In short

To divide two complex numbers, multiply the numerator and denominator by the conjugate of the denominator. The denominator becomes a real number — the sum of squares of its parts — and the result falls into the standard form a + bi. Division is the last of the four arithmetic operations for complex numbers, and it rests entirely on the conjugate.

What is \dfrac{3 + 5i}{1 - 2i}? It is supposed to be a complex number -- something of the form a + bi. But staring at the fraction does not tell you what a and b are. The imaginary unit is tangled up in the denominator, and you cannot read off the answer.

The trick is one you have seen before with surds: multiply top and bottom by the conjugate of the denominator. The denominator collapses into a plain real number, the numerator becomes a standard complex product, and the answer falls right out.

Why the denominator is the problem

A complex number in standard form looks like a + bi, where a and b are both real. When you write \dfrac{3 + 5i}{1 - 2i}, the expression is not in standard form because the denominator is complex. You cannot read off the real and imaginary parts of the answer just by looking at it.

The goal of complex division is: given \dfrac{z_1}{z_2}, find a real number a and a real number b such that \dfrac{z_1}{z_2} = a + bi.

The conjugate does exactly this. Recall from Algebra of Complex Numbers that for any non-zero complex number z = x + yi, the product z \cdot \bar{z} = x^2 + y^2 is a positive real number. So if you multiply the denominator by its conjugate, the denominator becomes real. Multiply the numerator by the same thing to keep the fraction's value unchanged.

Strategy for dividing complex numbers: multiply by conjugate over conjugateA horizontal flowchart with three boxes connected by arrows. The first box shows the original fraction z1 over z2. The second box shows the multiplication by z2-bar over z2-bar, which equals 1. The third box shows the result: z1 times z2-bar over a real number (the sum of squares of the parts of z2).z₁z₂×z̄₂z̄₂= 1 (value unchanged)z₁ · z̄₂a² + b²denominator is now real
The division algorithm in one picture: multiply top and bottom by the conjugate of the denominator. The denominator simplifies to $a^2 + b^2$, a real number. The numerator is a complex product you can expand the usual way.

The division algorithm

Here is the procedure stated precisely.

Division of complex numbers

If z_1 = a + bi and z_2 = c + di \neq 0, then

\frac{z_1}{z_2} = \frac{(a + bi)(c - di)}{c^2 + d^2} = \frac{(ac + bd) + (bc - ad)i}{c^2 + d^2}
= \frac{ac + bd}{c^2 + d^2} + \frac{bc - ad}{c^2 + d^2}\,i.

You do not need to memorise this formula. The mechanical procedure is short enough to run from scratch every time:

  1. Write the conjugate of the denominator.
  2. Multiply both numerator and denominator by that conjugate.
  3. Expand the numerator as a binomial product (and use i^2 = -1).
  4. The denominator is c^2 + d^2 — compute it.
  5. Divide the real and imaginary parts of the numerator by the denominator.

That is the entire algorithm. Every complex division problem is these five steps.

Rationalising the denominator

The phrase "rationalise the denominator" comes from surd arithmetic. When you have \dfrac{1}{1 + \sqrt{2}}, you multiply top and bottom by 1 - \sqrt{2} to clear the surd from the denominator. The denominator becomes (1 + \sqrt{2})(1 - \sqrt{2}) = 1 - 2 = -1, a rational number.

Complex division uses the identical move. The imaginary unit i plays the role of \sqrt{2}. The conjugate c - di plays the role of 1 - \sqrt{2}. The product (c + di)(c - di) = c^2 + d^2 plays the role of the difference of squares.

The difference: with surds, you get a rational number in the denominator. With complex numbers, you get a real number. In both cases, the denominator is tamed — brought down to a simpler number system — and the result can be read off.

Comparison of surd rationalisation and complex rationalisation side by sideTwo boxes side by side. The left box is labelled Surd case and shows the fraction 1 over 1 plus root 2, multiplied by 1 minus root 2 over 1 minus root 2, giving negative 1 plus root 2. The right box is labelled Complex case and shows the fraction 1 over 1 plus i, multiplied by 1 minus i over 1 minus i, giving one-half minus one-half i. Both boxes have the key step highlighted: the denominator simplifies to a number in the simpler system.Surd case11 + √2multiply by (1 − √2)/(1 − √2)denominator: (1+√2)(1−√2)= 1 − 2 = −1rational numberresult: −1 + √2Complex case11 + imultiply by (1 − i)/(1 − i)denominator: (1+i)(1−i)= 1 + 1 = 2real numberresult: ½ − ½i
Surd rationalisation and complex rationalisation are the same idea. The conjugate kills the "awkward" part of the denominator — the surd or the $i$ — through the difference-of-squares identity. What remains is a number from the simpler system: rational in the surd case, real in the complex case.

