(3 + 4i)(1 − 2i)
- Operation
- Multiply z₁ × z₂
- z₁
- 3 + 4i
- z₂
- 1 - 2i
- Result
- 11 − 2i
- Real part
- 11
- Imaginary part
- -2
Checked against: 3 − 6i + 4i − 8i² = 11 − 2i
Add, subtract, multiply and divide complex numbers, raise them to powers, find nth roots, and convert a + bi to polar form, with an Argand diagram.
By AdminDrafted with AI writing toolsUpdated Checked examples: 7
z₁ × z₂ = 11 − 2i, which is 11.18034 ∠ −10.304846° in polar form.
A complex number z = a + bi has a real part a and an imaginary part b, where i² = −1. Addition, subtraction and multiplication work in rectangular form; division multiplies top and bottom by the conjugate of the divisor. Powers and roots use polar form, z = r(cos θ + i sin θ): De Moivre's formula raises the modulus r to the power n and multiplies the angle θ by n.
Electrical engineers use complex numbers for AC impedance and phasors, and algebra students meet them as roots of polynomials. The default multiplies (3 + 4i)(1 − 2i) = 3 − 6i + 4i − 8i² = 11 − 2i, whose modulus is √125 ≈ 11.1803.
Inputs can be rectangular (3 + 4i, with j accepted for i) or polar (2∠90). Angles follow the degree, radian or gradian setting, and the argument is given between −180° and 180°. The Argand diagram draws each number as an arrow from the origin.
Treat i as a number with i² = −1. Add or subtract the real and imaginary parts separately. To multiply, expand and replace i² with −1: (3 + 4i)(1 − 2i) = 3 − 6i + 4i − 8i² = 11 − 2i. To divide, multiply the top and bottom by the conjugate of the divisor.
| Operation | z₁ | z₂ or n | Result |
|---|---|---|---|
| Add | 3 + 4i | 1 - 2i | 4 + 2i |
| Subtract | 3 + 4i | 1 - 2i | 2 + 6i |
| Multiply | 3 + 4i | 1 - 2i | 11 − 2i |
| Divide | 3 + 4i | 1 - 2i | −1 + 2i |
| Power | 1 + i | 8 | 16 |
| nth roots | -4 | 2 | 2i, −2i |
Dividing by 1 − 2i uses its conjugate, 1 + 2i. The denominator becomes 1² + 2² = 5 and the numerator (3 + 4i)(1 + 2i) = −5 + 10i, so the quotient is −1 + 2i.
Multiply back to check a quotient: (−1 + 2i)(1 − 2i) = 3 + 4i. Division by 0 + 0i is undefined. The OpenStax complex numbers lesson covers these rules and the conjugate.
The modulus r = √(a² + b²) is the distance from the origin, and the argument θ = atan2(b, a) is the angle from the positive real axis. For 3 + 4i, r = √(3² + 4²) = 5 and θ ≈ 53.130102°, so 3 + 4i = 5∠53.130102°.
Use atan2, not arctan(b/a), because the signs of both parts decide the quadrant. −3 − 4i also has modulus 5, but its principal argument is −126.869898°, while arctan(4/3) gives 53.130102° for both numbers.
The argument lies between −180° and 180°, in your chosen angle unit: 53.130102° is 0.927295 rad or 59.033447 gradians. The exponential form re^(iθ) always uses radians, so 3 + 4i = 5e^(0.927295i).
In AC circuits, a 3 Ω resistance in series with 4 Ω of inductive reactance has impedance 3 + 4j Ω: magnitude 5 Ω at a phase angle of 53.13°.
The number 0 has no defined argument, so an angle shown for it is only a placeholder.
For a whole number n, zⁿ = rⁿ(cos nθ + i sin nθ): raise the modulus to the power and multiply the angle. (1 + i)⁸ has r = √2 and θ = 45°, so it equals (√2)⁸ = 16 at 8 × 45° = 360°, the real number 16.
Negative powers give reciprocals: (1 + i)⁻¹ = 0.5 − 0.5i. A non-integer power returns one principal value, using the argument between −180° and 180°, so (−4)^0.5 gives 2i alone. The exponent must be a real number.
A nonzero number has n distinct nth roots. They share the modulus |z|^(1/n) and are spaced 360°/n apart, starting from the principal root at angle θ/n. For z = 0, every root is 0.
The cube roots of 8 are 2 and −1 ± √3i (about −1 ± 1.732051i), at 0°, 120° and 240°; 240° is the same direction as −120°. The square roots of −4 are 2i and −2i, and 2i is the principal one.
Asking for (−4)^0.5 and asking for the square roots of −4 are different questions: the power gives one value, the roots give both. The OpenStax polar form lesson derives both from polar coordinates.
3 + 4i, 3 - 4i, -2i or 1/2 + 3/4i; j works in place of i.2∠90°.2∠90 in radian mode means 90 radians, a different point from 2i.For the roots of a polynomial, use the polynomial root finder, or the quadratic calculator for ax² + bx + c = 0. For real-number arithmetic and functions, use the scientific calculator.
Checked against: 3 − 6i + 4i − 8i² = 11 − 2i
Checked against: Independent calculation
Checked against: atan2(4, 3) in degrees via Machin-series arctangent
Checked against: (√2)⁸ = 16 at angle 8 × 45° = 360°
Expand the brackets and replace i² with −1: (a + bi)(c + di) = (ac − bd) + (ad + bc)i. For (3 + 4i)(1 − 2i) that is (3 + 8) + (−6 + 4)i = 11 − 2i. In polar form the rule is shorter: multiply the moduli and add the angles, so 5∠53.13° × 2.236∠−63.43° = 11.18∠−10.30°.
Multiply the top and bottom by the conjugate of the divisor, which makes the denominator a real number. (3 + 4i)/(1 − 2i) = (3 + 4i)(1 + 2i)/(1² + 2²) = (3 + 6i + 4i + 8i²)/5 = (−5 + 10i)/5 = −1 + 2i. Division by 0 + 0i is undefined.
The modulus is r = √(a² + b²) and the argument is θ = atan2(b, a), which picks the correct quadrant. For 3 + 4i, r = √(9 + 16) = 5 and θ ≈ 53.130°, so 3 + 4i = 5∠53.130°. Plain arctan(b/a) gives the wrong angle when the real part is negative: −3 − 4i has an argument of about −126.870°, not 53.130°.
For z = r(cos θ + i sin θ) and a whole number n, zⁿ = rⁿ(cos nθ + i sin nθ). So (1 + i)⁸, with r = √2 and θ = 45°, equals (√2)⁸ = 16 at an angle of 8 × 45° = 360°, which is the real number 16. The same idea gives n evenly spaced nth roots: the cube roots of 8 are 2 and −1 ± 1.732051i, 120° apart.
The imaginary unit i, defined by i² = −1; −i is the other square root. Every negative number has two imaginary square roots, so √−4 = ±2i, and the principal root, the one with argument 90°, is 2i. Powers of i repeat every four steps: i, −1, −i, 1, then i again.
This calculator includes 7 worked examples with independently sourced answers. They run in the test suite, and you can run them here too.