Class 11 Maths NCERT Solutions for Chapter 5 Complex Numbers and Quadratic Equations Miscellaneous Exercise
![Class 11 Maths NCERT Solutions for Chapter 5 Complex Numbers and Quadratic Equations Miscellaneous Exercise Class 11 Maths NCERT Solutions for Chapter 5 Complex Numbers and Quadratic Equations Miscellaneous Exercise](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhxhazMuncXX7vTMAl3y5N6Pfvevpja7pJgBKRdcP_9sYGiN3sfix8KAyfcTM1qWVIyWJTp1rODH5Xcbt-vOmO72HPkIu-H9X_tG0wFWwyHo5X9jvmNlZd-8FTlUg9Cr0XqnL6vXX2SDVxCawWVr3mOmS6_GX8xG1muGK3WkzwNQRs7O5RQkxbEP_Pi/w655-h309-rw/ncert-solutions-for-class11-maths-complex-numbers-and-quadratic-equation-miscellaneous-exercise.jpg)
Complex Numbers and Quadratic Equations Miscellaneous Exercise Solutions
1. Evaluate : [i18 + (1/i)25 ]3 .
Solution
= [-1 - i]3
= (-1)3 [1 +i]3
= -[13 + i3 + 3.1.i(1 + i)]
= -[1 + i3 + 3i + 3i2 ]
= -[1 - i + 3i -3]
= -[-2+2i]
= 2 -2i
2. For any two complex numbersz1 and z2, prove that
Re (z1z2)= Re z1 Re z2 – Im z1 Im z2
Solution
Let z1 = x1 + iy1 and z2 = x2 + iy2
∴ z1z2 = (x1 + iy1)(x2 + iy2)
= x1 (x2 + iy2) + iy1 (x2 + iy2)
= x1x2 + ix1y2 + iy1x2 + i2 y1y2
= x1x2 + ix1y2 + iy1x2 + y1y2 [i2 = -1]
= (x1x2 - y1y2) + i(x1y2 + y1x2)
⇒ Re (z1z2) = x1x2 - y1x2
⇒ Re (z1z2) = Re z1 Re z2 - Im z1 Im z2
Hence, proved.
∴ z1z2 = (x1 + iy1)(x2 + iy2)
= x1 (x2 + iy2) + iy1 (x2 + iy2)
= x1x2 + ix1y2 + iy1x2 + i2 y1y2
= x1x2 + ix1y2 + iy1x2 + y1y2 [i2 = -1]
= (x1x2 - y1y2) + i(x1y2 + y1x2)
⇒ Re (z1z2) = x1x2 - y1x2
⇒ Re (z1z2) = Re z1 Re z2 - Im z1 Im z2
Hence, proved.
3. Reduce [(1/(1-4i) - 2/(1 + i)][(3 - 4i)/(5 +i)] to the standard form.
Solution
4. If x - iy = √(a - ib)/(c-id) prove that (x2 + y2)2 = (a2 + b2)/(c2 + d2)
Solution
5. Convert the following in the polar form :
(i) (1 + 7i)/(2 - i)2 ,
(ii) (1 + 3i)/(1 - 2i)
(i) (1 + 7i)/(2 - i)2 ,
(ii) (1 + 3i)/(1 - 2i)
Solution
Let r cos θ = –1 and r sin θ = 1
On squaring and adding, we obtain
r2 (cos2 θ + sin2 θ) = 1 + 1
⇒ r2 (cos2 θ + sin2 θ) = 2
⇒ r2 = 2 [cos2 θ + sin2 θ = 1]
⇒ r = √2 [Conventionally, r > 0]
∴ √2 cosθ = -1 and √2 sinθ = 1
⇒ cos θ = -1/√2 and sin θ = 1/√2
∴ θ = Ï€ - Ï€/4 = 3Ï€/4 [As θ lies in II quadrant]
∴ z = r cos θ + i r sin θ
= √2 cos (3Ï€/4) + i √2sin (3Ï€/4) = √2[ cos(3Ï€/4) + i sin(3Ï€/4)
This is the required polar form.