A concrete walk-through

Take \dfrac{7 + i}{3 + 4i}. The conjugate of the denominator is 3 - 4i.

Numerator: (7 + i)(3 - 4i) = 21 - 28i + 3i - 4i^2 = 21 - 25i + 4 = 25 - 25i.

Denominator: (3 + 4i)(3 - 4i) = 9 + 16 = 25.

Result: \dfrac{25 - 25i}{25} = 1 - i.

Check: (1 - i)(3 + 4i) = 3 + 4i - 3i - 4i^2 = 3 + i + 4 = 7 + i. Correct.

The check is always the same: multiply the quotient by the divisor and confirm you recover the dividend. This is the definition of division — \dfrac{z_1}{z_2} = w means w \cdot z_2 = z_1.

Dividing when the denominator is purely imaginary

A special case: the denominator has no real part, like \dfrac{5 + 3i}{2i}.

The conjugate of 2i = 0 + 2i is -2i. Multiplying top and bottom by -2i:

\frac{(5 + 3i)(-2i)}{(2i)(-2i)} = \frac{-10i - 6i^2}{-4i^2} = \frac{-10i + 6}{4} = \frac{6}{4} - \frac{10}{4}i = \frac{3}{2} - \frac{5}{2}i.

There is a shortcut for purely imaginary denominators: dividing by ki is the same as multiplying by \dfrac{1}{ki} = \dfrac{-i}{k} (since \dfrac{1}{i} = -i). So \dfrac{5 + 3i}{2i} = (5 + 3i) \cdot \dfrac{-i}{2} = \dfrac{-5i - 3i^2}{2} = \dfrac{3 - 5i}{2}. Same answer, fewer steps.

Dividing by a purely imaginary number on the complex planeA complex plane showing three points. The point 5 plus 3i is at coordinates (5, 3) in the first quadrant. The divisor 2i is on the positive imaginary axis at (0, 2). The quotient 3/2 minus 5/2 i is at (1.5, negative 2.5) in the fourth quadrant. An arrow from the origin to 5 plus 3i is solid, and an arrow from the origin to the quotient is in accent colour. The divisor is on the imaginary axis.ReIm01234512−1−25 + 3i2i3/2 − 5/2 idividing by 2i rotates 90° CW and scales by ½
Dividing $5 + 3i$ by $2i$. On the complex plane, dividing by $i$ rotates the point $90°$ clockwise, and dividing by $2$ halves the distance from the origin. The quotient $\frac{3}{2} - \frac{5}{2}i$ is in the fourth quadrant — a clockwise quarter-turn from the first-quadrant dividend.

The multiplicative inverse

The special case z_1 = 1 gives the multiplicative inverse (reciprocal) of z_2:

z_2^{-1} = \frac{1}{z_2} = \frac{\bar{z_2}}{|z_2|^2} = \frac{c - di}{c^2 + d^2}.

For example, \dfrac{1}{3 + 4i} = \dfrac{3 - 4i}{9 + 16} = \dfrac{3 - 4i}{25} = \dfrac{3}{25} - \dfrac{4}{25}i.

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

The inverse exists whenever z_2 \neq 0 — that is, whenever c^2 + d^2 > 0. Every non-zero complex number has a reciprocal in \mathbb{C}.

Properties of division

Division of complex numbers inherits a clean set of properties from the field structure of \mathbb{C}.

1. Division is not commutative. In general, \dfrac{z_1}{z_2} \neq \dfrac{z_2}{z_1}. For instance, \dfrac{1}{i} = -i but \dfrac{i}{1} = i.

2. Conjugate of a quotient. The conjugate distributes over division:

\overline{\left(\frac{z_1}{z_2}\right)} = \frac{\bar{z_1}}{\bar{z_2}}.

This follows from the conjugate distributing over multiplication and the fact that conjugation is an involution.

3. Modulus of a quotient. The modulus of a quotient is the quotient of the moduli:

\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}.

