ComplexLab

Complex Number Calculator

Complex arithmetic without the pain.

Re(z₁)

Im(z₁)

Re(z₂)

Im(z₂)

z₁ + z₂: 4 + 6i
z₁ − z₂: 2 + 2i
z₁ · z₂: -5 + 10i
|z₁|: 5

What you entered

z₁: 3 + 4iz₂: 1 + 2i

1
Add: combine real and imaginary parts separately
(3+1) + (4+2)i= 4 + 6i
2
Subtract: same, but subtracting
(3−1) + (4−2)i= 2 + 2i
3
Multiply: expand with i² = −1
(3×1 − 4×2) + (3×2 + 4×1)i= -5 + 10i
4
Modulus of z₁ (distance from origin)
√(3² + 4²)= 5

Result

z₁ · z₂: -5 + 10i

Multiplying complex numbers uses i² = −1 to fold the imaginary terms back into a single real + imaginary result: -5 + 10i.

Complex arithmetic: sum, difference, product and modulus

Addition and subtraction work component by component. Multiplication uses FOIL with i squared replaced by -1.

z1	z2	Sum	Difference	Product	Modulus of z1
3 + 4i	1 + 2i	4 + 6i	2 + 2i	-5 + 10i	5.0000
1 + 1i	1 - 1i	2 + 0i	0 + 2i	2 + 0i	1.4142
2 + 3i	4 - 1i	6 + 2i	-2 + 4i	11 + 10i	3.6056
5 + 0i	0 + 5i	5 + 5i	5 - 5i	0 + 25i	5.0000
-2 + 3i	1 + 4i	-1 + 7i	-3 - 1i	-14 - 5i	3.6056

Row two multiplies a number by its own conjugate and the imaginary part vanishes, leaving a purely real 2 - that is always true, and it is the trick used to divide complex numbers by rationalising the denominator. Row one shows why multiplication is not component-wise: 3 times 1 is 3, yet the real part of the product is -5, because the 4i times 2i term contributes -8. The modulus is the distance from the origin, so 3 + 4i has modulus exactly 5 by the same Pythagorean triple that appears everywhere else.

Numbers with two parts

A complex number a + bi has a real part (a) and an imaginary part (b), where i is defined as √−1 — a number that doesn't exist on the ordinary number line but is essential for solving equations like x² = −1.

Not just theoretical

Electrical engineers use complex numbers to model AC circuits, and they're fundamental to signal processing and quantum mechanics — 'imaginary' doesn't mean 'not useful.'

Frequently asked questions

How do I multiply (3 + 2i) by (1 − 4i)?

Use FOIL: (3)(1) + (3)(−4i) + (2i)(1) + (2i)(−4i) = 3 − 12i + 2i − 8i². Since i² = −1: 3 − 10i − 8(−1) = 3 − 10i + 8 = 11 − 10i.

What is the modulus of a complex number and when is it useful?

The modulus of a + bi is √(a² + b²) — the distance from the origin in the complex plane. For 3 + 4i, |z| = √(9 + 16) = 5. In AC circuits, the modulus of impedance gives the total opposition to current flow.

How are complex numbers related to the quadratic formula?

When the discriminant (b² − 4ac) is negative, the quadratic formula produces complex roots. For x² + 1 = 0, the discriminant is −4, giving roots x = ±i. The polynomial roots calculator shows this connection with full discriminant analysis.