Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
314 views
in Complex Numbers by (95.5k points)
closed by
For a non-zero complex number `z` , let `arg(z)` denote the principal argument with `pi lt arg(z)leq pi` Then, which of the following statement(s) is (are) FALSE? `arg(-1,-i)=pi/4,` where `i=sqrt(-1)` (b) The function `f: R->(-pi, pi],` defined by `f(t)=arg(-1+it)` for all `t in R` , is continuous at all points of `RR` , where `i=sqrt(-1)` (c) For any two non-zero complex numbers `z_1` and `z_2` , `arg((z_1)/(z_2))-arg(z_1)+arg(z_2)` is an integer multiple of `2pi` (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(((z-z_1)(z_2-z_3))/((z-z_3)(z_2-z_1)))=pi` , lies on a straight line
A. `arg (-1 -i) = pi/4 "where" i= sqrt(-1)`
B. The function f: R `rArr (- pi ,pi]` defined by f(t) = arg (-1 it) for all `t ne R` is continuous at all points of R, where `i=sqrt(-1)`
C. For any two non-zero complex numbers `z_1 and z_2 arg (z_1)+ arg (z_2)` is an interger multiple of arg `(z_1)/(z_2)-arg (z_1)+arg (z_2)` is an interger multiple of `2 pi`
D. For any three given distinct complex number and `z_1 ,z_2 and z_3` the locus of the point z satisfying the condition arg `((z-z_1)(z_2-z_3))/((z-z_3)(z_2-z_1))=pi`

1 Answer

0 votes
by (94.8k points)
selected by
 
Best answer
Correct Answer - A::B::D
Let z=-1-I and arg(z) = `theta`
Now `tan theta =|(lim (z))/(Re (z) )|=|(-1)/(-1)|=1`
`rArr theta =pi/4`
`Since x lt 0, y gt 0 `
`therefore (z) =- (pi-pi/4)=-(3pi)/(4)`
(b) We have ,f(t) =arg (-1 + it )
`arg(-1+it)={:(pi- tan^(-1) t"," t ge 0 ),-(pi + tan ^(-1))","t lt 0 ):}`
This function is discontinuous at t = 0.
(c ) We have
`arg (z_1/z_2)-arg (z_1)+arg(z_2)+2n pi `
`therefore arg(z_1/z_2)-arg (z_1)+arg(z_2)`
`therefore are (z_1/z_2)-arg (z_1)+arg (z_2)`
`= arg (z_1)-arg (z_2)+2 n pi -arg (z_1)+arg (z_2)`
`=2 n pi`
So given expression is multiple of `2 pi `
(d) We have arg `(((z-z_1)(z_2-z_3))/((z-z_3)(z_2-z_1)))=pi`
`rArr ((z-z_1)/(z-z_3))((z_2-z3)/(z_2-z_1))` is purely real
Thus , the points `A(z_1)B(z_2),C(z_3)` and D(z) taken in order would be concylic if purely real .
Hence, it is a circle
image
`threrefore (a),(b),(d)` are false statment.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...