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

Complex numbers are an essential concept in mathematics providing the solutions to equations that cannot be solved within the real number system. In Class 11, RD Sharma’s Chapter 13 on Complex Numbers focuses on understanding and performing operations with complex numbers including addition, subtraction, multiplication, division, and their geometric representation. Exercise 13.2 specifically deals with practicing these operations allowing students to strengthen their grasp of the complex numbers.

Complex Numbers

A complex number is expressed in the form z=a+bi, where:

• a is the real part.
• b is the imaginary part.
• i represents the imaginary unit defined as i²=−1.

The set of complex numbers includes both real numbers and purely imaginary numbers. These numbers extend the number system beyond real numbers helping the solve quadratic equations and represent the geometric transformations.

Question 1. Express the following complex numbers in the standard form a + ib:

(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) \frac{3+2i}{−2+i}

Solution:

We have, z =\frac{3+2i}{−2+i}

=\frac{(3+2i)(-2-i)}{(-2+i)(-2-i)}

=\frac{3(-2-i) + 2i (-2-i)}{(-2)^2–(i)^2}

=\frac{-6-3i–4i-2i^2}{4-i^2}

=\frac{-6-7i+2}{4+1}

=\frac{-4 -7i}{5}

Therefore, the standard form is\frac{-4 -7i}{5}  where a = −4/5 and b = −7/5.

(iii)\frac{1}{(2 + i)^2}

Solution:

We have, z =\frac{1}{(2 + i)^2}

=\frac{1}{4+i^2+4i}

=\frac{1}{3+4i}

=\frac{3-4i}{(3+4i)(3-4i)}

=\frac{3-4i}{9+16}

=\frac{3-4i}{25}

Therefore, the standard form is\frac{3-4i}{25}  where a = 3/25 and b = −4/25.

(iv)\frac{1–i}{1+i}

Solution:

We have, z =\frac{1–i}{1+i}

=\frac{(1–i)(1-i)}{(1+i)(1-i)}

=\frac{1+i^2-2i}{1-i^2}

=\frac{-2i}{2}

= −i

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

(v)\frac{(2+i)^3}{2+3i}

Solution:

We have, z =\frac{(2+i)^3}{2+3i}

=\frac{8+i^3+12i+6i^2}{2+3i}

=\frac{8-i+12i-6}{2+3i}

=\frac{2+11i}{2+3i}

=\frac{(2+11i)(2-3i)}{(2+3i)(2-3i)}

=\frac{4–6i+22i–33i^2}{4+9}

=\frac{37+16i}{13}

Therefore, the standard form is\frac{37+16i}{13}  where a = 37/13 and b = 16/13.

(vi)\frac{(1+i)(1+\sqrt{3}i)}{1–i}

Solution:

We have, z =\frac{(1+i)(1+\sqrt{3}i)}{1–i}

=\frac{1+\sqrt{3}i+i+\sqrt{3}i^2}{1–i}

=\frac{(1-\sqrt{3})+(1+\sqrt{3})i}{(1-i)}

=\frac{[(1-\sqrt{3})+(1+\sqrt{3})i](1+i)}{(1-i)(1+i)}

=\frac{[1-\sqrt{3}+(1-\sqrt{3})i+(1+\sqrt{3})i+(1+\sqrt{3})i^2]}{(1-(-1))}

=\frac{[(1-\sqrt{3})+(1-\sqrt{3}+1+\sqrt{3})i+(1+\sqrt{3})(-1)]}{2}

=\frac{-2\sqrt{3}+2i}{2}

= –√3 + i

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

(vii)\frac{2+3i}{4+5i}

Solution:

We have, z =\frac{2+3i}{4+5i}

=\frac{(2+3i)(4-5i)}{(4+5i)(4-5i)}

=\frac{8–10i+12i–15i^2}{16+25}

=\frac{23+2i}{41}

Therefore, the standard form is\frac{23+2i}{41}  where a = 23/41 and b = 2/41.

(viii)\frac{(1–i)^3}{1–i^3}

Solution:

We have, z =\frac{(1–i)^3}{1–i^3}

=\frac{1–3i+3i^2–i^3}{1–(-1)i}

=\frac{-2-4i}{1+i}

=\frac{(-2-4i)(1-i)}{(1+i)(1-i)}

=\frac{-2+2i-4i+4i^2}{1–(-1)}

=\frac{-6-2i}{2}

= –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

=\frac{1}{1+6i+4i^2+8i^3}

=\frac{1}{1+6i-4-8i}

=\frac{-1}{3+2i}

=\frac{-1(3-2i)}{(3+2i)(3-2i)}

=\frac{-3+2i}{9+4}

=\frac{-3+2i}{13}

Therefore, the standard form is\frac{-3+2i}{13}  where a = –3/13 and b = 2/13.

