## Complex Numbers Quiz-4

As per analysis for previous years, it has been observed that students preparing for JEE MAINS find Mathematics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Mathematics due to their ENGINEERING background.

Furthermore, sections such as Mathematics are dominantly based on theories, laws, numerical in comparison to a section of Engineering which is more of fact-based, Physics, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced.

Q1. The general value of Î¸ which satisfies the equation (cos⁡Î¸+i sin⁡Î¸ )(cos⁡3Î¸+i sin⁡3Î¸ )(cos⁡5Î¸+i sin⁡5Î¸ )… (cos⁡(2n-1)Î¸+i sin⁡(2n-1)Î¸=1) is
•  (r Ï€)/n2
•  (r-1)Ï€/n2
•  (2r+1)Ï€/n3
•  2rÏ€/n2
Solution
(d) We have, (cos⁡Î¸+ i sin⁡Î¸ )(cos⁡3 Î¸ + i sin⁡3 Î¸ ) …[cos⁡(2 n-1)Î¸+i sin⁡(2n-1)Î¸ ]=1+i 0
⇒cos⁡[Î¸+3 Î¸+5 Î¸+⋯+(2 n-1)Î¸]+i sin⁡[Î¸+3 Î¸+5 Î¸+⋯+(2n-1)Î¸]=1+i0
⇒cos⁡(n2 Î¸)+i sin⁡(n2 Î¸)=1+i0
⇒cos⁡n2 Î¸=1 and sin⁡ n2 Î¸=0
⇒n2 Î¸=2 r Ï€
⇒Î¸=(2 r Ï€)/n2

Q2.If n is a positive integer, then (1+i)n+(1-i)n is equal to
•  (√2)(n-2) cos⁡(nÏ€/4)
•  (√2)(n-2) sin⁡(nÏ€/4)
•  (√2)(n+2) cos⁡(nÏ€/4)
•  (√2)(n+2) sin⁡(nÏ€/4)
Solution
(c) (1+i)n+(1-i)n =(√2 (1/√2+i/√2))n+(√2 (1/√2-i/√2))n
=2n/2 (cos⁡Ï€/4+i sin⁡Ï€/4 )n+2n/2 (cos⁡Ï€/4-i sinÏ€/4)n
=2n/2 (cos⁡n Ï€/4+sin nÏ€/4+cos nÏ€/4-i sin⁡ nÏ€/4)
=2n/2+1 cos⁡(nÏ€/4)
=(√2)n+2 cos⁡(nÏ€/4)
Q3.  Let a=e(i 2Ï€/3 ). Then, the equation whose roots are a+a(-2) and a2+a(-4) is
•   x2-2x+4=0
•  x2-x+1=0
•  x2+x+4=0
•  x2+2x+4=0
Solution
(d) Here, a=e(i 2Ï€/3)=Ï‰
∴ a+1/a2 =Ï‰+1/Ï‰2 =Ï‰+Ï‰=2Ï‰
Similarly, a2+1/a2 =Ï‰2+1/Ï‰4 =2Ï‰2
∴ a+1/a2 +a2+1/a4 =2Ï‰+2Ï‰2=-2
And (a+1/a2 )(a2+1/a4 )=2Ï‰.2Ï‰2=4
∴ required equation is x2+2x+4=0

Q4. If z2+z|z|+|z|2=0, then the locus of z is
•  A circle
•  A straight line
•  A pair of straight lines
•  None of these
Solution
(c) We have, z2+z|z|+|z|2=0
⇒ (z/|z| )2+z/|z| +1=0
This is a quadratic equation in z/|z| , therefore roots are z/|z| =Ï‰,Ï‰2
⇒ z=Ï‰|z| or z=Ï‰2 |z|
Let z=x+iy
⇒ x+iy=|z|((-1)/2+(i√3)/2) or x+iy=|z|((-1)/2-(i√3)/2)
⇒ x=-1/2 |z|,y=|z| √3/2 or x=-|z|/2,y=-(|z| √3)/2
⇒ y+√3 x=0 or y-√3 x=0
⇒ y2-3x2=0
⇒ It represents a pair of straight lines

Q5.The points representing cube roots of unity
•  Are collinear
•  Lie on a circle of radius √3
•  From an equilateral triangle
•  None of these
Solution
(c) The cube roots or unity are 1,Ï‰,Ï‰2.
Let P,Q and R represent 1,Ï‰ and Ï‰2 respectively.
Clearly, PQ=|1-Ï‰|=√((3/2)2+(√3/2)2 )=√3 QR=|Ï‰-Ï‰2 |=√3, and RP=|1-Ï‰2 |=√3
∴PQ=QR=RP
Thus, points representing 1,Ï‰,Ï‰2 form an equilateral triangle.
ALITER
Let zi=1,z_2=Ï‰ and z_3=Ï‰2
Then, z12+z22+z23=z1 z2+z2 z3+z3 z1
Hence, points representing 1,Ï‰,Ï‰2 form an equilateral triangle

Q6. If (x-2) is a common factor of the expressions x2+ax+b and x2+cx+d, then (b-d)/(c-a) is equal to
•  -2
•  -1
• 1
•  2
Solution
(d) Since, (x-2) is a common factor of the expressions x2+ax+b and x2+cx+d ⇒4+2a+b=0 ...(i)
and 4+2c+d=0 ...(ii)
⇒2a+b=2c+d
⇒b-d=2(c-a)
⇒ (b-d)/(c-a)=2

Q7.The centre and the radius of the circle z.zc+(2+3i) zc+(2-3i)z+12=0 are respectively
•  -(2+3i),(1)
•  (3+2i),(1)
•  (3+6i),(3)
•  None of these
Solution
(a) Given equation of circle is zzc+(2+3i) zc+(2-3i)z+12=0
Here, centre is {-(2+3i)} and radius =√(|2+3i|2-12)=√(13-12)=1

Q8.For any complex number z, the minimum value of |z|+|z-1| is
•  0
•  1
•  2
•  -1
Solution
(b) Since, |-z|=|z| And |z1+z2 |≤|z1 |+|z2
Now, |z|+|z-1|=|z|+|1-z|≥|z+(1-z)|=1

Q9.

•  √2(1+i)/10
•  √2(1-i)/10
•  10(1-i)/√2
•  10(1+i)/√2
Solution

Q10. Number of solutions of the equation z2+|z|2=0, where z∈C is
•  1
•  2
•  3
• Infinitely many
Solution
(d) Let z=x+i y.
Then, z2+|z|2=0
⇒x2-y2+2 ixy+x2+y2=0
⇒2x2+2ixy=0
⇒x2=0,2xy=0
⇒x=0,y∈R
Hence, there are infinitely many solution

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