# Class 11 NCERT Solutions – Chapter 5 Complex Numbers And Quadratic Equations – Miscellaneous Exercise on Chapter 5 | Set 2

**Question 11. If a + ib = ****, prove that a**^{2} + b^{2} =

^{2}+ b

^{2}=

**Solution:**

Given:

a + ib =

On comparing the real and imaginary parts, we have

a = and b =

Therefore,

a

^{2}+ b^{2}=Hence, proved,

a

^{2}+ b^{2}=

**Question 12. Let z**_{1} = 2 – **i****, z**_{2} = -2 + **i****. Find**

_{1}= 2 –

**i**

_{2}= -2 +

**i**

**(i) **

**(ii) **

**Solution:**

(i)Given:z

_{1}= 2 – i, z_{2}= -2 + i(i) z

_{1}z_{2}= (2 – i)(-2 + i) = -4 + 2i + 2i – i^{2}= -4 + 4i – (-1) = -3 + 4i= 2 + i

Therefore,

On multiplying numerator and denominator by (2 – i), we get

On comparing the real parts, we have

(ii)On comparing the imaginary part, we get

= 0

**Question 13. Find the modulus and argument of the complex number **

**Solution:**

Let, z = , then

z =

Let z = r cosθ + ir sinθ

So,

r cosθ = and r sinθ =

On squaring and adding, we get

r2(cos2θ + sin2θ) =

r

^{2}=r =

Now,

cosθ = and sinθ =

= cosθ = and sinθ =

Therefore,

θ = [As θ lies in the II quadrant]

**Question 14. Find the real numbers ****x**** and ****y**** if (****x**** – ****iy****) (3 + 5****i****) is the conjugate of – 6 – 24****i****.**

**x**

**y**

**x**

**iy**

**i**

**i**

**Solution:**

Let us assume z = (

x–iy) (3 + 5i)z = 3x + xi – 3yi – 5yi

^{2}= 3x + 5xi – 3yi + 5y = (3x + 5y) + i(5x – 3y)Therefore,

=(3x + 5y) – i(5x – 3y)

Also given, = -6 – 24i

And,

(3x + 5y) –

i(5x – 3y) = -6 -24iAfter equating real and imaginary parts, we get

3x + 5y = -6 …… (i)

5x – 3y = 24 …… (ii)

After doing (i) x 3 + (ii) x 5, we have

(9x + 15y) + (25x – 15y) = -18 + 120

34x = 102

x = 102/34 = 3

Putting the value of

xin equation (i), we get3(3) + 5y = -6

5y = -6 – 9 = -15

y = -3

Therefore, the values of

xandyare 3 and –3 respectively.

**Question 15. Find the modulus of **

**Solution:**

**Question 16. If (****x**** + ****iy****)**^{3} = **u**** + ****iv****, then show that ** = 4(x^{2} – y^{2})

**x**

**iy**

^{3}=

**u**

**iv**

**Solution:**

(x + iy)

^{3}= u + ivx

^{3}+ (iy)^{3 }+ 3 × x × iy(x + iy) = u + ivx

^{3}+ i^{3}y^{3}+ 3x^{2}yi + 3xy^{2}= u + ivx

^{3}– iy^{3}+ 3x^{2}yi – 3xy^{2}= u + iv(x

^{3}– 3xy^{2}) + i(3x^{2}y – y^{3}) = u + ivOn equating real and imaginary parts, we get

u = x

^{3}– 3xy^{2}, v = 3x^{2}y – y^{3}= x

^{2}– 3y^{2}+ 3x^{2}– y^{2}= 4x

^{2}– 4y^{2}= 4(x

^{2}– y^{2})Hence proved

**Question 17. If **α and β** are different complex numbers with** |β| = 1,** then find **

**Solution:**

Assume α = a + ib and β = x + iy

Given: |β| = 1

So,

= x2 + y2 = 1 ….(1)

= 1

**Question 18. Find the number of non-zero integral solutions of the equation |1 – i|**^{x} = 2^{x}

^{x}= 2

^{x}

**Solution:**

|1 – i|

^{x}= 2^{t}x = 2x

2x – x = 0

Thus, ‘0’ is the only integral solution of the given equation.

Therefore, the number of non-zero integral solutions of the given equation is 0.

**Question 19. If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that (a**^{2} + b^{2}) (c^{2} + d^{2}) (e^{2} + f^{2}) (g^{2} + h^{2}) = A^{2} + B^{2}.

^{2}+ b

^{2}) (c

^{2}+ d

^{2}) (e

^{2}+ f

^{2}) (g

^{2}+ h

^{2}) = A

^{2}+ B

^{2}.

**Solution:**

Given:

(a + ib)(c + id)(e + if)(g + ih) = A + iB

Therefore,

|(a + ib)(c + id)(e + if)(g + ih)| = |A + iB|

= |(a + ib)| × |(c + id)| × |(e + if)| × |(g + ih)| = |A + iB|

On squaring both sides, we get

(

a^{2}+b^{2}) (c^{2}+d^{2}) (e^{2}+f^{2}) (g^{2}+h^{2}) = A^{2}+ B^{2}Hence, proved.

**Question 20. If, then find the least positive integral value of **m.** **

**Solution:**

i

^{m}= 1Hence, m = 4k, where k is some integer.

Hence, the least positive integer is 1.

Thus, the least positive integral value of

mis 4 (= 4 × 1).