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Test: Modulus & Argument (May 15) - JEE MCQ


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10 Questions MCQ Test Daily Test for JEE Preparation - Test: Modulus & Argument (May 15)

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Test: Modulus & Argument (May 15) - Question 1

Find the multiplicative inverse of 1 + i?

Detailed Solution for Test: Modulus & Argument (May 15) - Question 1

Concept:
Multiplicative Inverse of a complex number, z  = 1/z
Application:
Multiplicative Inverse of z = 1 + i is z-1 i.e.
Multiplicative Inverse of z = 1/z
Puttng z = 1 + i, we get,
Multiplicative Inverse of 1 + i  = 1/1+i
Rationalizing, 

Test: Modulus & Argument (May 15) - Question 2

Find the conjugate of (6 + 5i)2?

Detailed Solution for Test: Modulus & Argument (May 15) - Question 2

Concept:

  • (a + b)2 = a2 + b2 + 2ab and
  • i2 = -1
  • The conjugate of the complex function is given by changing the sign of i.
  • If z = x + iy , then   = x - iy

Explanation:
Given complex number, z = (6 + 5i)2
i.e.  (6 + 5i) 2 = 62 + (5i)2 +2(6)(5i)
⇒(6 + 5i) 2 = 36 - 25 +60i
⇒(6 + 5i) 2 = 11 + 60i
⇒ z = 11 + 60i
Thus,
Conjugate of z = = 11 - 60i
ADDITIONAL INFORMATION :
If z = x + iy , then

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Test: Modulus & Argument (May 15) - Question 3

The multiplicative inverse of 2 - 3i is

Detailed Solution for Test: Modulus & Argument (May 15) - Question 3

Concept Used:
The multiplicative inverse of a complex number a + bi is given by:

Explanation:
Multiplicative inverse of the complex number 2 - 3i, is 

So, the multiplicative inverse of 2 - 3i is 

Test: Modulus & Argument (May 15) - Question 4

If z = 1 + 3i, find the polynomial generated by its roots?

Detailed Solution for Test: Modulus & Argument (May 15) - Question 4

Concept:
Quadratic Polynomial: A polynomial in which the highest degree monomial is of the second degree, A quadratic polynomial is also known as a second-order polynomial.
The general form of a quadratic polynomial is ax2 + bx + c, where a ≠ 0.
Conjugate of Complex Number: A conjugate of a complex number is another complex number which has the same real part as the original complex number and the imaginary part has the same magnitude but opposite sign.
Formula Used:
(a - b)2 = a2 + b2 - 2ab
(a2 - b2) = (a - b)(a + b)
Calculation:
We have,
z = 1 + 3i is one of the root.
Other root is conjugate of z = 1 - 3i
So the polynomial is quadratic polynomial with two roots. 
we can write the expression for our quadratic polynomial as follows:
⇒ x = (z - z1)(z - z2)
⇒ x = (z - (1 + 3i)) (z - (1 - 3i))
⇒ x = (z - 1 - 3i)(z - 1 + 3i)
⇒ x = [(z - 1) - 3i] [(z - 1) + 3i]
⇒ x = (z - 1)2 - (3i)2
⇒ x = z2 - 2z + 1 - 9i2
⇒ x = z2 - 2z + 1 - 9(-1)
⇒ x = z2 - 2z + 1 + 9
⇒ x = z2 - 2z + 10
∴ If z = 1 + 3i, then the polynomial generated by its roots is z2 - 2z + 10.

Test: Modulus & Argument (May 15) - Question 5

If z is a complex number such that  is purely imaginary, then what is |z| equal to ? 

Detailed Solution for Test: Modulus & Argument (May 15) - Question 5

Concept:
If a number ω = a + ib is purely imaginary, then

  • a = 0 
  • and ω =

Calculation:
Given, z is a complex number such that  is purely imaginary,


∴ The correct option is (3).

Test: Modulus & Argument (May 15) - Question 6

If (1 + i) (x + iy) = 2 + 4i then "5x" is

Detailed Solution for Test: Modulus & Argument (May 15) - Question 6

Equality of complex numbers.
Two complex numbers z1 = x1 + iy1 and z2 = x2 + iy2 are equal if and only if x1 = x2 and y1 = y2
Or Re (z1) = Re (z2) and Im (z1) = Im (z2).
Calculation:
Given: (1 + i) (x + iy) = 2 + 4i
⇒ x + iy + ix + i2y = 2 + 4i
⇒ (x – y) + i(x + y) = 2 + 4i
Equating real and imaginary part,
x - y = 2         …. (1)
x + y = 4        …. (2)
Adding equation 1 and 2, we get
x = 3
Now,
5x = 5 × 3 = 15

Test: Modulus & Argument (May 15) - Question 7

What is the modulus of 

Detailed Solution for Test: Modulus & Argument (May 15) - Question 7

Concept:
Let z = x + iy be a complex number, Where x is called real part of the complex number or Re (z) and y is called Imaginary part of the complex number or Im (z)
Modulus of z = 
Calculations:

As we know i2 = -1 

As we know that if z = x + iy be any complex number, then its modulus is given by,|z| = 

Test: Modulus & Argument (May 15) - Question 8

Find the conjugate of (i - i2)3

Detailed Solution for Test: Modulus & Argument (May 15) - Question 8

1 Concept:
Let z = x + iy be a complex number.

  • Modulus of z = 
  • arg (z) = arg (x + iy) = 
  • For calculating the conjugate, replace i with -i.
  • Conjugate of z = x – iy

Calculation:
Let z = (i - i2)3
⇒ z = i3 (1 - i) 3  = - i (1 - i)3
For calculating the conjugate, replace i with -i.
⇒ z̅  =  -(- i) (1 - (- i))3
⇒ z̅  =  i(1 + i)3
Using (a + b)3 = a3 + b+ 3a2b + 3ab2
⇒ z̅  =  i(1 + i3 +3 ×12 × i + 3 × i2 × 1 ) 
⇒ z̅  =  i(1 - i + 3i - 3) 
⇒ z̅  =  i(-2 + 2i)
⇒ z̅  = -2i + 2i2
⇒ z̅  = -2 - 2 i
So, the conjugate of  (i - i2)3 is -2 - 2i

Test: Modulus & Argument (May 15) - Question 9

What is the value of (i2 + i4 + i6 +... + i2n), Where n is even number.

Detailed Solution for Test: Modulus & Argument (May 15) - Question 9

Concept:
i2 = -1
i3 = - i
i4 = 1
i4n = 1
Calculation:
We have to find the value of (i2 + i4 + i6 +... + i2n)
(i2 + i4 + i6 +... + i2n) = (i2 + i4) + (i6 + i8) + …. + (i2n-2 + i2n)
= (-1 + 1) + (-1 + 1) + …. (-1 + 1)
= 0 + 0 + …. + 0
= 0

Test: Modulus & Argument (May 15) - Question 10

The conjugate of the complex number  is:

Detailed Solution for Test: Modulus & Argument (May 15) - Question 10

Concept: 
Let z = x + iy be a complex number.

  • Modulus of z = 
  • arg (z) = arg (x + iy) = 
  • Conjugate of z = z̅ = x – iy

Calculation:
Given complex number is z = 

Conjugate of z = 

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