If the dividend has modulus 5 and the divisor has modulus 2, the quotient has modulus \frac{5}{2}. Division scales distances from the origin.

4. Dividing conjugates. If z = a + bi with b \neq 0, then \dfrac{z}{\bar{z}} = \dfrac{a + bi}{a - bi}. Multiplying top and bottom by a + bi:

\frac{(a + bi)^2}{a^2 + b^2} = \frac{a^2 - b^2 + 2abi}{a^2 + b^2}.

The result always has modulus 1 (since |z| = |\bar{z}|, the modulus quotient is 1), and it encodes twice the argument angle of z. You will see this when you study the argument of a complex number.

5. Division by a real number. If the denominator z_2 = r is real (so d = 0), division is trivial: \dfrac{a + bi}{r} = \dfrac{a}{r} + \dfrac{b}{r}i. No conjugate trick is needed.

An interactive division explorer

Drag the two points z_1 (dividend) and z_2 (divisor) around the complex plane. The readout shows the quotient z_1 / z_2 in real time. Notice how the quotient's distance from the origin is the ratio of the two distances, and its angle is the difference of the two angles.

Interactive complex division explorerA coordinate plane representing the complex plane with real axis horizontal and imaginary axis vertical, both ranging from negative 5 to 5. Two red draggable points represent z1 and z2. A readout panel shows the values of z1, z2, and their quotient z1 divided by z2, updated in real time as the reader drags either point.ReIm1234−1−2−3−4↔ drag either red point
Drag $z_1$ and $z_2$ around the complex plane. The readout shows the quotient. Try placing $z_2$ at $i$ (the point $(0, 1)$) — dividing by $i$ rotates $z_1$ by $90°$ clockwise. Place $z_2$ at $-1$ and the quotient is $-z_1$. Complex division is inverse rotation-and-scaling.

Worked examples

Example 1: Compute $\dfrac{2 + 3i}{1 + i}$ in the form $a + bi$

A straightforward division with small numbers. Follow the five-step algorithm.

Step 1. Write the conjugate of the denominator.

\overline{1 + i} = 1 - i

Why: the conjugate flips only the sign of the imaginary part. The denominator 1 + i becomes 1 - i.

Step 2. Multiply numerator and denominator by 1 - i.

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

Why: multiplying by \frac{1-i}{1-i} = 1 preserves the value. The denominator will become real.

Step 3. Expand the numerator.

(2 + 3i)(1 - i) = 2 - 2i + 3i - 3i^2 = 2 + i - 3(-1) = 2 + i + 3 = 5 + i

Why: binomial expansion, then i^2 = -1 turns the -3i^2 into +3, contributing to the real part.

Step 4. Compute the denominator.

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

Why: the product z\bar{z} = a^2 + b^2. Here a = 1, b = 1, so the denominator is 1 + 1 = 2.

Step 5. Divide.

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

Result: \dfrac{2 + 3i}{1 + i} = \dfrac{5}{2} + \dfrac{1}{2}i.

Division of 2 plus 3i by 1 plus i shown on the complex planeA complex plane with three labelled points. The dividend 2 plus 3i is at coordinates (2, 3). The divisor 1 plus i is at (1, 1). The quotient 5/2 plus 1/2 i is at (2.5, 0.5). Dashed lines from the origin to each point show their positions. The quotient is highlighted with a larger red dot.ReIm0123412−12 + 3i1 + i5/2 + 1/2 i|z₁| = √13, |z₂| = √2|quotient| = √13/√2 = √(13/2)
The dividend $2 + 3i$, divisor $1 + i$, and quotient $\frac{5}{2} + \frac{1}{2}i$ on the complex plane. The quotient's distance from the origin is $|z_1|/|z_2| = \sqrt{13}/\sqrt{2}$. The angle has been reduced: the dividend's angle minus the divisor's angle gives the quotient's angle.

Verify: \left(\frac{5}{2} + \frac{1}{2}i\right)(1 + i) = \frac{5}{2} + \frac{5}{2}i + \frac{1}{2}i + \frac{1}{2}i^2 = \frac{5}{2} + 3i - \frac{1}{2} = 2 + 3i. The quotient times the divisor recovers the dividend.

Example 2: Simplify $\dfrac{(1 + i)^2}{1 - i}$

This problem combines a power with a division. The numerator needs to be computed first.

