Complex Number Multiplication
Enter the real and imaginary parts of z1 and z2 to get the product (ac − bd) + (ad + bc)i, the expansion that applies i squared = -1.
z₁ × z₂
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Product of complex numbers: (a + bi)(c + di) = (ac − bd) + (ad + bc)i
To multiply two complex numbers you distribute the terms and apply i² = −1: (a + bi)(c + di) = ac + adi + bci + bdi² = (ac − bd) + (ad + bc)i. Try (1 + i)(1 − i) = 1 − i² = 1 − (−1) = 2; multiplying conjugates lands you back on the real line. The polar form z = |z|·e^(iθ) makes the whole thing easier, since z₁ · z₂ = |z₁||z₂| · e^(i(θ₁ + θ₂)), so the moduli multiply while the arguments (phases) add. There is a picture behind this. Multiplying by a complex number amounts to scaling and rotating a point in the plane, and multiplying by i (that is, e^(iπ/2)) spins a vector exactly 90°. Euler's identity, e^(iπ) + 1 = 0, often called the most beautiful in mathematics, ties together five fundamental constants (0, 1, π, e, i) and drops right out of this representation.
Applications
Computer graphics and game engines use it for 2D rotations, where it makes a tidy stand-in for 2x2 matrices. The FFT depends on it too: the Cooley-Tukey algorithm is built on multiplications by roots of unity e^(2πik/N). In electronics, complex impedance Z = R + jX bundles resistance and reactance, and series or parallel circuits multiply those impedances. Quantum physics writes its evolution operators as complex exponentials, and AC signal analysis cares about exactly this when phase differences come into play.
FAQ
Why does multiplying by i rotate by 90°? Because i = e^(iπ/2). In polar form, multiplying by i tacks π/2 (90°) onto the argument and leaves the modulus alone. Work it out and (x + yi) · i = −y + xi, which is just the original vector turned 90° counter-clockwise.
What is the conjugate and why is it useful? The conjugate of z = a + bi is z̄ = a − bi. The product z · z̄ = a² + b² = |z|² is always real, which is what makes it handy for rationalising denominators and working out moduli.
How do I divide complex numbers? Multiply the top and bottom by the conjugate of the denominator: (a + bi) / (c + di) = (a + bi)(c − di) / (c² + d²). That leaves a real number downstairs, which is the whole point.
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