Complex Numbers and Quadratic Equations

1Polar form, modulus and argument

A complex number z=x+iy is a point (x,y) in the Argand plane. Its modulus is the distance from the origin, |z|=\sqrt{x^2+y^2}, and its argument \theta=\arg z is the directed angle that Oz makes with the positive real axis. Together they give the polar and exponential forms z=r(\cos\theta+i\sin\theta)=re^{i\theta}, with r=|z|.

The conjugate \bar{z}=x-iy reflects z across the real axis, and the identity |z|^2=z\bar{z} turns modulus questions into algebra — never into a square root you have to keep carrying.

Figure. In the Argand plane z is the tip of a vector from the origin. The length of that vector is the modulus r; the angle it makes with the positive real axis is the argument \theta.

Reading a complex number

1. CartesianWrite z=x+iy and mark the point (x,y) in the Argand plane.
2. ModulusCompute r=|z|=\sqrt{x^2+y^2}. This is the length of the segment from O to z.
3. Argument and polarRead \theta=\arg z from the correct quadrant, then write z=r(\cos\theta+i\sin\theta)=re^{i\theta}.

2De Moivre's theorem

On the unit circle, raising a complex number to a positive integer power is a pure rotation of its argument. De Moivre's theorem states (\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta for integer n. In exponential form the same fact is (e^{i\theta})^n=e^{in\theta} — multiply the argument by n, leave the modulus alone when |z|=1.

For a general z=re^{i\theta} the modulus scales as r^n while the argument still scales as n\theta. That is the whole computational rule behind roots of unity and most polar-form powering questions.

Figure. On the unit circle, z\mapsto z^n is rotation by a factor of n on the argument. The figure marks e^{i\theta} and its image e^{in\theta} for n=3 with \theta=\pi/6.

Powering in polar form

1. Write polarExpress z as r(\cos\theta+i\sin\theta) or re^{i\theta}, with \theta in the correct quadrant.
2. ScaleReplace r by r^n and \theta by n\theta (reduce n\theta modulo 2\pi if a principal argument is asked).
3. Expand if neededConvert back to x+iy only when the question demands Cartesian form.

3Cube roots of unity

The solutions of z^3=1 are the cube roots of unity 1,\omega,\omega^2, where \omega=e^{2\pi i/3}=-\frac{1}{2}+i\frac{\sqrt{3}}{2} and \omega^2=e^{-2\pi i/3}. They sit at the vertices of an equilateral triangle on the unit circle, and they satisfy the two identities that unlock every exam manipulation: \omega^3=1 and 1+\omega+\omega^2=0.

The n-th roots of unity likewise form a regular n-gon on the unit circle. For cube roots, any three consecutive powers sum to zero: \omega^k+\omega^{k+1}+\omega^{k+2}=0.

Figure. The three cube roots of unity are equally spaced on the unit circle. Their vector sum is the zero vector — that is the geometric content of 1+\omega+\omega^2=0.

Collapsing an \omega expression

1. Replace sumsUse 1+\omega+\omega^2=0 to rewrite 1+\omega=-\omega^2 and 1+\omega^2=-\omega at the first chance.
2. Reduce powersReplace \omega^3 by 1 (and \omega^4 by \omega, \omega^5 by \omega^2, \ldots) before expanding.
3. CollectFactor out common powers of \omega; consecutive triples that remain sum to zero.

4Quadratic formula and Vieta

For ax^2+bx+c=0 with a\neq 0 the roots are x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}. Independently of solving, Vieta's relations give the sum and product at once: if the roots are \alpha and \beta, then \alpha+\beta=-b/a and \alpha\beta=c/a.

Those two symmetric sums are usually all a JEE question needs. To build the monic quadratic with a known sum s and product p, write x^2-sx+p=0 directly — do not solve for each root and multiply back.

No figure — the content is the two identities \alpha+\beta=-b/a and \alpha\beta=c/a, and the monic template x^2-sx+p=0 built from them.

From sum and product to the equation

1. Read s and pFrom the roots, or from the question, set s=\alpha+\beta and p=\alpha\beta.
2. Write monicThe unique monic quadratic is x^2-sx+p=0.
3. Scale if neededIf a leading coefficient a is required, multiply through: ax^2-asx+ap=0.

5Discriminant and nature of roots

The discriminant D=b^2-4ac decides the nature of the roots before you extract them. For real coefficients a,b,c: D>0 gives two distinct real roots, D=0 gives equal real roots, and D<0 gives a conjugate pair of complex roots.

Check the sign of D first. A negative discriminant is not a stop sign — it is the announcement that the answer will be a conjugate pair, which the quadratic formula still produces once \sqrt{D} is written as i\sqrt{|D|}.

Figure. One family y=x^2-4x+c: c=3 gives D=4>0 and two crossings at x=1 and x=3; c=4 gives D=0 and a single touch at x=2; c=13 gives D=-36<0 and the curve never meets the axis (minimum 9), so its roots are the conjugate pair 2\pm 3i of the worked example. Axes are not to equal scale.

6Conjugate root pairs

Complex roots of a real-coefficient equation always occur in conjugate pairs. If p+iq (q\neq 0) is a root of a quadratic with real coefficients, then p-iq is automatically the other root — you never need a second independent condition to force it.

With both roots known, Vieta builds the equation at once: sum 2p and product p^2+q^2 give x^2-2px+(p^2+q^2)=0.

Figure. Real coefficients send 2+3i to its conjugate 2-3i. Vieta then builds the monic quadratic from sum 2(2)=4 and product 2^2+3^2=13: x^2-4x+13=0.

From one complex root to the equation

1. Pair itIf p+iq is a root and coefficients are real, write the conjugate p-iq as the other root.
2. Sum and productSum =2p, product =p^2+q^2.
3. Monic quadraticWrite x^2-2px+(p^2+q^2)=0.

7Loci in the Argand plane

Three standard equations describe curves in the Argand plane. |z-z_0|=r is the circle of radius r centred at z_0. |z-z_1|=|z-z_2| is the perpendicular bisector of the segment joining z_1 and z_2 — the set of points equidistant from two fixed points. And \arg\left(\dfrac{z-z_1}{z-z_2}\right)=\text{constant} is an arc of a circle through z_1 and z_2.

Translate the equation into geometry before algebra. Most JEE locus questions are one of these three templates with different letters.

Figure. The circle |z-z_0|=r with z_0=0 and r=1 is the unit circle. The equal-modulus locus is a straight bisector (no arc needed); the constant-argument locus is an arc and is carried by the loci table rather than this figure.