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# 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 a2 + b2 =

Solution:

Given:

a + ib =

On comparing the real and imaginary parts, we have

a =  and b =

Therefore,

a2 + b2

Hence, proved,

a2 + b2

### Question 12. Let z1 = 2 â€“ i, z2 = -2 + i. Find

(i)

(ii)

Solution:

(i) Given:

z1 = 2 – i, z2 = -2 + i

(i) z1z2 = (2 – i)(-2 + i) = -4 + 2i + 2i – i2 = -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 cosÎ¸ + ir sinÎ¸

So,

r cosÎ¸ =  and r sinÎ¸ =

On squaring and adding, we get

r2(cos2Î¸ + sin2Î¸) =

r2

r =

Now,

cosÎ¸ =  and sinÎ¸ =

= cosÎ¸ =  and sinÎ¸ =

Therefore,

Î¸ =      [As Î¸ lies in the II quadrant]

### Question 14. Find the real numbers x and y if (x â€“ iy) (3 + 5i) is the conjugate of â€“ 6 â€“ 24i.

Solution:

Let us assume z = (x â€“ iy) (3 + 5i)

z = 3x + xi – 3yi – 5yi2 = 3x + 5xi – 3yi + 5y = (3x + 5y) + i(5x – 3y)

Therefore,

=(3x + 5y) – i(5x – 3y)

Also given,  = -6 – 24i

And,

(3x + 5y) â€“ i(5x â€“ 3y) = -6 -24i

After equating real and imaginary parts, we get

3x + 5y = -6 â€¦â€¦ (i)

5x â€“ 3y = 24 â€¦â€¦ (ii)

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

(9x + 15y) + (25x â€“ 15y) = -18 + 120

34x = 102

x = 102/34 = 3

Putting the value of x in equation (i), we get

3(3) + 5y = -6

5y = -6 â€“ 9 = -15

y = -3

Therefore, the values of and y are 3 and â€“3 respectively.

Solution:

### Question 16. If (x + iy)3 = u + iv, then show that  = 4(x2 – y2)

Solution:

(x + iy)3 = u + iv

x3 + (iy) + 3 Ă— x Ă— iy(x + iy) = u + iv

x3 + i3y3 + 3x2yi + 3xy2 = u + iv

x3 – iy3 + 3x2yi – 3xy2 = u + iv

(x3 – 3xy2) + i(3x2y – y3) = u + iv

On equating real and imaginary parts, we get

u = x3 – 3xy2, v = 3x2y – y3

= x2 – 3y2 + 3x2 – y2

= 4x2 – 4y2

= 4(x2 – y2)

Hence proved

### Question 17. If Î± and Î˛ are different complex numbers with |Î˛| = 1, then find

Solution:

Assume Î± = a + ib and Î˛ = x + iy

Given: |Î˛| = 1

So,

= x2 + y2 = 1            ….(1)

= 1

### Question 18. Find the number of non-zero integral solutions of the equation |1 â€“ i|x = 2x

Solution:

|1 – i|x = 2t

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 (a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = A2 + B2.

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)| Ă— |(c + id)| Ă— |(e + if)| Ă— |(g + ih)| = |A + iB|

On squaring both sides, we get

(a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = A2 + B2

Hence, proved.

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

Solution:

im = 1

Hence, m = 4k, where k is some integer.

Hence, the least positive integer is 1.

Thus, the least positive integral value of m is 4 (= 4 Ă— 1).

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