(x)\frac{3–4i}{(4–2i)(1+i)}

Solution:

We have, z =\frac{3–4i}{(4–2i)(1+i)}

=\frac{3-4i}{4+4i-2i-2i^2}

=\frac{3-4i}{6+2i}

=\frac{(3-4i)(6-2i)}{(6+2i)(6-2i)}

=\frac{18–6i–24i+8i^2}{36–4i^2}

=\frac{10-30i}{40}

=\frac{1–3i}{4}

Therefore, the standard form is\frac{1–3i}{4}  where a = –1/4 and b = –3/4.

(xi)\left(\frac{1}{1-4i}-\frac{2}{1+i}\right)\frac{3-4i}{5+i}

Solution:

We have, z =\left(\frac{1}{1-4i}-\frac{2}{1+i}\right)\frac{3-4i}{5+i}

=\left(\frac{1+i-2+8i}{(1-4i)(1+i)}\right)\frac{3-4i}{5+i}

=\left(\frac{-1+9i}{1+i-4i-4i^2}\right)\frac{3-4i}{5+i}

=\frac{(-1+9i)(3-4i)}{(5-3i)(5+i)}

=\frac{-3+4i+27i-9i^2}{25+5i-15i-3i^2}

=\frac{6+31i}{28-10i}

=\frac{(6+31i)(28+10i)}{(28-10i)(28+10i)}

=\frac{168+60i+868i+310i^2}{784+100}

=\frac{478+928i}{884}

Therefore, the standard form is\frac{478+928i}{884}  where a = 478/884 and b = 928/884.

(xii)\frac{5+\sqrt{2}i}{1-\sqrt{2}i}

Solution:

We have, z =\frac{5+\sqrt{2}i}{1-\sqrt{2}i}

=\frac{(5+\sqrt{2}i)(1+\sqrt{2}i)}{(1-\sqrt{2}i)(1+\sqrt{2}i)}

=\frac{5+5\sqrt{2}i+\sqrt{2}i+2i^2}{1–2i^2}

=\frac{3+6\sqrt{2}i}{3}

= 1+ 2√2i

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

Question 2. Find the real values of x and y, if

(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 =\frac{10+10i}{3+4i}

=> 3x – 2yi =\frac{(10+10i)(3-4i)}{(3+4i)(3-4i)}

=> 3x – 2yi =\frac{30-40i+30i-40i^2}{9+16}

=> 3x – 2yi =\frac{70-10i}{25}

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)\frac{(1+i)x-2i}{3+i}+\frac{(2-3i)y+i}{3-i}=i

Solution:

We have,

=>\frac{(1+i)x-2i}{3+i}+\frac{(2-3i)y+i}{3-i}=i

=>\frac{((1+i)(3-i)x)-2i(3-i)+((2-3i)(3+i)y)+i(3+i)}{(3+i)(3-i)}=i

=>\frac{(3-i+3i-i^2)x-6i+2i^2+(6+2i-9i-3i^2)y+3i+i^2}{9+1}=i

=> (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 =\frac{2–5i}{1+i}

=> x + iy =\frac{(2–5i)(1-i)}{(1+i)(1-i)}

=> x + iy =\frac{2-i-5i+5i^2}{1+1}

=> x + iy =\frac{-3-7i}{2}

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.

Question 3. Find the conjugates of the following complex numbers:

(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)\frac{1}{3+5i}

Solution:

We have, z =\frac{1}{3+5i}

=\frac{3-5i}{(3+5i)(3-5i)}

=\frac{3-5i}{9+25}

=\frac{3-5i}{34}

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

Therefore, the conjugate of\frac{1}{3+5i}  is\frac{3+5i}{34}  .

(iii)\frac{1}{1+i}

Solution:

We have, z =\frac{1}{1+i}

=\frac{1-i}{(1+i)(1-i)}

=\frac{1-i}{1+1}

=\frac{1-i}{2}

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

Therefore, the conjugate of\frac{1}{1+i}  is\frac{1+i}{2}  .

(iv)\frac{(3–i)^2}{2+i}

Solution:

We have, z =\frac{(3–i)^2}{2+i}

=\frac{9+i^2-6i}{2+i}

=\frac{8-6i}{2+i}

=\frac{(8-6i)(2-i)}{(2+i)(2-i)}

=\frac{16-8i-12i+6i^2}{4-i^2}

=\frac{10-20i}{5}

= 2 – 4i

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

Therefore, the conjugate of\frac{(3–i)^2}{2+i}  is 2 + 4i.

