# 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) 2x**^{3} + 2x^{2} − 7x + 72, when x = (3−5i)/2

^{3}+ 2x

^{2}− 7x + 72, when x = (3−5i)/2

**Solution:**

We have, x = (3−5i)/2

=> 2x = 3 − 5i

=> 2x − 3 = −5i

=> (2x − 3)

^{2}= 25i^{2}=> 4x

^{2}+ 9 − 12x = −25=> 4x

^{2}− 12x + 34 = 0=> 2x

^{2}− 6x + 17 = 0Now, 2x

^{3}+ 2x^{2}− 7x + 72 = x (2x^{2}− 6x + 17) + 6x^{2}− 17x + 2x^{2}− 7x + 72= x (0) + 8x

^{2}− 24x + 72= 4 (2x

^{2}− 6x + 17) + 4= 4 (0) + 4

= 4

Therefore, the value of 2x^{3}+ 2x^{2}− 7x + 72 is 4.

**(ii) x**^{4} − 4x^{3} + 4x^{2} +8x +44, when x = 3 + 2i

^{4}− 4x

^{3}+ 4x

^{2}+8x +44, when x = 3 + 2i

**Solution:**

We have, x = 3 + 2i

=> x − 3 = 2i

=> (x − 3)

^{2}= (2i)^{2}=> x

^{2}+ 9 − 6x = 4i^{2}=> x

^{2}− 6x + 9 + 4 = 0=> x

^{2}− 6x + 13 = 0Now, x

^{4}− 4x^{3}+ 4x^{2}+ 8x + 44 = x^{2}(x^{2}− 6x + 13) + 6x^{3}− 13x^{2}− 4x^{3}+ 4x^{2}+ 8x + 44= 2x

^{3}− 9x^{2}+ 8x + 44= 2x (x

^{2}− 6x + 13) + 12x^{2}− 26x − 9x^{2}+ 8x + 44= 3x

^{2}− 18x + 44= 3 (x

^{2}− 6x + 13) + 5= 5

Therefore, the value of x^{4}− 4x^{3}+ 4x^{2}+ 8x + 44 is 5.

**(iii) x**^{4} + 4x^{3} + 6x^{2} + 4x + 9, when x = −1 + i√2

^{4}+ 4x

^{3}+ 6x

^{2}+ 4x + 9, when x = −1 + i√2

**Solution:**

We have, x = −1 + i√2

=> x + 1 = i√2

=> (x + 1)

^{2}= 2i^{2}=> x

^{2}+ 1 + 2x = −2=> x

^{2}+ 2x + 3 = 0Now, x

^{4}+ 4x^{3}+ 6x^{2}+ 4x + 9 = x^{2}(x^{2}+ 2x + 3) − 2x^{3}− 3x^{2}+ 4x^{3}+ 6x^{2}+ 4x + 9= 2x

^{3}+ 3x^{2}+ 4x + 9= 2x (x

^{2}+ 2x + 3) − 4x^{2}− 6x + 3x^{2}+ 4x + 9= − x

^{2}− 2x + 9= − (x

^{2}+ 2x + 3) + 3 + 9= 3 + 9

= 12

Therefore, the value of x^{4}+ 4x^{3}+ 6x^{2 }+ 4x + 9 is 12.

**(iv) x**^{6} + x^{4} + x^{2} + 1, when x = (1+i)/√2

^{6}+ x

^{4}+ x

^{2}+ 1, when x = (1+i)/√2

**Solution:**

We have, x = (1+i)/√2

=> √2x = 1 + i

=> 2x

^{2}= 1 + i^{2}+ 2i=> 2x

^{2}= 2i=> 4x

^{4}= 4i^{2}=> x4 = −1

=> x

^{4}+ 1 = 0Now, x

^{6}+ x^{4 }+ x^{2}+ 1 = (x^{6}+ x^{2}) + (x^{4}+1)= x

^{6}+ x^{2}= x

^{2}(x^{4}+ 1)= 0

Therefore, the value of x^{6 }+ x^{4}+ x^{2 }+ 1 is 0.

**(v) 2x**^{4} + 5x^{3} + 7x^{2} − x + 41, when x = −2 − √3i

^{4}+ 5x

^{3}+ 7x

^{2}− x + 41, when x = −2 − √3i

**Solution:**

We have, x = −2 − √3i

x

^{2}= (−2 − √3i)^{2}= 4 + 4√3i + 3i^{2}= 1 + 4√3ix

^{3}= (1 + 4√3i) (−2 − √3i) = −2 − √3i − 8√3i −12i^{2}= 10 − 9√3ix

^{4}= (1 + 4√3i)^{2}= 1 + 8√3i + 48i^{2}= −47 + 8√3iNow, 2x

^{4}+ 5x^{3}+ 7x^{2}− 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 2x^{4}+ 5x^{3}+ 7x^{2}− x + 41 is 6.

