Complex Numbers: PYQ Walkthrough (10)

easy • 2 min read
Tags Complex Numbers

Question

If z=1+i3z = 1 + i\sqrt{3}, find the modulus and argument of zz, and write zz in polar form. (JEE Main 2022 pattern)

Solution — Step by Step

z=1+i3z = 1 + i\sqrt{3} has real part a=1a = 1 and imaginary part b=3b = \sqrt{3}.

 ∣z∣=a2+b2=1+3=2|z| = \sqrt{a^2 + b^2} = \sqrt{1 + 3} = 2

tan⁡θ=b/a=3/1=3\tan\theta = b/a = \sqrt{3}/1 = \sqrt{3}. Both a,b>0a, b > 0, so zz is in the first quadrant.

θ=tan⁡−1(3)=π3\theta = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} z=∣z∣(cos⁡θ+isin⁡θ)=2(cos⁡π3+isin⁡π3)z = |z|(\cos\theta + i\sin\theta) = 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)

In Euler form: z=2eiπ/3z = 2 e^{i\pi/3}.

Final answer: ∣z∣=2|z| = 2, arg⁡(z)=π/3\arg(z) = \pi/3, polar form z=2(cos⁡π3+isin⁡π3)z = 2(\cos\tfrac{\pi}{3} + i\sin\tfrac{\pi}{3}).

Why This Works

Polar form turns multiplication into addition of angles, and powers into multiplication of angles (De Moivre’s theorem). Whenever a JEE problem asks for znz^n, z1/nz^{1/n}, or roots of unity, convert to polar form first.

For 1+i31 + i\sqrt{3}, the angle π/3\pi/3 is recognisable from the standard 3030-6060-9090 triangle. Always check the quadrant before reporting the argument.

Alternative Method

Plot zz in the Argand plane. From (0,0)(0,0) to (1,3)(1, \sqrt{3}) is a vector of length 22 at angle 60°60° above the real axis. Same answer, geometrically.

For JEE Main, memorise the polar form of standard numbers: 1+i=2eiπ/41 + i = \sqrt{2}e^{i\pi/4}, 1+i3=2eiπ/31 + i\sqrt{3} = 2e^{i\pi/3}, 3+i=2eiπ/6\sqrt{3} + i = 2e^{i\pi/6}. These appear repeatedly in PYQs.

Common Mistake

Reporting the argument as tan⁡−1(3)=60°\tan^{-1}(\sqrt{3}) = 60° without checking the quadrant. For z=−1−i3z = -1 - i\sqrt{3}, the same tan⁡\tan value gives 60°60° from the calculator, but the actual argument is −120°-120° or 240°240°. Always plot first, compute second.

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