(v)\frac{(1+i)(2+i)}{3+i}

Solution:

We have, z =\frac{(1+i)(2+i)}{3+i}

=\frac{2+i+2i+i^2}{3+i}

=\frac{1+3i}{3+i}

=\frac{(1+3i)(3-i)}{(3+i)(3-i)}

=\frac{3-i+9i-3i^2}{9+1}

=\frac{6+8i}{10}

=\frac{3+4i}{5}

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

The conjugate of\frac{(1+i)(2+i)}{3+i}  is\frac{3-4i}{5}  .

(vi)\frac{(3–2i)(2+3i)}{(1+2i)(2–i)}

Solution:

We have, z =\frac{(3–2i)(2+3i)}{(1+2i)(2–i)}

=\frac{6+9i-4i-6i^2}{2-i+4i-2i^2}

=\frac{12+5i}{4+3i}

=\frac{(12+5i)(4-3i)}{(4+3i)(4-3i)}

=\frac{48-36i+20i-15i^2}{16+9}

=\frac{63-16i}{25}

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

Therefore, the conjugate of\frac{(3–2i)(2+3i)}{(1+2i)(2–i)}  is\frac{63+16i}{25}  .

Question 4. Find the multiplicative inverse of the following complex numbers:

(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,

=\frac{1}{1-i}

=\frac{1+i}{(1-i)(1+i)}

=\frac{1+i}{1-i^2}

=\frac{1+i}{2}

Therefore, the multiplicative inverse of (1 – i) is\frac{1+i}{2}  .

(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,

=\frac{1}{−2 + 2 i\sqrt{3}}

=\frac{−2-2i\sqrt{3}}{(−2+2i\sqrt{3})(−2-2i\sqrt{3})}

=\frac{−2-2i\sqrt{3}}{4-12i^2}

=\frac{−2-2i\sqrt{3}}{16}

=\frac{−1-i\sqrt{3}}{8}

Therefore, the multiplicative inverse of (1 + i √3)² is\frac{−1-i\sqrt{3}}{8}  .

(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,

=\frac{1}{4–3i}

=\frac{4+3i}{(4–3i)(4+3i)}

=\frac{4+3i}{16+9}

=\frac{4+3i}{25}

Therefore, the multiplicative inverse of 4 – 3i is\frac{4+3i}{25}  .

(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,

=\frac{1}{\sqrt{5}+3i}

=\frac{\sqrt{5}-3i}{(\sqrt{5}+3i)(\sqrt{5}-3i)}

=\frac{\sqrt{5}-3i}{5+9}

=\frac{\sqrt{5}-3i}{14}

Therefore, the multiplicative inverse of √5 + 3i is\frac{\sqrt{5}-3i}{14}  .

Question 5. If z₁ = 2 − i, z₂ = 1 + i, find|\frac{z_1+z_2+1}{z_1-z_2+i}|  .

Solution:

Given z₁ = 2 − i, z₂ = 1 + i, we get,

|\frac{z_1+z_2+1}{z_1-z_2+i}|  =|\frac{2-i+1+i+1}{2-i-1-i+i}|

=\frac{|4|}{|1-i|}

=\frac{\sqrt{4^2+0^2}}{\sqrt{1^2+(-1)^2}}

=\frac{4}{\sqrt{2}}

= 2√2

Therefore, the value of|\frac{z_1+z_2+1}{z_1-z_2+i}|  is 2√2.

Question 6. If z₁ = (2 – i), z₂ = (–2 + i), find

(i) Re(\frac{z_1z_2}{\bar{z_1}})

Solution:

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

\frac{z_1z_2}{\bar{z_1}}  =\frac{z^2_1z_2}{\bar{z_1}z_1}

=\frac{z^2_1z_2}{|z_1|^2}

=\frac{(2-i)^2(-2+i)}{2^2+(-1)^2}

=\frac{(4+i^2-4i)(-2+i)}{4+1}

=\frac{(3-4i)(-2+i)}{5}

=\frac{-6+3i+8i+4}{5}

=\frac{-2+11i}{5}

Therefore, Re(\frac{z_1z_2}{\bar{z_1}})  =\frac{-2}{5}  .

(ii) Im(\frac{1}{z_1\bar{z_1}})

Now,\frac{1}{z_1\bar{z_1}}  =\frac{1}{|z_1|^2}

=\frac{1}{|2-i|^2}

=\frac{1}{2^2+(-1)^2}

=\frac{1}{4+1}

=\frac{1}{5}+0i

Therefore, Im(\frac{1}{z_1\bar{z_1}})  = 0.

