NCERT Solutions for Class 11 Maths Chapter 5 – Complex Numbers and Quadratic Equations Miscellaneous Exercise
Page No 112:
Question 1:
Evaluate: ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5170/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_m7acf44a4.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5170/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_m7acf44a4.gif)
Answer:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5170/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_73835564.gif)
Question 2:
For any two complex numbers z1 and z2, prove that
Re (z1z2) = Re z1 Re z2 – Im z1 Im z2
Answer:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5171/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_37b5fc68.gif)
Question 3:
Reduce
to the standard form.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5172/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_m44175856.gif)
Answer:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5172/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_3b0080ad.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5172/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_m6daa0528.gif)
Question 4:
If x – iy =
prove that
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5544/Chapter%205_html_m15935963.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5544/Chapter%205_html_1ea9f5cc.gif)
Answer:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5544/Chapter%205_html_68ee037a.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5544/Chapter%205_html_21b1bb0d.gif)
Question 5:
Convert the following in the polar form:
(i)
, (ii) ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5174/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_m3e7bd60e.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5174/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_m59b98ab0.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5174/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_m3e7bd60e.gif)
Answer:
(i) Here, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5174/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_me2a1ee0.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5174/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_me2a1ee0.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5174/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_39f19a8.gif)
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]
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5174/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_38425a1c.gif)
∴z = r cos θ + i r sin θ
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5174/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_m1d873d38.gif)
This is the required polar form.
(ii) Here, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5174/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_m5f85a9ef.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5174/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_m5f85a9ef.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5174/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_m21c6be01.gif)
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]
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5174/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_m20ff383b.gif)
∴z = r cos θ + i r sin θ
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5174/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_m1d873d38.gif)
This is the required polar form.
Question 6:
Solve the equation![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5175/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_5e07f3aa.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5175/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_5e07f3aa.gif)
Answer:
The given quadratic equation is ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5175/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_5e07f3aa.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5175/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_5e07f3aa.gif)
This equation can also be written as ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5175/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_3a5b444b.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5175/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_3a5b444b.gif)
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://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5175/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_2c7fa35e.gif)
Question 7:
Solve the equation![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5176/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_2ba079b4.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5176/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_2ba079b4.gif)
Answer:
The given quadratic equation is![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5176/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_2ba079b4.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5176/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_2ba079b4.gif)
This equation can also be written as ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5176/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_m1de30ce0.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5176/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_m1de30ce0.gif)
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://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5176/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_m2fa8dcd.gif)
Question 8:
Solve the equation 27x2 – 10x + 1 = 0
Answer:
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://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5177/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_m698c631.gif)
Page No 113:
Question 9:
Solve the equation 21x2 – 28x + 10 = 0
Answer:
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
Therefore, the required solutions are
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5178/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_m718fef22.gif)
Question 10:
If
find
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5179/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_5ec39a83.gif)
![](https://img-nm.mnimgs.com/img/study_content/editlive_ncert/33/2013_06_27_09_59_04/NCERT_10-10-08_Khushboo_11_Math_MiscellaneousCh-5_20_SU_SS_html_49f41ce6_9050829863458491646.png)
Answer:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5179/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_49f41ce6.gif)
Question 11:
If a + ib =
, prove that a2 + b2 = ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5546/Chapter%205_html_m26128272.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5546/Chapter%205_html_53f4fbc0.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5546/Chapter%205_html_m26128272.gif)
Answer:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5546/Chapter%205_html_1906e2dd.gif)
On comparing real and imaginary parts, we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5546/Chapter%205_html_5574b6ca.gif)
Hence, proved.
Question 12:
Let
. Find
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5547/Chapter%205_html_3c99b1f5.gif)
(i)
, (ii) ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5547/Chapter%205_html_5f46c0a8.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5547/Chapter%205_html_5dfb6105.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5547/Chapter%205_html_5f46c0a8.gif)
Answer:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5547/Chapter%205_html_3c99b1f5.gif)
(i) ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5547/Chapter%205_html_4724c2b5.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5547/Chapter%205_html_4724c2b5.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5547/Chapter%205_html_745329e4.gif)
On multiplying numerator and denominator by (2 – i), we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5547/Chapter%205_html_m46a446f5.gif)
On comparing real parts, we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5547/Chapter%205_html_m1d2a11d0.gif)
(ii) ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5547/Chapter%205_html_m4a2467b7.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5547/Chapter%205_html_m4a2467b7.gif)
On comparing imaginary parts, we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5547/Chapter%205_html_m59e6ebdf.gif)
Question 13:
Find the modulus and argument of the complex number
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5182/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_m64c2f37f.gif)
Answer:
Let
, then
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5182/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_m6edea852.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5182/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_m731d2e93.gif)
On squaring and adding, we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5182/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_13b55a9a.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5182/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_499d1a01.gif)
Therefore, the modulus and argument of the given complex number are
respectively.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5182/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_m66ec51bd.gif)
Question 14:
Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of –6 – 24i.
Answer:
Let ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5183/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_b9d2fc7.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5183/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_b9d2fc7.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5183/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_b76a676.gif)
It is given that, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5183/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_61946830.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5183/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_61946830.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5183/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_5bff58df.gif)
Equating real and imaginary parts, we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5183/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_67e1525c.gif)
Multiplying equation (i) by 3 and equation (ii) by 5 and then adding them, we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5183/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_m74e563d6.gif)
Putting the value of x in equation (i), we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5183/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_583f7903.gif)
Thus, the values of x and y are 3 and –3 respectively.
Question 15:
Find the modulus of
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5184/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_581765e2.gif)
Answer:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5184/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_470aed60.gif)
Question 16:
If (x + iy)3 = u + iv, then show that
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5185/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_2f726302.gif)
Answer:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5185/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_m2d17c8bc.gif)
On equating real and imaginary parts, we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5185/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_m459b0122.gif)
Hence, proved.
Question 17:
If α and β are different complex numbers with
= 1, then find
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5548/Chapter%205_html_94a9e3a.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5548/Chapter%205_html_m3a2ebd87.gif)
Answer:
Let α = a + ib and β = x + iy
It is given that, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5548/Chapter%205_html_m1bc79ade.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5548/Chapter%205_html_m1bc79ade.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5548/Chapter%205_html_m2dae3f80.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5548/Chapter%205_html_m279538f2.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5548/Chapter%205_html_6417e530.gif)
Question 18:
Find the number of non-zero integral solutions of the equation
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5187/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_71021ba5.gif)
Answer:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5187/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_1a52f543.gif)
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.
Answer:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5188/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_c628180.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5188/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_m25070e56.gif)
On squaring both sides, we obtain
(a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = A2 + B2
Hence, proved.
Question 20:
If
, then find the least positive integral value of m.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5189/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_m52369c9c.gif)
Answer:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/165/5189/NCERT_10-10-08_Khushboo_11_Math_Miscellaneous%20Ch-5_20_SU_SS_html_79424024.gif)
Therefore, the least positive integer is 1.
Thus, the least positive integral value of m is 4 (= 4 × 1).