Complex Numbers

Roots of quadratic equations

For a quadratic equation ax^2+bx+c=0 with a\neq 0, let the roots be \alpha and \beta. Vieta's formulas give \alpha+\beta=-\dfrac{b}{a} and \alpha\beta=\dfrac{c}{a}. These two relations let you read the sum and product of the roots without solving the equation, and most questions on this topic reduce to imposing conditions on these two quantities.

If one root is known, the other root follows at once from the sum or the product. For example, if 2 is a root of x^2-5x+6=0, the other root is 5-2=3.

The nature of the roots is decided by the discriminant D=b^2-4ac. If D>0 the roots are real and distinct, if D=0 they are real and equal, and if D<0 they are a complex conjugate pair. For real roots the condition is D\ge 0, and the boundary case D=0 is included. For x^2-6x+m=0 to have real roots, 36-4m\ge 0, so m\le 9, and m=9 gives equal roots.

When the coefficients are real, non-real roots always occur in conjugate pairs. If 2-i\sqrt{3} is a root of a real quadratic, then 2+i\sqrt{3} is also a root, and the sum 4 and product 7 give the quadratic x^2-4x+7=0.

A common error is to treat (\alpha-\beta) as known from (\alpha-\beta)^2 alone. The identity (\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta fixes the square, and the sign of the difference must be decided from the ordering of the roots. If (p-q)^2=9c with p>q and c>0, then p-q=3\sqrt{c}. For x^2-4x+c=0 with roots differing by 2, 16-4c=4 gives c=3.

Symmetric expressions in the roots should be rewritten in terms of \alpha+\beta and \alpha\beta. Useful forms are \alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta, \alpha^2-\beta^2=(\alpha-\beta)(\alpha+\beta), and \alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta). Applying the last form to the roots of x^2+px+q=0 gives \alpha^3+\beta^3=-p^3+3pq. Moreover, the roots \alpha^3 and \beta^3 have sum \alpha^3+\beta^3 and product (\alpha\beta)^3, which is how a new equation is built from powers of the old roots.

To form a new quadratic with roots u and v, write u+v and uv in terms of the given roots and then use x^2-(u+v)x+uv=0. If the roots of x^2-(a+b)x+(ab-c)=0 are \alpha and \beta, then a and b have sum \alpha+\beta and product \alpha\beta+c, so the required equation is x^2-(\alpha+\beta)x+(\alpha\beta+c)=0.

Equal roots require D=0. The discriminant of x^2-(2+m)x+2m=0 is (2+m)^2-8m=(m-2)^2, which vanishes only for m=2. In general, if the discriminant factorises as a perfect square, its zero gives the condition directly, as in (a+b+c)^2-4(bc+ca)=(a+b-c)^2. Note that a discriminant written as a square plus a non-zero square is strictly positive, so the roots are real and unequal.

If the sum of the roots is zero, the roots are \pm r and the product is -r^2. If the product is positive with zero sum, the roots are non-real. For ax^2+bx+c=0, the condition that the sum equals the product is -\dfrac{b}{a}=\dfrac{c}{a}, that is, b+c=0. Any a\neq 0 and any b with c=-b satisfy this, so there are infinitely many such quadratics.

When the roots themselves are the coefficients, as in x^2+px+q=0 with roots p and q, write both Vieta relations and solve the resulting system. Cancel a common factor only after checking it is non-zero.

For roots of unit modulus with \alpha+\beta=1 and \alpha\beta=1, as in x^2-x+1=0, the roots are \dfrac{1\pm i\sqrt{3}}{2}. Each is a primitive sixth root of unity, so \alpha^6=1, and powers can be reduced modulo 6. Also \alpha-\beta=i\sqrt{3} has modulus \sqrt{3}.

For a quadratic with roots in an interval (m,n), the conditions are D\ge 0, f(m)>0, f(n)>0 (when a>0), and the vertex -\dfrac{b}{2a} lies in (m,n). For x^2-3x+m=0 with both roots in (0,3): 9-4m\ge 0 gives m\le\dfrac94, f(0)=m>0, and f(3)=m>0. Hence 0<m\le\dfrac94, which contains the integers 1 and 2.

When a statement is tested for sufficiency, check whether the given conditions force a single value. A zero sum gives roots \pm r, so an extra condition must break the symmetry. For x^2+ax+b=0 with a=0, the roots satisfy \beta=-\alpha, and a condition such as \alpha\beta^2=-1 then gives \alpha^3=-1 and a unique \alpha.