(ii) Here, z = (1 + 3i)/(1 - 2i)
Let r cos θ = –1 and r sin θ = 1
On squaring and adding, we obtain
r2 (cos2 θ + sin2 θ) = 1 + 1
⇒r2 (cos2 θ + sin2 θ) = 2
⇒ r2 = 2 [cos2 θ + sin2 θ = 1]
⇒ r = √2 [Conventionally, r > 0]
∴ √2 cosθ = -1 and √2sinθ = 1
⇒ cos θ = -1/√2 and sinθ = 1/√2
∴ θ = Ï€ - Ï€4 = 3Ï€/4 [As θ lies in II quadrant]
∴ z = r cos θ + i r sin θ
= √2 cos 3Ï€/4 + i√2 sin3Ï€/4 = √2 [cos (3Ï€/4 + i sin(3Ï€/4)]
This is the required polar form.
r2 (cos2 θ + sin2 θ) = 1 + 1
⇒r2 (cos2 θ + sin2 θ) = 2
⇒ r2 = 2 [cos2 θ + sin2 θ = 1]
⇒ r = √2 [Conventionally, r > 0]
∴ √2 cosθ = -1 and √2sinθ = 1
⇒ cos θ = -1/√2 and sinθ = 1/√2
∴ θ = Ï€ - Ï€4 = 3Ï€/4 [As θ lies in II quadrant]
∴ z = r cos θ + i r sin θ
= √2 cos 3Ï€/4 + i√2 sin3Ï€/4 = √2 [cos (3Ï€/4 + i sin(3Ï€/4)]
This is the required polar form.
6. Solve the equation 3x2 - 4x + 20/3 = 0
Solution
The given quadratic equation is 3x2 - 4x + 20/3 = 0
This equation can also be written as 9x2 - 12x + 20 = 0
On comparing this equation with ax2 + bx + c = 0 , we obtain
a = 9 b = -12 and c = 20
Therefore, the discriminant of the given equation is
D = b2 - 4ac = (-12)2 - 4× 9 × 20 = 144 - 720 = -576
Therefore, the required solutions are
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiAX4F8ll-ZpCxGJbgszCI25D6E3PjY_OYh8cwUvEv8KfKtJdshczrNvbsyOFNYwRc0oCnM9tro1yKL1MCQecv4CNejzfi8pqgptVsmA__eBDDJK0aabpl3DxoLFSL0scw3dZFuOIXq7LF-J1YQN9Jj581u3teSRXWRPHFKg8u7RnvuJE643sB6TuZ1/w436-h98-rw/NCERT%20Solution%20For%20Class%2011%20Math%20Chapter%205%20Complex%20Numbers%20Misc.%20Exercise%20img%206.JPG)
This equation can also be written as 9x2 - 12x + 20 = 0
On comparing this equation with ax2 + bx + c = 0 , we obtain
a = 9 b = -12 and c = 20
Therefore, the discriminant of the given equation is
D = b2 - 4ac = (-12)2 - 4× 9 × 20 = 144 - 720 = -576
Therefore, the required solutions are
7. Solve the equation x2 - 2x + 3/2 = 0
Solution
The given quadratic equation is x2 - 2x + 3/2 = 0
This equation can also be written as 2x2 - 4x +3 = 0
On comparing this equation with ax2 + bx + c = 0 we obtain
a = 2, b = -4 and c = 3
Therefore, the discriminant of the given equation is
D = b2 - 4ac = (-4)2 - 4 × 2 × 3 = 16 - 24 = -8
Therefore, the required solutions are
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiqdu55G5fLNsmTDgA-mhYTSTa2ayVOYD5JwyoXZAMk_D9NgeigYa-SS59wnWcXDFWy4pwJfDuMcFVRFqRuucURumAQbvRtSYCNeiLl9W98FqUzlYGOkLBHBDmDCsbVnmTu9SkNb8BcSxM8vc-Czktbdl3mV0Hsw0AHo360ckgQ3MxdBAL8TJKL75p_/w408-h105-rw/NCERT%20Solution%20For%20Class%2011%20Math%20Chapter%205%20Complex%20Numbers%20%20Misc.%20Exercise%20img%207.JPG)
On comparing this equation with ax2 + bx + c = 0 we obtain
a = 2, b = -4 and c = 3
Therefore, the discriminant of the given equation is
D = b2 - 4ac = (-4)2 - 4 × 2 × 3 = 16 - 24 = -8
Therefore, the required solutions are
8. Solve the equation 27x2 - 10x + 1 = 0
The given quadratic equation is 27x2 - 10x + 1 = 0
On comparing the given equation with ax2 + bx + c = 0, we obtain
a = 27, b = -10, and c = 1
Therefore, the discriminant of the given equation is
D = b2 - 4ac = (-10)2 - 4 × 27 × 1 = 100 - 108 = -8
Therefore, the required solutions are
