Complex number operations — addition to De Moivre's theorem progression

medium CBSE JEE-MAIN • 3 min read
Tags Complex Numbers

Question

If z=1+i3z = 1 + i\sqrt{3}, find z6z^6 using De Moivre’s theorem. Also, find the cube roots of unity and show they sum to zero.

Solution — Step by Step

z=1+i3z = 1 + i\sqrt{3}. Modulus: ∣z∣=1+3=2|z| = \sqrt{1 + 3} = 2

Argument: θ=tan⁡−1(3/1)=π/3\theta = \tan^{-1}(\sqrt{3}/1) = \pi/3

Polar form: z=2(cos⁡π3+isin⁡π3)=2eiπ/3z = 2(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}) = 2e^{i\pi/3}

 z6=[2(cos⁡π3+isin⁡π3)]6=26(cos⁡6π3+isin⁡6π3)z^6 = [2(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3})]^6 = 2^6(\cos\frac{6\pi}{3} + i\sin\frac{6\pi}{3}) =64(cos⁡2π+isin⁡2π)=64(1+0)=64= 64(\cos 2\pi + i\sin 2\pi) = 64(1 + 0) = \mathbf{64}

We need z3=1z^3 = 1, or z3−1=0z^3 - 1 = 0. Using De Moivre: z=ei⋅2kπ/3z = e^{i \cdot 2k\pi/3} for k=0,1,2k = 0, 1, 2.

ω0=1\omega_0 = 1, ω1=e2πi/3=−12+32i\omega_1 = e^{2\pi i/3} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i, ω2=e4πi/3=−12−32i\omega_2 = e^{4\pi i/3} = -\frac{1}{2} - \frac{\sqrt{3}}{2}i

Sum: 1+(−12+32i)+(−12−32i)=1−1+0=01 + (-\frac{1}{2} + \frac{\sqrt{3}}{2}i) + (-\frac{1}{2} - \frac{\sqrt{3}}{2}i) = 1 - 1 + 0 = \mathbf{0} ✓

Why This Works

graph TD
    A["Complex Number Operation"] --> B["Addition/Subtraction"]
    B --> C["Use Cartesian: a+bi ± c+di"]
    A --> D["Multiplication/Division"]
    D --> E["Cartesian works, but polar is often easier"]
    A --> F["Powers"]
    F --> G["Convert to polar, use De Moivre"]
    A --> H["Roots"]
    H --> I["Convert to polar, use nth root formula"]
    G --> J["De Moivre: r,θ ^n = r^n, nθ"]
    I --> K["nth roots: r^{1/n}, θ+2kπ / n"]

De Moivre’s theorem says: (cos⁡θ+isin⁡θ)n=cos⁡(nθ)+isin⁡(nθ)(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta). This converts a hard problem (raising a complex number to a power in Cartesian form requires repeated multiplication) into a simple one (multiply the angle, raise the modulus to the power).

The cube roots of unity (1,ω,ω21, \omega, \omega^2) are vertices of an equilateral triangle inscribed in the unit circle. Their sum is zero because the three vectors from the origin to these vertices cancel out (by symmetry). This generalises: the nnth roots of unity always sum to zero for n≥2n \geq 2.

Alternative Method

For the cube roots of unity, you do not need De Moivre at all. Factor: z3−1=(z−1)(z2+z+1)=0z^3 - 1 = (z-1)(z^2 + z + 1) = 0. So z=1z = 1 or z=−1±3i2z = \frac{-1 \pm \sqrt{3}i}{2}.

The standard notation is ω=−1+i32\omega = \frac{-1 + i\sqrt{3}}{2}. Key properties: ω3=1\omega^3 = 1, 1+ω+ω2=01 + \omega + \omega^2 = 0, and ω2=ωˉ\omega^2 = \bar{\omega}. These appear in JEE problems on determinants, matrices, and series.

Common Mistake

Getting the argument wrong for complex numbers in Q2, Q3, or Q4. The formula θ=tan⁡−1(y/x)\theta = \tan^{-1}(y/x) gives the correct angle only in Q1. For −1+i-1 + i, students compute tan⁡−1(1/(−1))=−π/4\tan^{-1}(1/(-1)) = -\pi/4 and write the argument as −π/4-\pi/4, but the point is in Q2 so the argument is 3π/43\pi/4. Always plot the point first and adjust the angle based on the quadrant.

Polar form: z=r(cos⁡θ+isin⁡θ)=reiθz = r(\cos\theta + i\sin\theta) = re^{i\theta}

De Moivre: (reiθ)n=rneinθ(r e^{i\theta})^n = r^n e^{in\theta}

nnth roots: zk=r1/nei(θ+2kπ)/nz_k = r^{1/n} e^{i(\theta + 2k\pi)/n}, k=0,1,...,n−1k = 0, 1, ..., n-1

Euler’s formula: eiθ=cos⁡θ+isin⁡θe^{i\theta} = \cos\theta + i\sin\theta

∣z1z2∣=∣z1∣∣z2∣|z_1 z_2| = |z_1||z_2|, arg⁡(z1z2)=arg⁡(z1)+arg⁡(z2)\arg(z_1 z_2) = \arg(z_1) + \arg(z_2)

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