Finally, when a quadratic is given in factorised form, the factors can be checked against the sum and product. The equation x^2-5x+6=0 factorises as (x-2)(x-3), and 2+3=5, 2\cdot 3=6 confirm it. For (ux-v)(x-1) the roots are 1 and \dfrac{v}{u}.

The reverse of Vieta is also useful. The roots of 4x^2-2x-1=0 are \cos 36^\circ and -\cos 72^\circ, so their sum is \tfrac12 and their product is -\tfrac14. Questions asking for p+q or pq after identifying the numbers need only these two quantities.

When a quantity is expressed as a rational function of a parameter, first simplify it and then read off its product or zeros. If p=\dfrac{-8}{3q}, then pq=-\dfrac{8}{3} for every admissible q. A zero of \dfrac1p must be found from the simplified form, but the points must also lie in the domain of the original expression. Restricted domains matter when counting solutions.

For a quadratic with logarithmic coefficients, put L=\log_{0.5}(a^2), so that \log_{a^2}(0.5)=\dfrac{1}{L}. Relations between the roots, such as the product \alpha\beta=L^4 together with \alpha=L\beta^2, determine \beta by elimination. The sum of roots then gives the other root directly, since \alpha+\beta=-L.

Cube roots of unity

The cube roots of unity are the three roots of z^3=1. They are 1, \omega=\dfrac{-1+i\sqrt{3}}{2}, and \omega^2=\dfrac{-1-i\sqrt{3}}{2}, where \omega is the non-real root. Since z^3-1=(z-1)(z^2+z+1), the two non-real roots satisfy z^2+z+1=0, and this gives the basic identity 1+\omega+\omega^2=0.

The key rules are \omega^3=1, \omega^2=\overline{\omega}, and \omega^{n} depends only on the remainder of n on division by 3. For example, \omega^{47}=\omega^{2}, because 47=3\cdot 15+2. The same reduction applies to any power of \omega.

The sum \omega^m+\omega^n+1 vanishes exactly when m and n leave remainders 1 and 2 on division by 3, since then the three powers are \omega, \omega^2 and 1. Therefore \omega^{1985}+\omega^{100}+1=0, because 1985\equiv 2 and 100\equiv 1 \pmod 3. In particular \omega is a root of z^{1985}+z^{100}+1=0, and the other two options give nonzero values.

When comparing two equations in \omega, substitute the roots of one into the other. The roots of z^3+2z^2+2z+1=0 are -1, \omega and \omega^2, because z^3+2z^2+2z+1=(z+1)(z^2+z+1). The root -1 does not satisfy z^{1985}+z^{100}+1=0, since (-1)^{1985}+(-1)^{100}+1=1. So the common roots are exactly \omega and \omega^2.

To solve (z-a)^3=-c^3, write w=z-a and use the three cube roots of -c^3, namely -c, -c\omega and -c\omega^2. Then z=a-c\omega^k for k=0,1,2. For example, (z-5)^3=-8 gives z=3, z=5-2\omega and z=5-2\omega^2. In the same way, (z-100)^3+1000=0 gives z=90 and z=100+10\cdot(-\omega) and 100+10\cdot(-\omega^2).

For the three distinct cube roots 1, \omega, \omega^2, the pairwise products have sum equal to the coefficient of t in t^3-1 by Vieta's formulas, which is 0. Hence 1\cdot\omega+\omega\cdot\omega^2+\omega^2\cdot 1=0. The same method works for any set of distinct cube roots.

For expressions built from a, b and powers of \omega, expand the squares and collect terms. Using \omega^3=1, \omega^4=\omega, and 1+\omega+\omega^2=0, the a^2 and b^2 terms cancel in the sum over the three conjugate combinations, and each cross term contributes 2ab. With x=a+b, y=a\omega+b\omega^2 and z=a\omega^2+b\omega, the total is x^2+y^2+z^2=6ab.

For the modulus of 1-\omega, use \overline{\omega}=\omega^2 to write |1-\omega|^2=(1-\omega)(1-\omega^2)=1-(\omega+\omega^2)+\omega^3=3. Since \omega+\omega^2=-1, the expression \left|\dfrac{1-\omega}{\omega+\omega^2}\right| equals \sqrt{3}.

In a polynomial in \alpha=\omega, reduce every power using the remainder rule and then use 1+\omega+\omega^2=0. For instance 1+\omega-\omega^2=-2\omega^2 and 1-3\omega+\omega^2=-4\omega. Then (-2\omega^2)^{100}=2^{100}\omega^{200}=2^{100}\omega^2 and (-4\omega)^{50}=4^{50}\omega^{50}=2^{100}\omega^2, so their difference is zero.