Argument of complex number (3+4i)/(1-2i) — convert to polar form

easy JEE-MAIN • JEE Main 2023 • 3 min read
Tags Complex Numbers

Question

Find the argument of z=3+4i1−2iz = \dfrac{3 + 4i}{1 - 2i} and express zz in polar form.

(JEE Main 2023, similar pattern)

Solution — Step by Step

Multiply numerator and denominator by the conjugate of the denominator:

z=3+4i1−2i×1+2i1+2i=(3+4i)(1+2i)(1−2i)(1+2i)z = \frac{3+4i}{1-2i} \times \frac{1+2i}{1+2i} = \frac{(3+4i)(1+2i)}{(1-2i)(1+2i)}

Numerator: (3+4i)(1+2i)=3+6i+4i+8i2=3+10i−8=−5+10i(3+4i)(1+2i) = 3 + 6i + 4i + 8i^2 = 3 + 10i - 8 = -5 + 10i

Denominator: (1)2+(2)2=1+4=5(1)^2 + (2)^2 = 1 + 4 = 5 (using ∣a−bi∣2=a2+b2|a - bi|^2 = a^2 + b^2)

z=−5+10i5=−1+2iz = \frac{-5 + 10i}{5} = -1 + 2i

Modulus: ∣z∣=(−1)2+22=1+4=5|z| = \sqrt{(-1)^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}

For the argument, z=−1+2iz = -1 + 2i lies in the second quadrant (negative real, positive imaginary).

tan⁡α=∣2−1∣=2  ⟹  α=tan⁡−1(2)\tan \alpha = \left|\frac{2}{-1}\right| = 2 \implies \alpha = \tan^{-1}(2)

Since the point is in Q2: arg⁡(z)=π−tan⁡−1(2)\arg(z) = \pi - \tan^{-1}(2)

 z=5[cos⁡(π−tan⁡−12)+isin⁡(π−tan⁡−12)]z = \sqrt{5}\left[\cos\left(\pi - \tan^{-1}2\right) + i\sin\left(\pi - \tan^{-1}2\right)\right]

Or equivalently: z=5 ei(π−tan⁡−12)z = \sqrt{5}\, e^{i(\pi - \tan^{-1}2)}

The argument is π−tan⁡−1(2)\mathbf{\pi - \tan^{-1}(2)} (approximately 2.0342.034 radians or about 116.6°116.6°).

Why This Works

Dividing complex numbers in Cartesian form is messy because of the ii in the denominator. Multiplying by the conjugate converts the denominator to a real number (a2+b2a^2 + b^2), making the division clean.

The argument (angle) tells us the direction of zz in the Argand plane. For a complex number x+iyx + iy:

  • Q1: θ=tan⁡−1(y/x)\theta = \tan^{-1}(y/x)
  • Q2: θ=π−tan⁡−1(∣y/x∣)\theta = \pi - \tan^{-1}(|y/x|)
  • Q3: θ=−π+tan⁡−1(∣y/x∣)\theta = -\pi + \tan^{-1}(|y/x|)
  • Q4: θ=−tan⁡−1(∣y/x∣)\theta = -\tan^{-1}(|y/x|)

Getting the quadrant right is critical. The principal argument lies in (−π,π](-\pi, \pi].

Alternative Method

You can find the argument without simplifying the fraction: arg⁡(z1/z2)=arg⁡(z1)−arg⁡(z2)\arg(z_1/z_2) = \arg(z_1) - \arg(z_2).

arg⁡(3+4i)=tan⁡−1(4/3)\arg(3+4i) = \tan^{-1}(4/3) and arg⁡(1−2i)=−tan⁡−1(2)\arg(1-2i) = -\tan^{-1}(2) (Q4).

arg⁡(z)=tan⁡−1(4/3)−(−tan⁡−12)=tan⁡−1(4/3)+tan⁡−1(2)\arg(z) = \tan^{-1}(4/3) - (-\tan^{-1}2) = \tan^{-1}(4/3) + \tan^{-1}(2)

Using tan⁡−1a+tan⁡−1b=π+tan⁡−1a+b1−ab\tan^{-1}a + \tan^{-1}b = \pi + \tan^{-1}\frac{a+b}{1-ab} when ab>1ab > 1:

=π+tan⁡−14/3+21−8/3=π+tan⁡−110/3−5/3=π+tan⁡−1(−2)=π−tan⁡−1(2)= \pi + \tan^{-1}\frac{4/3 + 2}{1 - 8/3} = \pi + \tan^{-1}\frac{10/3}{-5/3} = \pi + \tan^{-1}(-2) = \pi - \tan^{-1}(2). Same answer.

For JEE MCQs, the conjugate multiplication method is faster. The argument subtraction method is elegant but involves inverse tangent addition formulas that can be error-prone under time pressure.

Common Mistake

The most common error: using θ=tan⁡−1(y/x)\theta = \tan^{-1}(y/x) blindly without checking the quadrant. For z=−1+2iz = -1 + 2i, the calculator gives tan⁡−1(2/(−1))=tan⁡−1(−2)≈−63.4°\tan^{-1}(2/(-1)) = \tan^{-1}(-2) \approx -63.4°. This is Q4, but our point is in Q2. You must add π\pi (or 180°180°) to get the correct argument. Always plot the point first to identify the quadrant.

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