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhhxAqx6Zt9y28XvisKh_aC_f8kt9Pu4LZOngS_iMVzbt1U4wFtbD8E8CnlKLOhM7ldA3R00vTPxlMZDKYpuLn1ELTjYZySO-8pLGJkh8sY2ZrhrJIeKIcUrWxr0jCjQatyhC0WKmLa96Upbvjl3FYVtQbXsjzmANSmIsMbppXEuVBXqeWLfx3BcMEu/w377-h100-rw/NCERT%20Solution%20For%20Class%2011%20Math%20Chapter%205%20Complex%20Numbers%20%20Misc.%20Exercise%20img%208.JPG)
On comparing the given equation with ax2 + bx + c = 0, we obtain
a = 27, b = -10, and c = 1
Therefore, the discriminant of the given equation is
D = b2 - 4ac = (-10)2 - 4 × 27 × 1 = 100 - 108 = -8
Therefore, the required solutions are
9. Solve the equation 21x2 - 28x + 10 = 0
Solution
The given quadratic equation is 21x2 - 28x + 10 = 0
On comparing the given equation with ax2 + bx + c = 0 , we obtain
a = 21, b = -28 and c = 10
therefore, the discriminant of the given equation is
D = b2 - 4ac = (-28)2 - 4×21 ×10 = 784 - 840 = -56
On comparing the given equation with ax2 + bx + c = 0 , we obtain
a = 21, b = -28 and c = 10
therefore, the discriminant of the given equation is
D = b2 - 4ac = (-28)2 - 4×21 ×10 = 784 - 840 = -56
Therefore, the required solutions are
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEij8jFdIY-gYcLKWL9T2V4NLkLDAh4Pm0SxWkf4bQBXcmzxDWzaf2hbJHrwqa1YlnehPnGoUH04jz5JXF-HnFJv4amwQGDS4exmOrL4Q-LXw-ZbCEqJabB1R0e8yqbr6cx_Qwiu5Cw9qppLrHcBTguw-m0VSsjnxKG8ZLx_k__qzRe766eSzMFtLZja/w278-h100-rw/NCERT%20Solution%20For%20Class%2011%20Math%20Chapter%205%20Complex%20Numbers%20%20Misc.%20Exercise%20img%209.JPG)
10. If z1 = 2 – i , z2 = 1+ i , find |(z1 + z2 + 1)/(z1 – z2 + 1)|
Solution
z1 = 2 – I , z2 = 1+ i ,
11. If (a + ib) = (x + 1)2 /(2x2 + 1) , prove that a2 + b2 = (x2 + 1)2 /(2x + 1)2
Solution
12. Let z1 = 2 – i , z2 = -2+ i . find
(i) Re(z1 z2 /z1)
(ii) Im(1/z1 z1)
(i) Re(z1 z2 /z1)
(ii) Im(1/z1 z1)
Solution
z1 = 2 – i , z2 = -2+ i
(i) z1 z2 = (2 - i)(-2 + i) = -4 + 2i +2i - i2 = -4 + 4i -(-1) = -3 + 4i
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjnXG6MyjxsHWU84xurYAAUcIJS3FUnfX7VKGx36S8gVGJiT-1JnK7wpZZCJ98nA_KV_XInIeuVx6MCc38NG7Q-QICWjOWkH4wg6KFxAKl0Z-ahSdL21ckS5_nYqPyze3gEwDoFyTVymJ6f1rfMChFU2mH2j9ibwT0pTfXZELLBOB3KlVoi3ajH5upF/w422-h406-rw/NCERT%20Solution%20For%20Class%2011%20Math%20Chapter%205%20Complex%20Numbers%20%20Misc.%20Exercise%20img%2012.JPG)
13. Find the modulus and argument of the complex number (1 + 2i)/(1 - 3i).
Solution
Let z = (1 + 2i)/(1 - 3i) , then
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgwt03WaT1Ta6tKRkM0tFJUjH5oCRvHcSytWqvvkGGD9FGZuJApOk7dmhlYFI0ynZhL8k96IWAY5y5X_gH2aCgbTUknZ1Gyvw7PVy5o5uIN5t3x7IuODokmTxfY044qvs8DN9fEEILSL8rSx_K52pTFNDMA-LdiFM3xRZ2tolt_BKLaGa0-iKUHf4u5/w384-h98-rw/NCERT%20Solution%20For%20Class%2011%20Math%20Chapter%205%20Complex%20Numbers%20%20Misc.%20Exercise%20img%2013.JPG)
Let z = r cos θ = -1/2 and r sin θ = 1/2
On squaring and adding, we obtain
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgNT6H8EGljGV9P9fy1FmJK6fb4xJGMMezAkfMLFFjphpu8fuOseYnrsR0kF6BYnHXxnVQlLLx3bUCl9p8JjZEkdZSKUGYsBJEJeQucweFm6LDwIueHZWoNFOcagEq-isspCrlA2XcCbge9h9Y5JlIpw_sUIL6DsbkoWDEqVev2HOdfVZJhtPF3JmFh/w364-h299-rw/NCERT%20Solution%20For%20Class%2011%20Math%20Chapter%205%20Complex%20Numbers%20%20Misc.%20Exercise%20img%2014.JPG)
Let z = r cos θ = -1/2 and r sin θ = 1/2
On squaring and adding, we obtain
14. Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of –6 – 24i.
