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

For any two non-zero complex numbers z 1 and z 2 , arg (frac z 1 z 2 )-arg (z 1 )+arg (z 2 ) is an integer multiple of 2 π

arg (-1-i)=π/4 , where i =√(-1)

The function f: R rightarrow(-π, π] , defined by f(t)=arg (-1+i t) for all t in R , is continuous at all points of R , where i=√(-1)

For any three given distinct complex numbers z 1 , z 2 and z 3 , the locus of the point z satisfying the condition arg (frac (z-z 1 ) (z 2 -z 3 ) (z-z 3 ) (z 2 -z 1 ) )=π , lies on a straight line

Mathematics · Complex Numbers

JEE Advanced 2018 — Paper 1 — Question 27

For a non-zero complex number $z$ , let $\arg (z)$ denote the principal $\operatorname{argument}$ with $-\pi<\arg (z) \leq \pi$ . Then, which of the following statement(s) is (are) FALSE ?

Option A:
$\arg (-1-i)=\frac{\pi}{4}$ , where $\mathrm{i}=\sqrt{-1}$
Option B:
The function $f: R \rightarrow(-\pi, \pi]$ , defined by $f(t)=\arg (-1+i t)$ for all $t \in R$ , is continuous at all points of $R$ , where $i=\sqrt{-1}$
Option C:
For any two non-zero complex numbers $z_{1}$ and $z_{2}, \arg \left(\frac{z_{1}}{z_{2}}\right)-\arg \left(z_{1}\right)+\arg \left(z_{2}\right)$ is an integer multiple of $2 \pi$
Correct
Option D:
For any three given distinct complex numbers $z_{1}, z_{2}$ and $z_{3}$ , the locus of the point $z$ satisfying the condition $\arg \left(\frac{\left(z-z_{1}\right)\left(z_{2}-z_{3}\right)}{\left(z-z_{3}\right)\left(z_{2}-z_{1}\right)}\right)=\pi$ , lies on a straight line

Answer: C

Step-by-step solution

(A) $\operatorname{Arg}(-1-i)=-\frac{3 \pi}{4}$

(B) $\operatorname{Arg}(-1+$ it $)=\left\{\begin{array}{lll}\pi-\tan ^{-1}(\mathrm{t}) & \text { if } & \mathrm{t} \geq 0\\ \tan ^{-1}(\mathrm{t})-\pi & \text { if } & \mathrm{t}<0\end{array}\right.$

Not continuous at $\mathrm{t}=0$

(C) $\operatorname{Arg}\left(\frac{\mathrm{z}_{1}}{\mathrm{z}_{2}}\right)-\arg \left(\mathrm{z}_{1}\right)+\arg \left(\mathrm{z}_{2}\right)=0$

(D) $\operatorname{Arg}\left(\frac{z-z_{1}}{z_{2}-z_{1}}\right)+\operatorname{Arg}\left(\frac{z_{2}-z_{3}}{z-z_{3}}\right)=\pi$

$\Rightarrow \mathrm{z}, \mathrm{z}_{1}, \mathrm{z}_{2}, \mathrm{z}_{3}$ form a cyclic quadrilateral

Hence locus of $z$ is circle

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2018
Paper: Paper 1
Subject: Mathematics
Chapter: Complex Numbers
Topic: Properties of Complex Numbers