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Complex Numbers question

2021 · 20 Jul · Shift 1 · Q27
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  5. /2021 · 20 Jul · Shift 1 · Q27

Complex Numbers question

2021 · 20 Jul · Shift 1 · Q27

JEE MainMathematicsComplex NumbersMCQ+4 / −1
If z and ω\omegaω are two complex numbers such that ∣zω∣=1\left| {z\omega } \right| = 1∣zω∣=1 and arg⁡(z)−arg⁡(ω)=3π2\arg (z) - \arg (\omega ) = {{3\pi } \over 2}arg(z)−arg(ω)=23π​, then arg⁡(1−2z‾ω1+3z‾ω)\arg \left( {{{1 - 2\overline z \omega } \over {1 + 3\overline z \omega }}} \right)arg(1+3zω1−2zω​) is : (Here arg(z) denotes the principal argument of complex number z)
  1. A
    π4{\pi \over 4}4π​
  2. B
    −3π4- {{3\pi } \over 4}−43π​
  3. C
    −π4- {\pi \over 4}−4π​
  4. D
    3π4{{3\pi } \over 4}43π​
View written solutionFree

Correct answer: B

  1. Given conditions

We are given: ∣zω∣=1|z\omega|=1∣zω∣=1 and arg⁡(z)−arg⁡(ω)=3π2.\arg(z)-\arg(\omega)=\frac{3\pi}{2}.arg(z)−arg(ω)=23π​.

We need to find arg⁡(1−2z‾ ω1+3z‾ ω).\arg\left(\frac{1-2\overline z\,\omega}{1+3\overline z\,\omega}\right).arg(1+3zω1−2zω​).

  1. Find arg⁡(z‾ ω)\arg(\overline z\,\omega)arg(zω) and its modulus

Let z=r1eiθ1,ω=r2eiθ2.z=r_1 e^{i\theta_1},\qquad \omega=r_2 e^{i\theta_2}.z=r1​eiθ1​,ω=r2​eiθ2​.

Then z‾ ω=r1r2ei(θ2−θ1).\overline z\,\omega = r_1 r_2 e^{i(\theta_2-\theta_1)}.zω=r1​r2​ei(θ2​−θ1​).

Now, ∣zω∣=∣z∣∣ω∣=r1r2=1.|z\omega|=|z||\omega|=r_1r_2=1.∣zω∣=∣z∣∣ω∣=r1​r2​=1. So, ∣z‾ ω∣=∣z∣∣ω∣=1.|\overline z\,\omega|=|z||\omega|=1.∣zω∣=∣z∣∣ω∣=1.

Also, arg⁡(z)−arg⁡(ω)=θ1−θ2=3π2.\arg(z)-\arg(\omega)=\theta_1-\theta_2=\frac{3\pi}{2}.arg(z)−arg(ω)=θ1​−θ2​=23π​. Hence, θ2−θ1=−3π2.\theta_2-\theta_1=-\frac{3\pi}{2}.θ2​−θ1​=−23π​. Modulo 2π2\pi2π, −3π2≡π2(mod2π).-\frac{3\pi}{2}\equiv \frac{\pi}{2} \pmod{2\pi}.−23π​≡2π​(mod2π).

Therefore, z‾ ω=eiπ/2=i.\overline z\,\omega = e^{i\pi/2}=i.zω=eiπ/2=i.

  1. Substitute into the expression

We need

=\arg\left(\frac{1-2i}{1+3i}\right).$$ Now simplify: $$\frac{1-2i}{1+3i}\cdot\frac{1-3i}{1-3i} =\frac{(1-2i)(1-3i)}{1+9}.$$ Compute the numerator: $$(1-2i)(1-3i)=1-3i-2i+6i^2=1-5i-6=-5-5i.$$ So, $$\frac{1-2i}{1+3i}=\frac{-5-5i}{10}=-\frac12-\frac12 i.$$ --- 4. **Find the principal argument** The complex number $$-\frac12-\frac12 i$$ is in the third quadrant, with equal negative real and imaginary parts. Hence its argument is $$-\frac{3\pi}{4}$$ (as principal argument lies in $(-\pi,\pi]$). Therefore, $$\arg\left(\frac{1-2\overline z\,\omega}{1+3\overline z\,\omega}\right)=-\frac{3\pi}{4}.$$ --- 5. **Compare with options** This matches **Option B**. --- 6. **Compare with stored correct answer** Stored correct answer: **B** Our derived answer: **B** So they agree.
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