Class 11 RD Sharma Solutions – Chapter 13 Complex Numbers – Exercise 13.2 | Set 2

Question 14. If, find (a, b).

Solution:

We have,

=>

=>

=>

=>

=> (−i)¹⁰⁰ = a + ib

=> a + ib = 1

On comparing real and imaginary parts on both sides, we get,

=> (a, b) = (1, 0)

Question 15. If a = cos θ + i sin θ, find the value of.

Solution:

Given a = cos θ + i sin θ, we get,

=

=

=

=

=

=

=

=

=

=

Therefore, the value ofis.

Question 16. Evaluate the following :

(i) 2x³ + 2x² − 7x + 72, when x = (3−5i)/2

Solution:

We have, x = (3−5i)/2

=> 2x = 3 − 5i

=> 2x − 3 = −5i

=> (2x − 3)² = 25i²

=> 4x² + 9 − 12x = −25

=> 4x² − 12x + 34 = 0

=> 2x² − 6x + 17 = 0

Now, 2x³ + 2x² − 7x + 72 = x (2x² − 6x + 17) + 6x² − 17x + 2x² − 7x + 72

= x (0) + 8x² − 24x + 72

= 4 (2x² − 6x + 17) + 4

= 4 (0) + 4

= 4

Therefore, the value of 2x³ + 2x² − 7x + 72 is 4.

(ii) x⁴ − 4x³ + 4x² +8x +44, when x = 3 + 2i

Solution:

We have, x = 3 + 2i

=> x − 3 = 2i

=> (x − 3)² = (2i)²

=> x² + 9 − 6x = 4i²

=> x² − 6x + 9 + 4 = 0

=> x² − 6x + 13 = 0

Now, x⁴ − 4x³ + 4x² + 8x + 44 = x² (x² − 6x + 13) + 6x³ − 13x² − 4x³ + 4x² + 8x + 44

= 2x³ − 9x² + 8x + 44

= 2x (x² − 6x + 13) + 12x² − 26x − 9x² + 8x + 44

= 3x² − 18x + 44

= 3 (x² − 6x + 13) + 5

= 5

Therefore, the value of x⁴ − 4x³ + 4x² + 8x + 44 is 5.

(iii) x⁴ + 4x³ + 6x² + 4x + 9, when x = −1 + i√2

Solution:

We have, x = −1 + i√2

=> x + 1 = i√2

=> (x + 1)² = 2i²

=> x² + 1 + 2x = −2

=> x² + 2x + 3 = 0

Now, x⁴ + 4x³ + 6x² + 4x + 9 = x² (x² + 2x + 3) − 2x³ − 3x² + 4x³ + 6x² + 4x + 9

= 2x³ + 3x² + 4x + 9

= 2x (x² + 2x + 3) − 4x² − 6x + 3x² + 4x + 9

= − x² − 2x + 9

= − (x² + 2x + 3) + 3 + 9

= 3 + 9

= 12

Therefore, the value of x⁴ + 4x³ + 6x² + 4x + 9 is 12.

(iv) x⁶ + x⁴ + x² + 1, when x = (1+i)/√2

Solution:

We have, x = (1+i)/√2

=> √2x = 1 + i

=> 2x² = 1 + i² + 2i

=> 2x² = 2i

=> 4x⁴ = 4i²

=> x4 = −1

=> x⁴ + 1 = 0

Now, x⁶ + x⁴ + x² + 1 = (x⁶ + x²) + (x⁴ +1)

= x⁶ + x²

= x² (x⁴ + 1)

= 0

Therefore, the value of x⁶ + x⁴ + x² + 1 is 0.

(v) 2x⁴ + 5x³ + 7x² − x + 41, when x = −2 − √3i

Solution:

We have, x = −2 − √3i

x² = (−2 − √3i)² = 4 + 4√3i + 3i² = 1 + 4√3i

x³ = (1 + 4√3i) (−2 − √3i) = −2 − √3i − 8√3i −12i² = 10 − 9√3i

x⁴ = (1 + 4√3i)² = 1 + 8√3i + 48i² = −47 + 8√3i

Now, 2x⁴ + 5x³ + 7x² − x + 41 becomes,

= 2(−47 + 8√3i) + 5(10 − 9√3i) + 7(1 + 4√3i) − (−2 − √3i) + 41

= −94 + 16√3i + 50 − 45√3i + 7 + 28√3i + 2 + √3i + 41

= 6

Therefore, the value of 2x⁴ + 5x³ + 7x² − x + 41 is 6.

Question 17. For a positive integer n, find the value of (1−i)ⁿ (1−1/i)ⁿ.

Solution:

We have,

(1−i)ⁿ (1−1/i)ⁿ = (1−i)ⁿ

=

=

=

= 2ⁿ

Therefore, the value of (1−i)ⁿ (1−1/i)ⁿ is 2ⁿ.