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

^{n}(1−1/i)

^{n}.

**Solution:**

We have,

(1−i)

^{n}(1−1/i)^{n}= (1−i)^{n}=

=

=

= 2

^{n}

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

**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(z**^{2}) = 0, |z| = 2.

^{2}) = 0, |z| = 2.

**Solution:**

Let z = x + iy.

Now z

^{2}= (x + iy)^{2}= x

^{2}+ i^{2}y^{2}+ 2xyi= x

^{2}− y^{2}+ 2xyiWe have, Re(z

^{2}) = 0=> x

^{2}− y^{2}= 0 . . . . (1)Also, it is given, |z| = 2.

=>= 2

=> x

^{2}+ y^{2}= 4 . . . . (2)Solving (1) and (2), we get, x = ±√2 and y = ±√2.

Therefore, x + iy = ±√2 ± √2i .

**Question 20. If****is 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

=> x

^{2}+ y^{2}= 1=>= 1

=> |z| = 1

Therefore, the value of |z| is 1.

**Question 21. If z**_{1} is a complex number other than −1 such that |z_{1}| = 1 and z_{2} =**,then show that the real parts of z**_{2} is zero.

_{1}is a complex number other than −1 such that |z

_{1}| = 1 and z

_{2}=

_{2}is zero.

**Solution:**

Given |z| = 1

=> |z|

^{2}= 1=> x

^{2}+ y^{2}= 1 . . . . (1)Let z

_{1}= x + iy and z_{2}= a + ib.According to the question, we have,

=> z

_{2}==> 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 z_{2}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 }+ y^{2}=> x

^{2}+ 4 + 4x = x^{2}+ 2x + 1+ y^{2}=> 2x = y

^{2}− 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

=> x

^{2}+ y^{2}= (x+1)^{2}+ (y+2)^{2}i^{2}+ 2 (x+1) (y+2)i=> x

^{2}+ y^{2}= x^{2}+1 + 2x − y^{2}− 1 + 2y + 2 (x+1) (y+2)i=> 2y

^{2}− 2x + 4y + 4 = 2i (x+1) (y+2)=> y

^{2}− 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, y

^{2}− x + 2y + 2 = 0Taking x = −1, we get y

^{2}− (−1) + 2y + 2 = 0=> y

^{2}+ 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}?

^{2n}= (1−i)

^{2n}?

**Solution:**

We are given,

=> (1+i)

^{2n}= (1−i)^{2n}=>= 1

=>= 1

=>= 1

=>= 1

=> i

^{2n}= 1=> i

^{2n}= i^{4}=> 2n = 4

=> n = 2

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

**Question 25. If z**_{1}, z_{2}, z_{3} are complex numbers such that |z_{1}| = |z_{2}| = |z_{3}| =**= 1, then find the value of |z**_{1} + z_{2} + z_{3}|.

_{1}, z

_{2}, z

_{3}are complex numbers such that |z

_{1}| = |z

_{2}| = |z

_{3}| =

_{1}+ z

_{2}+ z

_{3}|.

**Solution:**

We are given,

|z

_{1}| = |z_{2}| = |z_{3}| == 1Now, |z

_{1}+ z_{2}+ z_{3}| ==

=

= 1

Therefore, the value of |z_{1}+ z_{2}+ z_{3}| is 1.

**Question 26. Find the number of solutions of z**^{2} + |z|^{2} = 0.

^{2}+ |z|

^{2}= 0.

**Solution:**

Let z = x + iy. We have,

=> z

^{2}+ |z|^{2}= 0=> (x + iy)

^{2}+ |x + iy|^{2}= 0=> x

^{2}+ i^{2}y^{2}+ 2xyi + x^{2}+ y^{2}= 0=> x

^{2}− y^{2}+ 2xyi + x^{2}+ y^{2}= 0=> 2x

^{2}+ 2xyi = 0On comparing the real and imaginary parts on both sides, we get

=> 2x

^{2}= 0 and 2xy = 0=> x = 0 and y ∈ R

Therefore, z = 0 + iy, where y ∈ R.

Attention reader! Don’t stop learning now. Join the **First-Step-to-DSA Course for Class 9 to 12 students ****, **specifically designed to introduce data structures and algorithms to the class 9 to 12 students