}\,0 = 1

.

Why it happensFor real numbers the inverse is just the reciprocal, so the step is applied to the length alone. The arguments of real numbers,

0

and

\pi

, give the same direction when negated, so the angle never had to change before.

How to checkMultiply the answer by

z

. Moduli multiply and arguments add, so the wrong answer gives

2 \cdot \frac{1}{2}\,\text{cis}\left(\frac{\pi}{3} + \frac{\pi}{3}\right) = \text{cis}\,\frac{2\pi}{3}

, not

1

.

2Dividing by the modulus, not its square

The conjugate method divides by

|z|^2

, but the denominator is written as

|z|

, the modulus itself. The result points in the right direction but has the wrong length.

Wrong turn

Error here

\frac{5 + 3i}{1 + 2i} = \frac{(5 + 3i)(1 - 2i)}{|1 + 2i|} = \frac{11 - 7i}{\sqrt{5}}

.

Right path

(1 + 2i)(1 - 2i) = 1 + 4 = 5 = |1 + 2i|^2

, so

\frac{5 + 3i}{1 + 2i} = \frac{(5 + 3i)(1 - 2i)}{5} = \frac{11 - 7i}{5} = \frac{11}{5} - \frac{7}{5}i

.

Why it happensThe modulus is the quantity usually named and computed, and the square is easy to lose next to

\sqrt{a^2 + b^2}

. Multiplying the denominator by its conjugate gives

a^2 + b^2

itself, with no square root.

How to checkMultiply the answer by the denominator; it must give the numerator back.

\left(\frac{11}{5} - \frac{7}{5}i\right)(1 + 2i) = \frac{11}{5} + \frac{22}{5}i - \frac{7}{5}i - \frac{14}{5}i^2 = 5 + 3i

, while the wrong answer gives

\frac{(11 - 7i)(1 + 2i)}{\sqrt{5}} = \frac{25 + 15i}{\sqrt{5}} = \sqrt{5}\,(5 + 3i)

.

Common pitfalls with the multiplicative inverse·/complex-numbers/multiplicative-inverseLearn Math Class

Both slips leave out half of the inverse. In polar form the modulus inverts and the argument changes sign, and in algebraic form the conjugate is divided by

|z|^2

, not by

|z|

. Multiplying the answer by

z

, which must give exactly

1

, catches both.

Summary: Worked Examples of z⁻¹

The formula

z^{-1} = \overline{z}/|z|^2

produces predictable patterns across different kinds of complex numbers. The table below works the inverse for several representative cases — a positive real, a pure imaginary, points off the unit circle, a point on the unit circle, and a point in the second quadrant — collecting

\overline{z}

,

|z|^2

,

z^{-1}

, and a brief geometric note for each.

z	z̄	\|z\|²	z⁻¹ = z̄ ⁄ \|z\|²	Geometric note
5 (positive real)	5	25	1⁄5	on the positive real axis; inverse on the same ray, closer to origin
2i (pure imaginary)	−2i	4	−(1⁄2)i	on positive imaginary axis; inverse on negative imaginary axis
1 + i	1 − i	2	(1 − i) ⁄ 2 = 1⁄2 − (1⁄2)i	modulus √2; inverse on circle of radius 1⁄√2
3 + 4i	3 − 4i	25	(3 − 4i) ⁄ 25 = 3⁄25 − (4⁄25)i	modulus 5; inverse on circle of radius 1⁄5
(3⁄5) + (4⁄5)i (on unit circle)	3⁄5 − (4⁄5)i	1	3⁄5 − (4⁄5)i (= z̄)	\|z\| = 1, so inverse equals conjugate; both remain on the unit circle
−1 + 2i	−1 − 2i	5	(−1 − 2i) ⁄ 5 = −1⁄5 − (2⁄5)i	second quadrant; inverse moves to third quadrant (reflection + rescale)

Multiplicative Inverse FAQ

What is the inverse of a product?

+

The product of the inverses, in either order, since complex multiplication commutes. This mirrors the real case and follows from the moduli multiplying and the arguments adding, so inverting reverses both. It is what makes dividing by a product the same as dividing by each factor in turn. Try it on the division explorer.Read more →

Is $\frac{1}{3 + 4i}$ equal to $\frac{1}{3} + \frac{1}{4}i$ ?