Question 18. If (1+i)z = (1−i), then show that z = −i.

Solution:

We have,

=> (1+i)z = (1−i)

=> z =

=> z =

=> z =

=> z =

=> z = −i

Hence proved.

Question 19. Solve the system of equations: Re(z²) = 0, |z| = 2.

Solution:

Let z = x + iy.

Now z² = (x + iy)²

= x² + i²y² + 2xyi

= x² − y² + 2xyi

We have, Re(z²) = 0

=> x² − y² = 0 . . . . (1)

Also, it is given, |z| = 2.

=>= 2

=> x² + y² = 4 . . . . (2)

Solving (1) and (2), we get, x = ±√2 and y = ±√2.

Therefore, x + iy = ±√2 ± √2i .

Question 20. Ifis purely imaginary number (z≠−1), find the value of |z|.

Solution:

Let z = x + iy

We have,

=

=

=

=

=

As the complex number is purely imaginary, therefore,

=> Re(z) = 0

=>= 0

=> x² + y² = 1

=>= 1

=> |z| = 1

Therefore, the value of |z| is 1.

Question 21. If z₁ is a complex number other than −1 such that |z₁| = 1 and z₂ =,then show that the real parts of z₂ is zero.

Solution:

Given |z| = 1

=> |z|² = 1

=> x² + y² = 1 . . . . (1)

Let z₁ = x + iy and z₂ = a + ib.

According to the question, we have,

=> z₂ =

=> a + ib =

=> a + ib =

=> a + ib =

=> a + ib =

Using (1) we get,

=> a + ib =

=> a + ib =

On comparing the real and imaginary parts on both sides, we get a = 0.

Therefore, the real parts of z₂ is 0. Hence proved.

Question 22. If |z+1| = z + 2(1+i), find z.

Solution:

Let z = x + iy. According to the question, we have,

=> |x + iy + 1| = x + iy + 2(1 + i)

=>= (x + 2) + i(y + 2)

On comparing the real and imaginary parts, we get

=> y + 2 = 0

=> y = −2

And also,

=> x + 2 =

=> (x + 2)² = (x+1)² + y²

=> x² + 4 + 4x = x² + 2x + 1+ y²

=> 2x = y² − 3

=> 2x = 4 − 3

=> 2x = 1

=> x = 1/2

Therefore, z = x + iy = 1/2 −2i.

Question 23. Solve the equation: |z| = z + 1 + 2i.

Solution:

Let z = x + iy. According to the question, we have,

=> |z| = z + 1 + 2i

=> |x + iy| = x + iy + 1 + 2i

=>= (x + 1) + (y + 2)i

=> x² + y² = (x+1)² + (y+2)²i² + 2 (x+1) (y+2)i

=> x² + y² = x²+1 + 2x − y² − 1 + 2y + 2 (x+1) (y+2)i

=> 2y² − 2x + 4y + 4 = 2i (x+1) (y+2)

=> y² − x + 2y + 2 = i (x+1) (y+2)

On comparing both sides, we get,

=> (x+1) (y+2) = 0

=> x = −1 and y = −2

Also, y² − x + 2y + 2 = 0

Taking x = −1, we get y² − (−1) + 2y + 2 = 0

=> y² + 2y + 3 = 0, which doesn’t have a solution as the roots are imaginary.

Taking y = −2, (4 − x −4 + 2) = 0

=> x = 2

Therefore, z = x + iy = 2 − 2i.

Question 24. What is the smallest positive integer n for which (1+i)²ⁿ = (1−i)²ⁿ?

Solution:

We are given,

=> (1+i)²ⁿ = (1−i)²ⁿ

=>= 1

=>= 1

=>= 1

=>= 1

=> i²ⁿ = 1

=> i²ⁿ = i⁴

=> 2n = 4

=> n = 2

Therefore, the smallest positive integer n for which (1+i)²ⁿ = (1−i)²ⁿ is 2.

Question 25. If z₁, z₂, z₃ are complex numbers such that |z₁| = |z₂| = |z₃| == 1, then find the value of |z₁ + z₂ + z₃|.

Solution:

We are given,

|z₁| = |z₂| = |z₃| == 1

Now, |z₁ + z₂ + z₃| =

=

=

= 1

Therefore, the value of |z₁ + z₂ + z₃| is 1.

Question 26. Find the number of solutions of z² + |z|² = 0.

Solution:

Let z = x + iy. We have,

=> z² + |z|² = 0

=> (x + iy)² + |x + iy|² = 0

=> x² + i²y² + 2xyi + x² + y² = 0

=> x² − y² + 2xyi + x² + y² = 0

=> 2x² + 2xyi = 0

On comparing the real and imaginary parts on both sides, we get

=> 2x² = 0 and 2xy = 0

=> x = 0 and y ∈ R

Therefore, z = 0 + iy, where y ∈ R.