Solution
Let z = (x - iy)(3 + 5i)
z = 3x + 5xi - 3yi - 5yi2 = 3x + 5xi - 3yi + 5y = (3x + 5y) + i(5x - 3y)
∴ z = (3x + 5y) - i(5x - 3y)
It is given that , z = -6-24i
∴ (3x +5y) - i(5x - 3y) = -6 - 24i
Equating real and imaginary parts, we obtain
3x + 5y = -6 ...(i)
5x - 3y = 24 ...(ii)
Multiplying equation (i) by 3 and equation (ii) by 5 and then adding them, we obtain
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi4Bm2KRoAN2e9wfOkdbV4DWjn2kvmidJcxkSsqOGGXzmIKkZqCfgRF4otly-TXFnHDqMZ3fhURfVNQ5K_XVAREnHXqLX0OBxUCdiBT79__HRSnKL1sQrJwqe3U-J1_CEHh-fdhb8cWhu9sFKPMaAnY6AE-Tfx_hOZ8ObLtGCWl9HWNr9MJsEY8HGH4/w124-h127-rw/NCERT%20Solution%20For%20Class%2011%20Math%20Chapter%205%20Complex%20Numbers%20%20Misc.%20Exercise%20img%2015.JPG)
z = 3x + 5xi - 3yi - 5yi2 = 3x + 5xi - 3yi + 5y = (3x + 5y) + i(5x - 3y)
∴ z = (3x + 5y) - i(5x - 3y)
It is given that , z = -6-24i
∴ (3x +5y) - i(5x - 3y) = -6 - 24i
Equating real and imaginary parts, we obtain
3x + 5y = -6 ...(i)
5x - 3y = 24 ...(ii)
Multiplying equation (i) by 3 and equation (ii) by 5 and then adding them, we obtain
Putting the value of x in equation (i), we obtain
3(3) + 5y = -6
⇒ 5y = -6 - 9 = -15
⇒ y = -3
Thus, the values of x and y are 3 and -3 respectively.
⇒ 5y = -6 - 9 = -15
⇒ y = -3
Thus, the values of x and y are 3 and -3 respectively.
16. Find the modulus of [(1+i)/(1-i)] -[(1 -i)/(1 +i)].
Solution
17. If α and β are different complex numbers with |β| = 1, then find |(β - α)/(1 - α β)|.
Solution
Let α = a + ib and β = x +iy
It is given that, |β| = 1
∴ √(x2 + y2 ) = 1
⇒ x2 + y2 = 1 ...(i)
It is given that, |β| = 1
∴ √(x2 + y2 ) = 1
⇒ x2 + y2 = 1 ...(i)
18. Find the number of non-zero integral solutions of the equation |1-i|x = 2x .
Solution
|1-x|x = 2x
⇒ 2x/2 = 2x
⇒ x/2 = x
⇒ x = 2x
⇒ 2x - x = 0
⇒ 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.
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.
(a2 + b2)(c2 + d2)(e2 + f2 )(g2 + h2) = A2 + B2.
Solution
(a + ib)(c + id)(e + if)(g + ih) = A + iB
∴ |(a + ib)(c + id)(e + if)(g + ih)| = |A + iB|
⇒ |(a + ib)|× |(c + id)|× |(e + if)| × |(g + ih)| = |A + B| [|z1 z2| z1||z2|]
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjojpqH2mI-93t_5DQYp170q4J4evXxq58wontMUnOsAEk3y-ScXy5z1h-Lv8YksxMCv6SH1D0d5DCGGWLE3zKDHqaM32W4u1hQYEc--K4fc0r2sV_FrORB-JZ_6GoGhd-BnaCO4JuhesSuBi7H8j2_WoQk2DT9ssepUIb-xarMRDSluW1wwrMcV59s/w453-h32-rw/NCERT%20Solution%20For%20Class%2011%20Math%20Chapter%205%20Complex%20Numbers%20%20Misc.%20Exercise%20img%2019.JPG)
On squaring both sides, we obtain
(a2 + b2 )(c2 + d2 )(e2 + f2 )(g2 + h2 ) = A2 + B2 .
Hence, proved.
∴ |(a + ib)(c + id)(e + if)(g + ih)| = |A + iB|
⇒ |(a + ib)|× |(c + id)|× |(e + if)| × |(g + ih)| = |A + B| [|z1 z2| z1||z2|]
On squaring both sides, we obtain
(a2 + b2 )(c2 + d2 )(e2 + f2 )(g2 + h2 ) = A2 + B2 .
Hence, proved.
20. If [(1 +i)/(1 - i)]m = 1, then find the least positive integral value of m.
Solution
∴ m = 4k, where k is some integer.
Therefore, the least positive integer is 1.
Thus, the least positive integral value of m is 4 (= 4× 1).