Complex numbers appear in Class 11 algebra, AC circuit analysis and signal processing. Enter two complex numbers and an operation to get the result, its modulus and argument, and the polar form.
Enter the real and imaginary parts of both numbers.
Choose the operation.
Read the result and its polar form.
Multiply: (a + bi)(c + di) = (ac − bd) + (ad + bc)i
Modulus: |z| = √(a² + b²)
A complex number has the form a + bi, where a is the real part, b is the imaginary part and i is a number whose square is −1. This calculator adds, subtracts, multiplies or divides two complex numbers, and gives the result's modulus, argument in degrees and polar form, along with the modulus, argument and conjugate of the first number.
Complex numbers are introduced in Class 11 to solve equations like x² + 1 = 0, but their biggest use is practical. Electrical engineers write AC voltages, currents and impedances as complex numbers, usually with j in place of i, so that resistors, capacitors and inductors can be handled with one algebra. Signal processing, control systems and quantum physics rely on them too.
1. Addition and subtraction: combine real parts and imaginary parts separately, (a + bi) ± (c + di) = (a ± c) + (b ± d)i.
2. Multiplication: expand the brackets and replace i² with −1, (a + bi)(c + di) = (ac − bd) + (ad + bc)i.
3. Division: multiply top and bottom by the conjugate c − di, giving [(ac + bd) + (bc − ad)i] ÷ (c² + d²).
4. Modulus: |z| = √(x² + y²) for z = x + yi.
5. Argument: θ = atan2(y, x), the angle from the positive real axis, taking the quadrant into account.
6. Polar form: z = |z|(cos θ + i sin θ).
Expanding (a + bi)(c + di) term by term gives ac + adi + bci + bdi². Since i² = −1, the last term becomes −bd, which joins the real part. So the real part is ac − bd and the imaginary part is ad + bc. The minus sign is the only surprise, and it is the whole reason complex numbers are useful: multiplying by i turns 1 into i and i into −1, which is a rotation by 90° on the Argand plane.
A fraction with a complex denominator is not in the form a + bi. Multiplying by the conjugate c − di fixes this, because (c + di)(c − di) = c² + d², a positive real number. The numerator becomes (a + bi)(c − di) = (ac + bd) + (bc − ad)i. Dividing each part by c² + d² gives the answer in standard form. Division by zero, where c and d are both zero, is undefined, and the calculator says so.
Plot z = x + yi as a point on the Argand plane. Its distance from the origin is the modulus and its angle from the positive real axis is the argument. In polar form, multiplication is simple: moduli multiply and arguments add; division divides moduli and subtracts arguments. This is the basis of De Moivre's theorem for powers and roots. The calculator uses atan2, which picks the correct quadrant, and reports the principal argument between −180° and 180°. Plain tan⁻¹(y/x) gives the wrong angle when x is negative.
Aditya, an electrical engineering student in Nagpur, needs to divide the complex quantity 5 + 2i by 1 + i while solving an AC circuit problem, and also wants the result in polar form.
Result: [(ac + bd) + (bc − ad)i] ÷ (c² + d²) = [7 + (-3)i] ÷ 2 = 3.5 − 1.5i
Modulus and argument of the result: |z| = √(3.5² + (-1.5)²) = 3.807887 arg z = atan2(-1.5, 3.5) = -23.1986°
z₁ in polar form: |z₁| = √(5² + 2²) = 5.385165, arg = 21.8014°, conjugate = 5 − 2i
Answer: Result 3.5 − 1.5i; Modulus |z| 3.807887; Argument -23.1986°
Forgetting that i² = −1 and adding bd instead of subtracting it when multiplying.
Multiplying only the denominator by the conjugate in division, instead of both top and bottom.
Finding the argument with tan⁻¹(y/x) for a number in the second or third quadrant, which puts it in the wrong quadrant.
Mixing degrees and radians; NCERT usually states the principal argument in radians, while this calculator uses degrees.
Writing the conjugate of a + bi as −a + bi instead of a − bi.
AC circuit analysis with complex impedances of resistors, inductors and capacitors.
Class 11 problems on algebra of complex numbers, modulus, argument and polar form.
Signal processing and Fourier analysis, where signals are rotating complex numbers.
Control engineering, where system poles are plotted on the complex plane.
Checking roots of quadratics with a negative discriminant.
Is the argument in degrees or radians?
Degrees, between −180° and 180° (the principal value).
What is i²?
−1, which is why bd is subtracted when multiplying.