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# Class 11 RD Sharma Solutions – Chapter 13 Complex Numbers – Exercise 13.2 | Set 1

• Last Updated : 30 Apr, 2021

### (i) (1 + i) (1 + 2i)

Solution:

We have, z = (1 + i) (1 + 2i)

= 1 (1 + 2i) + i (1 + 2i)

= 1 + 2i + i + 2i2

= 1 + 3i + 2(−1)

= 1 + 3i − 2

= −1 + 3i

Therefore, the standard form is −1 + 3i where a = −1 and b = 3.

### (ii)

Solution:

We have, z =

=

=

=

=

=

Therefore, the standard form iswhere a = −4/5 and b = −7/5.

### (iii)

Solution:

We have, z =

=

=

=

=

=

Therefore, the standard form iswhere a = 3/25 and b = −4/25.

### (iv)

Solution:

We have, z =

=

=

=

= −i

Therefore, the standard form is −i where a = 0 and b = −1.

### (v)

Solution:

We have, z =

=

=

=

=

=

=

Therefore, the standard form iswhere a = 37/13 and b = 16/13.

### (vi)

Solution:

We have, z =

=

=

=

=

=

=

= –√3 + i

Therefore, the standard form is –√3 + i where a = –√3 and b = 1.

### (vii)

Solution:

We have, z =

=

=

=

Therefore, the standard form iswhere a = 23/41 and b = 2/41.

### (viii)

Solution:

We have, z =

=

=

=

=

=

= –3 – i

Therefore, the standard form is –3 – i where a = –3 and b = – 1.

### (ix) (1 + 2i)-3

Solution:

We have z = (1 + 2i)-3

=

=

=

=

=

=

Therefore, the standard form iswhere a = –3/13 and b = 2/13.

### (x)

Solution:

We have, z =

=

=

=

=

=

=

Therefore, the standard form iswhere a = –1/4 and b = –3/4.

### (xi)

Solution:

We have, z =

=

=

=

=

=

=

=

=

Therefore, the standard form iswhere a = 478/884 and b = 928/884.

### (xii)

Solution:

We have, z =

=

=

=

= 1+ 2√2i

Therefore, the standard form is 1+ 2√2i where a = 1 and b = 2√2.

### (i) (x + iy) (2 – 3i) = 4 + i

Solution:

We have,

=> (x + iy) (2 – 3i) = 4 + i

=> 2x – 3xi + 2yi – 3yi2 = 4 + i

=> 2x + (–3x+2y)i + 3y = 4 + i

=> (2x+3y) + i(–3x+2y) = 4 + i

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

2x + 3y = 4 . . . . (1)

And –3x + 2y = 1 . . . . (2)

On multiplying (1) by 3 and (2) by 2 and adding, we get

=> 6x – 6x – 9y + 4y = 12 + 2

=> 13y = 14

=> y = 14/13

On putting y = 14/13 in (1), we get

=> 2x + 3(14/13) = 4

=> 2x = 4 – (42/13)

=> 2x = 10/13

=> x = 5/13

Therefore, the real values of x and y are 5/13 and 14/13 respectively.

### (ii) (3x – 2iy) (2 + i)2 = 10(1 + i)

Solution:

We have,

=> (3x – 2iy) (2 + i)2 = 10(1 + i)

=> (3x – 2yi) (4 + i2 + 4i) = 10 + 10i

=> (3x – 2yi) (3 + 4i) = 10+10i

=> 3x – 2yi =

=> 3x – 2yi =

=> 3x – 2yi =

=> 3x – 2yi =

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

=> 3x = 70/25 and –2y = –10/25

=> x = 70/75 and y = 1/5

Therefore, the real values of x and y are 70/75 and 1/5 respectively.

### (iii)

Solution:

We have,

=>

=>

=>

=> (4+2i) x − 3i − 3 + (9−7i)y = 10i

=> (4x+9y−3) + i(2x−7y−3) = 10i

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

4x + 9y − 3 = 0 . . . . (1)

And 2x − 7y − 3 = 10 . . . . (2)

On multiplying (1) by 7 and (2) by 9 and adding, we get,

=> 28x + 18x + 63y – 63y = 117 + 21

=> 46x = 117 + 21

=> 46x = 138

=> x = 3

On putting x = 3 in (1), we get

=> 4x + 9y − 3 = 0

=> 9y = −9

=> y = −1

Therefore, the real values of x and y are 3 and −1 respectively.

### (iv) (1 + i) (x + iy) = 2 – 5i

Solution:

We have,

=> (1 + i) (x + iy) = 2 – 5i

=> x + iy =

=> x + iy =

=> x + iy =

=> x + iy =

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

=> x = −3/2 and y = −7/2

Therefore, the real values of x and y are −3/2 and −7/2 respectively.

