Pages

Welcome to Ctanujit Institute of Statistics & Mathematics...An Initiative by 'ISI'ian

Urgent Notice:

Due to unavoidable reason, Ctanujit Classes will be closed from 1st November to 30th November 2018.
Postal Package will be available from 1st December 2018.

IIT JEE SOLVED PAPERS ON COMPLEX NUMBER

Solved problems on Complex Numbers from IIT-JEE Papers

Q. [IIT-JEE 2004] If ω (≠ 1) be n cube root of unity and (1+ ω2)n =(1+ ω4)n , then the least positive value of n is
(a) 2       (b) 3      (c) 5       (d) 6
Solution :(b)   Using   ω4 = ω3. ω= ω
Now 1+ ω2 = -ω, we get (-ω)n =(1+ ω)n = (-ω2)n implying  ωn = ω2n
ωn= 1 n = 3.
Q. [IIT-JEE 2005] If a, b, c are integers not all equal and ω is a cube root of unity (ω ≠ 1), then the minimum value of   a+bω+cω2 is
(a) 0      (b) 1      (c) √3/2      (d) ½
Solution :(b) Let   y = a+bω+cω2
y2 = a+bω+cω22  = a2+b2+c2– ab–bc –ca (by actual multiplication)
= ½[(a-b)2+(b-c)2+(c-a)2] --------------(1)
Let    a ≠ b, b ≠ c, c ≠ a, Then difference of two integers is an integer. So, (b-c)2≥ 1, (c-a)2≥ 1
From (1), we have  y2 ≥ 1 . The minimum value of y is 1.
Q. [IIT-JEE 2002] For all complex numbers z1, z2 satisfying z1=12 and   z2-3-4i = 5, the minimum value of  z1-z2 is
(a) 0      (b) 2      (c) 7     (d) 17
Solution :(b)  We know   z1-z2z1 - z2
Also,     z1-z2-z3z1 - z2 - z3
z1-z2 = z1- (z2-3-4i)-(3+4i)z1 - z2-3-4i - 3+4i = 12 – 5 – 5 = 2.