JEE Challenger
More from Complex Numbers and Quadratic Equations
MathMedAdv 2018 • Paper 1

Identify False Statements on Complex Numbers and Principal Arguments

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

Options

A

arg⁡(−1−i)=π4\arg(-1 - i) = \frac{\pi}{4}, where i=−1i = \sqrt{-1}

Correct
B

The function f:R→(−π,π]f: \mathbb{R} \to (-\pi, \pi], defined by f(t)=arg⁡(−1+it)f(t) = \arg(-1 + it) for all t∈Rt \in \mathbb{R}, is continuous at all points of R\mathbb{R}, where i=−1i = \sqrt{-1}

Correct
C

For any two non-zero complex numbers z1z_1 and z2z_2, arg⁡(z1z2)−arg⁡(z1)+arg⁡(z2)\arg\left(\frac{z_1}{z_2}\right) - \arg(z_1) + \arg(z_2) is an integer multiple of 2π2\pi

D

For any three given distinct complex numbers z1z_1, z2z_2 and z3z_3, the locus of the point zz satisfying the condition arg⁡((z−z1)(z2−z3)(z−z3)(z2−z1))=π,\arg\left(\frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}\right) = \pi, lies on a straight line

Correct

Topics & Concepts

Complex Numbers and Quadratic EquationsComplex NumbersArgand Plane

Step-by-Step Solution

To determine which of the given statements are FALSE, we analyze each option individually:

  1. Option (A): The complex number z=−1−iz = -1 - i lies in the third quadrant of the complex plane (Re(z)<0\text{Re}(z) < 0 and Im(z)<0\text{Im}(z) < 0). The principal argument arg⁡(z)∈(−π,π]\arg(z) \in (-\pi, \pi] is given by: arg⁡(−1−i)=−π+arctan⁡(∣−1−1∣)=−π+π4=−3π4≠π4\arg(-1 - i) = -\pi + \arctan\left(\left|\frac{-1}{-1}\right|\right) = -\pi + \frac{\pi}{4} = -\frac{3\pi}{4} \neq \frac{\pi}{4} Therefore, statement (A) is FALSE.

  2. Option (B): The function is defined as f(t)=arg⁡(−1+it)f(t) = \arg(-1 + it) for all t∈Rt \in \mathbb{R}.

    • For t>0t > 0, the complex number −1+it-1 + it lies in the second quadrant, so: f(t)=π−arctan⁡(t)f(t) = \pi - \arctan(t) Taking the right-hand limit as t→0+t \to 0^+: lim⁡t→0+f(t)=lim⁡t→0+(π−arctan⁡(t))=π\lim_{t \to 0^+} f(t) = \lim_{t \to 0^+} (\pi - \arctan(t)) = \pi
    • For t=0t = 0: f(0)=arg⁡(−1)=πf(0) = \arg(-1) = \pi
    • For t<0t < 0, the complex number −1+it-1 + it lies in the third quadrant, so: f(t)=−π−arctan⁡(t)f(t) = -\pi - \arctan(t) Taking the left-hand limit as t→0−t \to 0^-: lim⁡t→0−f(t)=lim⁡t→0−(−π−arctan⁡(t))=−π\lim_{t \to 0^-} f(t) = \lim_{t \to 0^-} (-\pi - \arctan(t)) = -\pi Since lim⁡t→0+f(t)≠lim⁡t→0−f(t)\lim_{t \to 0^+} f(t) \neq \lim_{t \to 0^-} f(t), the function f(t)f(t) is discontinuous at t=0t = 0. Therefore, statement (B) is FALSE.

  3. Option (C): For any two non-zero complex numbers z1z_1 and z2z_2, let their polar representations be z1=r1eiθ1z_1 = r_1 e^{i\theta_1} and z2=r2eiθ2z_2 = r_2 e^{i\theta_2}, where θ1=arg⁡(z1)\theta_1 = \arg(z_1) and θ2=arg⁡(z2)\theta_2 = \arg(z_2). Then: z1z2=r1r2ei(θ1−θ2)\frac{z_1}{z_2} = \frac{r_1}{r_2} e^{i(\theta_1 - \theta_2)} By definition of the principal argument: arg⁡(z1z2)=θ1−θ2+2kπ=arg⁡(z1)−arg⁡(z2)+2kπfor some k∈{−1,0,1}\arg\left(\frac{z_1}{z_2}\right) = \theta_1 - \theta_2 + 2k\pi = \arg(z_1) - \arg(z_2) + 2k\pi \quad \text{for some } k \in \{-1, 0, 1\} Rearranging gives: arg⁡(z1z2)−arg⁡(z1)+arg⁡(z2)=2kπ\arg\left(\frac{z_1}{z_2}\right) - \arg(z_1) + \arg(z_2) = 2k\pi which is indeed an integer multiple of 2π2\pi. Therefore, statement (C) is TRUE.

  4. Option (D): The condition: arg⁡((z−z1)(z2−z3)(z−z3)(z2−z1))=π\arg\left(\frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}\right) = \pi implies that the cross-ratio of the four points z,z2,z1,z3z, z_2, z_1, z_3 is a negative real number. In the Argand plane, if the cross-ratio of four distinct points is real, the points z,z1,z2,z3z, z_1, z_2, z_3 are concyclic (or collinear if all three given points are collinear). If z1,z2,z3z_1, z_2, z_3 form the vertices of a non-degenerate triangle, the locus of zz represents a circular arc of the circumcircle passing through these points, which is a circle and not a straight line. Therefore, statement (D) is FALSE.

Hence, the statements that are FALSE are A, B, and D.

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