Step 1. Compute (1 + i)^2.

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

Why: binomial square. The real parts cancel (1 - 1 = 0), leaving a purely imaginary number.

Step 2. The problem reduces to \dfrac{2i}{1 - i}. The conjugate of 1 - i is 1 + i.

Why: the conjugate flips the sign of -i to +i.

Step 3. Multiply numerator and denominator by 1 + i.

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

Why: the numerator gives 2i + 2(-1) = -2 + 2i. The denominator is 1^2 + 1^2 = 2.

Step 4. Divide.

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

Result: \dfrac{(1 + i)^2}{1 - i} = -1 + i.

Computation chain showing (1+i) squared equals 2i and the final quotient negative 1 plus iA complex plane with four points. The base number 1 plus i is at (1, 1). Its square 2i is at (0, 2) on the imaginary axis. The divisor 1 minus i is at (1, negative 1). The final quotient negative 1 plus i is at (negative 1, 1) in the second quadrant, highlighted in red.ReIm012−1−21−11 + i(1+i)² = 2i1 − i−1 + i= (1+i)²/(1−i)
The journey on the complex plane. Start at $1 + i$, square it to land at $2i$ on the imaginary axis, then divide by $1 - i$ to arrive at $-1 + i$ in the second quadrant. Each operation — squaring, dividing — rotates and scales the point. The final answer sits at distance $\sqrt{2}$ from the origin, at an angle of $135°$.

An alternative path: note that \dfrac{(1+i)^2}{1-i} = \dfrac{(1+i)(1+i)}{1-i}. Since (1+i)(1-i) = 2, you can write this as (1+i) \cdot \dfrac{1+i}{1-i} = (1+i) \cdot \dfrac{(1+i)^2}{(1-i)(1+i)}... but the direct conjugate method is faster. The algebraic shortcut is worth noting, though: spotting factors before expanding can save work.

Common confusions

Going deeper

If you can divide any two complex numbers (with non-zero denominator), rationalise any complex denominator, and know the properties of division — you have the complete picture for this topic. The material below is for readers who want to see the connection to geometry and fields.

Division as inverse rotation-and-scaling

In Algebra of Complex Numbers, multiplication by a complex number z_2 was described as a rotation-and-scaling operation. Specifically, multiplying by z_2 scales distances by |z_2| and rotates by the argument of z_2.

Division is the inverse operation. Dividing by z_2 scales distances by 1/|z_2| and rotates by minus the argument of z_2. If multiplying by z_2 = 1 + i (which has modulus \sqrt{2} and argument 45°) stretches and rotates counterclockwise by 45°, then dividing by 1 + i shrinks and rotates clockwise by 45°.

This is why the division algorithm works: the conjugate \bar{z_2} has the same modulus as z_2 but the opposite argument. Multiplying numerator and denominator by \bar{z_2} rotates the denominator to land on the positive real axis. That is the geometric meaning of "the denominator becomes real."

Why \mathbb{C} is a field

A field is a number system where addition, subtraction, multiplication, and division (by non-zero elements) all work and obey the standard rules: commutativity, associativity, distributivity, and the existence of identity and inverse elements. The real numbers \mathbb{R} are a field, and so are the complex numbers \mathbb{C}.

The division formula z^{-1} = \bar{z}/|z|^2 is the proof that every non-zero complex number has a multiplicative inverse. This is the piece that closes the field structure. With the inverse in hand, \mathbb{C} inherits all the algebraic machinery of \mathbb{R} — and gains the additional power that every polynomial equation has roots (the Fundamental Theorem of Algebra). The trade-off is that \mathbb{C} cannot be ordered: there is no consistent way to say i > 0 or i < 0.

Repeated division and complex fractions

Just as with real fractions, you can stack divisions:

\frac{\frac{z_1}{z_2}}{\frac{z_3}{z_4}} = \frac{z_1}{z_2} \cdot \frac{z_4}{z_3} = \frac{z_1 z_4}{z_2 z_3}.

The "flip and multiply" rule for compound fractions works identically in \mathbb{C}. This is because division is defined as multiplication by the inverse, and the inverse of a quotient is the reciprocal of the quotient: \left(\frac{z_3}{z_4}\right)^{-1} = \frac{z_4}{z_3}.

Where this leads next

Division is the last arithmetic operation. The articles below build on this foundation.