Question 7. Find the modulus of\frac{1+i}{1–i}–\frac{1–i}{1+i}  .

Solution:

We have, z =\frac{1+i}{1–i}–\frac{1–i}{1+i}

=\frac{(1+i)^2–(1-i)^2}{1–i^2}

=\frac{1+i^2+2i–1-i^2+2i}{1–(-1)}

=\frac{4i}{2}

= 2i

So, modulus of z =\sqrt{0^2+2^2}  = 2.

Therefore, the modulus of\frac{1+i}{1–i}–\frac{1–i}{1+i}  is 2.

Question 8. If x + iy =\frac{a+ib}{a−ib}  , prove that x² + y² = 1.

Solution:

We have,

=> x + iy =\frac{a+ib}{a−ib}

On applying modulus on both sides we get,

=> |x + iy| =|\frac{a+ib}{a−ib}|

=> |x + iy| =\frac{|a+ib|}{|a−ib|}

=>\sqrt{x^2+y^2}=\frac{\sqrt{a^2+b^2}}{\sqrt{a^2+b^2}}

=>\sqrt{x^2+y^2}  = 1

=> x² + y² = 1

Hence proved.

Question 9. Find the least positive integral value of n for which\left[\frac{1+i}{1-i}\right]^n  is real.

Solution:

We have, z =\left[\frac{1+i}{1-i}\right]^n

=\left[\frac{(1+i)^2}{(1-i)(1+i)}\right]^n

=\left[\frac{1+i^2+2i}{1-i^2}\right]^n

=\left[\frac{2i}{2}\right]^n

= iⁿ

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

Therefore, the least positive integral value of n for which\left[\frac{1+i}{1-i}\right]^n  is real is 2.

Question 10. Find the real values of θ for which the complex number\frac{1 + i cos θ}{1 – 2i cos θ}  is purely real.

Solution:

We have, z =\frac{1 + i cos θ}{1 – 2i cos θ}

=\frac{(1+icosθ)(1+2icosθ)}{(1–2icosθ)(1+2icosθ)}

=\frac{1+2icosθ+icosθ+2i^2cos^2θ}{1-4i^2cos^2θ}

=\frac{1-2cos^2θ+3icosθ}{1+4cos^2θ}

=\frac{1-2cos^2θ}{1+4cos^2θ}+\frac{3cosθi}{1+4cos^2θ}

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

So, we get,\frac{3cosθ}{1+4cos^2θ}  = 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 which\frac{(1+i)^n}{(1-i)^{n-2}}  is a real number.

Solution:

We have, z =\frac{(1+i)^n}{(1-i)^{n-2}}

=\frac{(1+i)^n(1-i)^2}{(1-i)^{n-2}(1-i)^2}

=\frac{(1+i)^n}{(1-i)^n}×(1-i)^2

=\left(\frac{1+i}{1-i}\right)^n×(1+i^2-2i)

=\left[\frac{(1+i)^2}{(1-i)(1+i)}\right]^n×(-2i)

=\left[\frac{1+i^2+2i}{1-i^2}\right]^n×(-2i)

=\left(\frac{2i}{2}\right)^n×(-2i)

= 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 number\frac{(1+i)^n}{(1-i)^{n-2}}  is 1.

Question 12. If\left(\frac{1+i}{1-i}\right)^3–\left(\frac{1-i}{1+i}\right)^3=x+iy  , find (x, y).

Solution:

We have,

=>\left[\frac{(1+i)^2}{(1-i)(1+i)}\right]^3–\left[\frac{(1-i)^2}{(1+i)(1-i)}\right]^3=x+iy

=>\left[\frac{1+i^2+2i}{1-i^2}\right]^3–\left[\frac{1+i^2-2i}{1-i^2}\right]^3=x+iy

=>\left[\frac{2i}{2}\right]^3–\left[\frac{-2i}{2}\right]^3=x+iy

=> 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\frac{(1+i)^2}{2-i} = x + iy  , find x + y.

Solution:

We have,

=>\frac{(1+i)^2}{2-i} = x + iy

=>\frac{(1+i^2+2i)(2+i)}{(2-i)(2+i)} = x + iy

=>\frac{2i(2+i)}{4+1} = x + iy

=>\frac{2i^2+4i}{5} = x + iy

=>\frac{-2}{5}+\frac{4i}{5} = x + iy

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.

Conclusion

Mastering the operations on complex numbers such as the addition, subtraction, multiplication and division is crucial for the solving many algebraic problems. Exercise 13.2 from RD Sharma Class 11 Mathematics helps students solidify their understanding of these basic concepts. Solving a variety of the problems enhances the conceptual grasp of the complex numbers and their properties.