# IIT JEE Main Solved Practice Paper Set 1.1: Complex Numbers

Find solved practice paper from the chapter Complex Numbers for IIT JEE Main Exam. This paper consists of 10 questions.

Updated: Jan 9, 2017 16:32 IST Find Solved Practice Paper for IIT JEE Main in this article. This practice paper consists of 10 questions. These questions are from the chapter complex numbers of Mathematics. We have tried to include all important concepts in these questions. Also, detailed solution of each question has been provided. All questions are very important from examination point of view.

Q1. If z1, z2, z3 ,. z4 are the four complex numbers represented by the vertices of a quadrilateral taken in order such that z1 – z4 =  z2  − z3

(a) rhombus

(b) square

(c) rectangle

Since two angle of the quadrilateral is 90° and the sum of opposite angles of a quadrilateral is 180° so, the quadrilateral ABCD is cyclic quadrilateral.

UPSEE 2017 Solved Sample Paper Set-1

Since all the elements of the second row of the determinant is 0, so the value of determinant is zero.

(a) 3ω

(b) 3ω (ω − 1)

(c) 3ω2

(d) 3ω (1 − ω)

Correct Option: (b)

Sol.

Applying C1C1 + C2 + C3 , then

JEE Main Mathematics Solved Sample Paper Set-VII

(a) Roots are purely imaginary for all negative realvalues of m

(b) Roots are not real for all complex numbers m

(c) Roots are purely irnaginary for all positive real values of m

(d) None of the above

Correct Option: (a)

Sol.

Given:

10z2 -4iz -m = 0

By using quadratic formula, we get

If m is negative, then D will also be negative.

So, the roots of z will be purely imaginary.

Hence, the correct option is (a).

JEE Main Physics Solved Sample Paper Set-VII

Q6.   Common roots of the equations z3 +2z2 + 2z + 1 = 0 and z1985 + z100 + 1 = 0 are

(a) ω, ω2

(b) ω, ω3

(c) ω2, ω3

(d) None of these

Sol.

Correct Option: (a)

We have equation,

z3 +2z2 + 2z + 1 = 0

(z + 1) (z2 + z + 1) = 0

Roots of the above equations are −1,  ω and ω2

Let   f(z) = z1985 + z100 + 1

Putting z = −1, ω and ω2 respectively, we get

f(−1) = (−1)1985 + (−1)100 + 1 ≠ 0

Therefore, −1 is not a root of the equation

Again, f(ω) = ω1985 + ω100+ 1

= (ω3)661 ω2 + (ω3)33 ω+1

= ω2 +ω + 1 = 0

Therefore, ω is a root of the equation f(z) = 0.

Similarly, f2) = 0

Hence, ω and ω2 are the common roots.

Q7.    The value of the expression 1×(2 - ω) (2 - ω2) + 2(3 - ω) (3 - ω2)+........+ (n - 1) (n - ω) (n - ω2), where ω is an imaginary cube root of unity is

According to the properties of cube root of unity:

Q8.      Let z be a complex number such that the imaginary part of z is non zero and

a = z2 + z + 1 is real. Then a cannot take the value

(a) - 1

(b) 1/3

(c) 1/2

(d) 3/4

Correct option: (c)

Sol.

We have,

(a) is equal to 5/2

(b) lies in the interval (1, 2)

(c) is strictly greater than 5/2

(d) is strictly greater than 3/2 but less than 5/2

Sol.

Correct option: (b)

JEE Main Chemistry Solved Sample Paper Set-VII

रोमांचक गेम्स खेलें और जीतें एक लाख रुपए तक कैश

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