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Pure · Complex numbers

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Complex roots of polynomial equations

Solving quadratics with a negative discriminant to get a conjugate pair of complex roots; the theorem that complex roots of a real-coefficient polynomial come in conjugate pairs; using one known complex root to find the others and factorise cubics and quartics; and finding the square roots of a + bi by equating real and imaginary parts.

Pure · Complex numbers

Complex roots of polynomial equations

Solving quadratics with a negative discriminant to get a conjugate pair of complex roots; the theorem that complex roots of a real-coefficient polynomial come in conjugate pairs; using one known complex root to find the others and factorise cubics and quartics; and finding the square roots of a + bi by equating real and imaginary parts.

Why it works

Once we have ii with i2=−1i^2 = -1, every quadratic can be solved. Take z2−4z+13=0z^2 - 4z + 13 = 0. The quadratic formula still applies: z=4±(−4)2−4(1)(13)2=4±−362.z = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(13)}}{2} = \frac{4 \pm \sqrt{-36}}{2}. The discriminant is negative, which is exactly the case that had no real solution. Now we handle it: −36=36 −1=6i\sqrt{-36} = \sqrt{36}\,\sqrt{-1} = 6i, so z=4±6i2=2±3i.z = \frac{4 \pm 6i}{2} = 2 \pm 3i. The two roots are 2+3i2 + 3i and 2−3i2 - 3i — the same real part, opposite imaginary parts. They are conjugates of each other. That is not a coincidence.

The conjugate-pair theorem. *If a polynomial has real coefficients and z=a+biz = a + bi is a root, then its conjugate zˉ=a−bi\bar z = a - bi is also a root.* The reason is that conjugation respects ++ and ×\times, so taking the conjugate of the whole equation p(z)=0p(z) = 0 turns every real coefficient into itself and turns zz into zˉ\bar z, giving p(zˉ)=0p(\bar z) = 0. Complex roots therefore always arrive in pairs. The "real coefficients" condition is essential — drop it and the theorem fails (e.g. z2−iz=0z^2 - iz = 0 has roots 00 and ii, which are not conjugates).

Why pairs are so useful. A conjugate pair a±bia \pm bi multiplies back to a real quadratic: (z−(a+bi))(z−(a−bi))=(z−a)2−(bi)2=z2−2az+(a2+b2).\big(z - (a+bi)\big)\big(z - (a-bi)\big) = (z-a)^2 - (bi)^2 = z^2 - 2az + (a^2 + b^2). Both the coefficient −2a-2a (twice the real part) and the constant a2+b2a^2 + b^2 (the squared modulus) are real. So knowing one complex root of a real polynomial hands you a real quadratic factor for free — and dividing it out reduces a cubic to a linear factor, or a quartic to a second quadratic.

Square roots of a complex number. To find a+bi\sqrt{a + bi} we look for x+yix + yi (with x,yx, y real) such that (x+yi)2=a+bi(x + yi)^2 = a + bi. Expanding, (x+yi)2=x2−y2+2xyi,(x+yi)^2 = x^2 - y^2 + 2xyi, so equating real and imaginary parts gives two simultaneous equations: x2−y2=aand2xy=b.x^2 - y^2 = a \qquad\text{and}\qquad 2xy = b. You need both. The second fixes the sign relationship between xx and yy; the first pins down their sizes. Every non-zero complex number has exactly two square roots, and they differ only by sign — that is why the answer is ±(x+yi)\pm(x + yi).