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

Question

If $z = 1 + i\sqrt{3}$ , find $z^6$ using De Moivre’s theorem.

(JEE Main 2024 pattern — converting to polar form is the key step)

Operation Hierarchy

flowchart TD
    A["Complex Number Problem"] --> B{Operation type?}
    B -->|Add/Subtract| C["Work in a + ib form"]
    B -->|Multiply/Divide| D{Simple or repeated?}
    B -->|Power/Root| E["Convert to polar form"]
    D -->|Simple product| D1["FOIL method or conjugate"]
    D -->|Repeated product| D2["Polar form is faster"]
    E --> F["z = r(cos theta + i sin theta)"]
    F --> G["De Moivre: z^n = r^n(cos n*theta + i sin n*theta)"]
    G --> H["Convert back to a + ib"]

Solution — Step by Step

$z = 1 + i\sqrt{3}$

Modulus: $r = |z| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$

Argument: $\theta = \tan^{-1}\left(\dfrac{\sqrt{3}}{1}\right) = \dfrac{\pi}{3}$

So $z = 2\left(\cos\dfrac{\pi}{3} + i\sin\dfrac{\pi}{3}\right)$

De Moivre’s theorem: $(r(\cos\theta + i\sin\theta))^n = r^n(\cos n\theta + i\sin n\theta)$

z^6 = 2^6\left(\cos\frac{6\pi}{3} + i\sin\frac{6\pi}{3}\right) = 64(\cos 2\pi + i\sin 2\pi)

$\cos 2\pi = 1$ , $\sin 2\pi = 0$

\boxed{z^6 = 64}

A complex number raised to the 6th power gives a real number — because the argument $\pi/3$ multiplied by 6 gives a full rotation ( $2\pi$ ), landing back on the real axis.

Why This Works

De Moivre’s theorem works because multiplication in polar form is elegant: multiply moduli, add arguments. Raising to the $n$ th power means multiplying the argument by $n$ and raising the modulus to the $n$ th power. This turns a messy algebraic expansion into simple arithmetic on angles.

Without polar form, computing $(1 + i\sqrt{3})^6$ would require expanding a binomial with 7 terms. With polar form, we just compute $2^6$ and $6 \times \pi/3$ .

Alternative Method — Step-by-Step Squaring

Compute $z^2$ , then $z^4 = (z^2)^2$ , then $z^6 = z^4 \cdot z^2$ :

$z^2 = (1 + i\sqrt{3})^2 = 1 + 2i\sqrt{3} - 3 = -2 + 2i\sqrt{3}$

$z^3 = z^2 \cdot z = (-2 + 2i\sqrt{3})(1 + i\sqrt{3}) = -2 - 2i\sqrt{3} + 2i\sqrt{3} - 6 = -8$

$z^6 = (z^3)^2 = (-8)^2 = 64$

Notice $z^3 = -8$ . This means $z$ is a cube root of $-8$ . In JEE, if the argument is a simple fraction of $\pi$ (like $\pi/3$ , $\pi/4$ , $\pi/6$ ), De Moivre’s theorem usually gives a clean answer. If you get a messy decimal, recheck your argument calculation.

Common Mistake

The most frequent error: getting the argument wrong. For $z = 1 + i\sqrt{3}$ , the point lies in the first quadrant, so $\theta = \pi/3$ directly. But for $z = -1 + i\sqrt{3}$ (second quadrant), the argument is $2\pi/3$ , not $\pi/3$ . Always check which quadrant the complex number lies in before computing the argument.