Verified Solution • 2025–26 Session moderate

QJEEAdv18Ma88cc: JEE Advanced 2018 Mathematics - Class 11 JEEAdv18Ma88cc Solution: For a non-zero complex number z, let arg(z) denote the…

For a non-zero complex number z, let arg(z) denote the principal argument with −- π\pi < arg(z) ≤\le π\pi . Then, which of the following statement(s) is (are) FALSE?

Multiple Choice Options

  • A arg( −- 1 −- i) = π4{\pi \over 4} , where i = −1\sqrt { - 1}
  • B The function f : R →\to ( −- π\pi , π\pi ), defined by f(t) = arg ( −- 1 + it) for all t ∈\in R, is continuous at all points of R, where i = −1\sqrt { - 1} .
  • C For any two non-zero complex numbers z1 and z2, arg (z1z2)\left( {{{{z_1}} \over {{z_2}}}} \right) −- arg (z1) + arg(z2) is an integer multiple of 2 π\pi .
  • D For any three given distinct complex numbers z1, z2 and z3, the locus of the point z satisfying the condition arg ((z−z1)(z2−z3)(z−z3)(z2−z1))=π\left( {{{(z - {z_1})({z_2} - {z_3})} \over {(z - {z_3})({z_2} - {z_1})}}} \right) = \pi , lies on a straight line.

Step-by-Step Verified Solution

(a) Let z= −- 1 −- i and arg(z) = θ\theta

Now, tan⁡θ=∣im(z)Re(z)∣=∣−1−1∣=1\tan \theta = \left| {{{im(z)} \over {{\mathop{\rm Re}\nolimits} (z)}}} \right| = \left| {{{ - 1} \over { - 1}}} \right| = 1

⇒\Rightarrow θ=π4\theta = {\pi \over 4}

Since, x < 0, y < 0

∴\therefore arg⁡(z)=−(π−π4)=−3π4\arg (z) = - \left( {\pi - {\pi \over 4}} \right) = - {{3\pi } \over 4}

(b) We have, f(t) = arg( −- 1 + it)

arg⁡(−1+it)={π−tan−1t,t≥0−(π+tan−1t),t<0

This function is discontinuous at t = 0.

(c) We have,

arg⁡(z1z2)=arg(z1)−arg(z2)+2nπ\arg \left( {{{{z_1}} \over {{z_2}}}} \right) = arg({z_1}) - arg({z_2}) + 2n\pi

∴\therefore arg⁡(z1z2)−arg(z1)+arg(z2)\arg \left( {{{{z_1}} \over {{z_2}}}} \right) - arg({z_1}) + arg({z_2})

= arg(z1) −- arg(z2) + 2n π\pi −- arg(z1) + arg(z2) = 2n π\pi

So, given expression is multiple of 2 π\pi .

(d) We have, arg⁡((z−z1)(z2−z3)(z−z3)(z2−z1))=π\arg \left( {{{(z - {z_1})({z_2} - {z_3})} \over {(z - {z_3})({z_2} - {z_1})}}} \right) = \pi

JEE Advanced 2018 Paper 1 Offline Mathematics - Complex Numbers Question 15 English Explanation
Now, arg⁡(z1−zz3−z)=θ\arg \left( {{{{z_1} - z} \over {{z_3} - z}}} \right) = \theta

and arg⁡(z3−z2z1−z2)=π−θ\arg \left( {{{{z_3} - {z_2}} \over {{z_1} - {z_2}}}} \right) = \pi - \theta

Therefore, arg⁡(z1−zz3−z)+arg⁡(z3−z2z1−z2)=θ+π−θ\arg \left( {{{{z_1} - z} \over {{z_3} - z}}} \right) + \arg \left( {{{{z_3} - {z_2}} \over {{z_1} - {z_2}}}} \right) = \theta + \pi - \theta

That is,

arg⁡(z1−zz3−z.z3−z2z1−z2)=arg⁡((z1−z)(z3−z2)(z3−z)(z1−z2))=π\arg \left( {{{{z_1} - z} \over {{z_3} - z}}.{{{z_3} - {z_2}} \over {{z_1} - {z_2}}}} \right) = \arg \left( {{{\left( {{z_1} - z} \right)\left( {{z_3} - {z_2}} \right)} \over {\left( {{z_3} - z} \right)\left( {{z_1} - {z_2}} \right)}}} \right) = \pi

This implies that for any three given distinct complex numbers z1 , z2 and z3 , the locus of the point z satisfying the condition
arg⁡((z1−z)(z3−z2)(z3−z)(z1−z2))=π\arg \left( {{{\left( {{z_1} - z} \right)\left( {{z_3} - {z_2}} \right)} \over {\left( {{z_3} - z} \right)\left( {{z_1} - {z_2}} \right)}}} \right) = \pi , lies on a circle.

∴\therefore (a), (b), (d) are false statement.

Hence, option (a), (b), (d) are correct answer.

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