## Pages

### 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.