+

No. Inverting each part on its own gives

(3 + 4i)\left(\frac{1}{3} + \frac{1}{4}i\right) = \frac{25}{12}i

, not

1

. A sum cannot be inverted piece by piece, just as

\frac{1}{2 + 3}

is

\frac{1}{5}

and not

\frac{1}{2} + \frac{1}{3}

. The inverse has to be found for the number as a whole:

\frac{1}{3 + 4i} = \frac{3 - 4i}{25} = \frac{3}{25} - \frac{4}{25}i

.Read more →

Which complex numbers are their own multiplicative inverse?

+

Only

1

and

-1

. A number with

z \cdot z = 1

solves

z^2 = 1

, whose only roots are

\pm 1

. Every other point of the unit circle keeps its distance from the origin under inversion but moves: there

z^{-1} = \bar{z}

, its mirror image across the real axis, so

i

goes to

-i

. Points inside the unit circle have inverses outside it, and points outside have inverses inside.Read more →

Can $\frac{1}{0}$ be defined as infinity for complex numbers?

+

Yes, at a cost. Complex analysis often adds one point

\infty

, sets

\frac{1}{0} = \infty

, and draws the result as a sphere with

\infty

at the top. On that sphere

z \mapsto \frac{1}{z}

is a half-turn about the axis through

1

and

-1

, swapping

0

and

\infty

. The extended system is not a field, though:

0 \cdot \infty

stays undefined, so ordinary algebra does not extend to

\infty

.Read more →

Why does $(3 + 2i)^{-1}$ need parentheses?

+

Because without them the exponent applies only to the

i

.

3 + 2i^{-1}

means

3 + 2 \cdot \frac{1}{i} = 3 + 2(-i) = 3 - 2i

, a different number, while

(3 + 2i)^{-1} = \frac{3 - 2i}{13}

. The slash sets the same trap in typed work:

1/3 + 2i

means

\frac{1}{3} + 2i

, so an inverse typed on one line is written

1/(3 + 2i)

.Read more →

Multiplicative Inverse of a Complex Number

The Number That Cancels Through Multiplication

Key Terms

Multiplicative Inverse in Real Numbers

Multiplicative Inverse of a Complex Number

Why the Formula Works

Multiplicative Inverse Notation

Multiplicative Inverse Notation

Geometric Interpretation

Properties of Multiplicative Inverse

Multiplicative Inverse vs. Complex Conjugate

Why Zero Has No Multiplicative Inverse

Connection to Division

Uniqueness

Formula Summary

Common Mistakes

Summary: Worked Examples of z⁻¹

Multiplicative Inverse FAQ

What is the inverse of a product?

Is $\frac{1}{3 + 4i}$ equal to $\frac{1}{3} + \frac{1}{4}i$ ?

Which complex numbers are their own multiplicative inverse?

Can $\frac{1}{0}$ be defined as infinity for complex numbers?

Why does $(3 + 2i)^{-1}$ need parentheses?

Multiplicative Inverse of a Complex Number

The Number That Cancels Through Multiplication

Key Terms

Multiplicative Inverse in Real Numbers

Multiplicative Inverse of a Complex Number

Why the Formula Works

Multiplicative Inverse Notation

Multiplicative Inverse Notation

Geometric Interpretation

Properties of Multiplicative Inverse

Multiplicative Inverse vs. Complex Conjugate

Why Zero Has No Multiplicative Inverse

Connection to Division

Uniqueness

Formula Summary

Common Mistakes

Summary: Worked Examples of z⁻¹

Multiplicative Inverse FAQ

What is the inverse of a product?

Is 13+4i\frac{1}{3 + 4i}3+4i1​ equal to 13+14i\frac{1}{3} + \frac{1}{4}i31​+41​i?

Which complex numbers are their own multiplicative inverse?

Can 10\frac{1}{0}01​ be defined as infinity for complex numbers?

Why does (3+2i)−1(3 + 2i)^{-1}(3+2i)−1 need parentheses?

Is $\frac{1}{3 + 4i}$ equal to $\frac{1}{3} + \frac{1}{4}i$ ?

Can $\frac{1}{0}$ be defined as infinity for complex numbers?

Why does $(3 + 2i)^{-1}$ need parentheses?