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

  • Last Updated : 30 Apr, 2021

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)

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= 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.




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