### (i) 4 – 5i

Solution:

We know the conjugate of a complex number (a + ib) is (a – ib).

Therefore, the conjugate of (4 – 5i) is (4 + 5i).

### (ii)

Solution:

We have, z =

=

=

=

We know the conjugate of a complex number (a + ib) is (a – ib).

Therefore, the conjugate ofis.

### (iii)

Solution:

We have, z =

=

=

=

We know the conjugate of a complex number (a + ib) is (a – ib).

Therefore, the conjugate ofis.

### (iv)

Solution:

We have, z =

=

=

=

=

=

= 2 – 4i

We know the conjugate of a complex number (a + ib) is (a – ib).

Therefore, the conjugate ofis 2 + 4i.

### (v)

Solution:

We have, z =

=

=

=

=

=

=

We know the conjugate of a complex number (a + ib) is (a – ib).

The conjugate ofis.

### (vi)

Solution:

We have, z =

=

=

=

=

=

We know the conjugate of a complex number (a + ib) is (a – ib).

Therefore, the conjugate ofis.

### (i) 1 – i

Solution:

We have z = 1 – i

We know the multiplicative inverse of a complex number z is 1/z. So, we get,

=

=

=

=

Therefore, the multiplicative inverse of (1 – i) is.

### (ii) (1 + i √3)2

Solution:

We have, z = (1 + i √3)2

= 1 + 3i2 + 2 i√3

= 1 + 3(−1) + 2 i√3

= 1 – 3 + 2 i√3

= −2 + 2 i√3

We know the multiplicative inverse of a complex number z is 1/z. So, we get,

=

=

=

=

=

Therefore, the multiplicative inverse of (1 + i √3)2 is.

### (iii) 4 – 3i

Solution:

We have z = 4 – 3i

We know the multiplicative inverse of a complex number z is 1/z. So, we get,

=

=

=

=

Therefore, the multiplicative inverse of 4 – 3i is.

### (iv) √5 + 3i

Solution:

We have z = √5 + 3i

We know the multiplicative inverse of a complex number z is 1/z. So, we get,

=

=

=

=

Therefore, the multiplicative inverse of √5 + 3i is.

### Question 5. If z1 = 2 − i, z2 = 1 + i, find.

Solution:

Given z1 = 2 − i, z2 = 1 + i, we get,

=

=

=

=

= 2√2

Therefore, the value ofis 2√2.

### (i) Re

Solution:

Given z1 = (2 – i), z2 = (–2 + i), we get,

=

=

=

=

=

=

=

Therefore, Re=.

### (ii) Im

Now,=

=

=

=

=

Therefore, Im= 0.

### Question 7. Find the modulus of.

Solution:

We have, z =

=

=

=

= 2i

So, modulus of z == 2.

Therefore, the modulus ofis 2.

### Question 8. If x + iy =, prove that x2 + y2 = 1.

Solution:

We have,

=> x + iy =

On applying modulus on both sides we get,

=> |x + iy| =

=> |x + iy| =

=>

=>= 1

=> x2 + y2 = 1

Hence proved.

### Question 9. Find the least positive integral value of n for whichis real.

Solution:

We have, z =

=

=

=

= in

For n = 2, we have in = i2 = −1, which is real

Therefore, the least positive integral value of n for whichis real is 2.

### Question 10. Find the real values of θ for which the complex numberis purely real.

Solution:

We have, z =

=

=

=

=

For a complex number to be purely real, the imaginary part should be equal to zero.

So, we get,= 0

=> cos θ = 0

=> cos θ = cos π/2

=> 2nπ ± π/2, for n ∈ Z

Therefore, the values of θ for the complex number to be purely real are 2nπ ± π/2, for n ∈ Z.

### Question 11. Find the smallest positive integer value of n for whichis a real number.

Solution:

We have, z =

=

=

=

=

=

=

= in × (−2i)

= −2in+1

For n = 1, we have z = −2i1+1

= −2i2

= 2, which is real

Therefore, the smallest positive integer value of n for which is a real numberis 1.

### Question 12. If,find (x, y).

Solution:

We have,

=>

=>

=>

=> i3 – (–i3) = x + iy

=> 2i3 = x + iy

=> x + iy = −2i

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

=> (x, y) = (0, −2)

### Question 13. If, find x + y.

Solution:

We have,

=>

=>

=>

=>

=>

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

=> x = −2/5 and y = 4/5

So, x + y = −2/5 + 4/5

= (−2+4)/5

= 2/5

Therefore, the value of (x + y) is 2/5.

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