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Pure · Complex numbers

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Complex numbers in Cartesian form

The imaginary unit i with i² = −1, real and imaginary parts, equality of complex numbers, and the four operations in the form a + bi — addition, subtraction, multiplication, and division by realising the denominator with the conjugate.

Pure · Complex numbers

Complex numbers in Cartesian form

The imaginary unit i with i² = −1, real and imaginary parts, equality of complex numbers, and the four operations in the form a + bi — addition, subtraction, multiplication, and division by realising the denominator with the conjugate.

Why it works

A quadratic like x2+1=0x^2 + 1 = 0 has no real solution, because no real number squares to a negative. We extend the number system by defining a new symbol ii with i2=−1.i^2 = -1. Everything else follows from ordinary algebra with this one extra rule. A complex number is written z=a+biz = a + bi, where aa and bb are real: a=Re⁡(z)a = \operatorname{Re}(z) is the real part and b=Im⁡(z)b = \operatorname{Im}(z) is the imaginary part (note the imaginary part is the real number bb, not bibi).

Equality. Two complex numbers are equal exactly when their real parts are equal and their imaginary parts are equal: a+bi=c+di  ⟺  a=c and b=d.a + bi = c + di \iff a = c \text{ and } b = d. So a single complex equation gives you two real equations — this is how you find unknown reals.

Addition and subtraction work component-wise, like collecting like terms: (a+bi)±(c+di)=(a±c)+(b±d)i(a+bi) \pm (c+di) = (a\pm c) + (b\pm d)i.

Multiplication is ordinary expansion, then replace i2i^2 by −1-1: (a+bi)(c+di)=ac+adi+bci+bdi2=(ac−bd)+(ad+bc)i.(a+bi)(c+di) = ac + adi + bci + bdi^2 = (ac - bd) + (ad + bc)i. The one thing that makes it "complex" is that last step — the i2i^2 term folds back into the real part with a sign change.

The conjugate. The conjugate of z=a+biz = a + bi is zˉ=a−bi\bar z = a - bi (flip the sign of the imaginary part only). Its point is that multiplying a number by its conjugate kills the ii: zzˉ=(a+bi)(a−bi)=a2−(bi)2=a2+b2,z\bar z = (a+bi)(a-bi) = a^2 - (bi)^2 = a^2 + b^2, a real, non-negative number.

Division uses exactly that. To write a+bic+di\dfrac{a+bi}{c+di} in the form x+yix + yi, multiply top and bottom by the conjugate of the denominator — this "realises" the denominator: a+bic+di=(a+bi)(c−di)(c+di)(c−di)=(a+bi)(c−di)c2+d2.\frac{a+bi}{c+di} = \frac{(a+bi)(c-di)}{(c+di)(c-di)} = \frac{(a+bi)(c-di)}{c^2+d^2}. Now the denominator is the real number c2+d2c^2 + d^2, and you just divide each part by it.