Class 11 NCERT Solutions Chapter 5 Complex Numbers And Quadratic Equations Miscellaneous Exercise on Chapter 5 | Set 2 (original) (raw)
Last Updated : 5 Sep, 2024
Chapter 5 of Class 11 NCERT Mathematics deals with Complex Numbers and Quadratic Equations. This chapter introduces the concept of complex numbers, which are essential for solving certain types of quadratic equations that do not have real solutions. The chapter also covers the algebraic operations on complex numbers and their geometrical representation in the Argand plane providing an understanding of the topic. Moreover, the quadratic equations section teaches the methods to find roots including the real and complex roots through various techniques like factorization and quadratic formulas.
Complex Numbers And Quadratic Equations
The Complex numbers extend the concept of real numbers by including the square root of the negative one denoted as i. A complex number is expressed in the form a+bi where a and b are real numbers. This concept is crucial in solving quadratic equations that do not have real solutions. The chapter also explains the addition, subtraction, multiplication, and division of complex numbers along with their graphical representation in the Argand plane. Understanding these concepts is vital for solving advanced mathematical problems and equations.
Question 11. If a + ib = \frac{(x+i)^2}{2x^2+1}, prove that a 2 + b 2 = \frac{(x^2+i)^2}{(2x^2+1)^2}
**Solution:
Given:
a + ib = \frac{(x+i)^2}{2x^2+1}\\ =\frac{x^2+i^2+2xi}{2x^2+1}\\ =\frac{x^2-1+i2x}{2x^2+1}\\ =\frac{x^2-1}{2x^2+1}+i\left(\frac{2x}{2x^2+1}\right)
On comparing the real and imaginary parts, we have
a = \frac{(x-1)}{2x^2+1} and b = \frac{2x}{2x^2+1}
Therefore,
a2 + b2 = \left(\frac{x^2-1}{2x^2+1}\right)^2+\left(\frac{2x}{2x^2+1}\right)^2\\ =\frac{x^4+1-2x^2+4x^2}{(2x+1)^2}\\ =\frac{x^4+1+2x^2}{(2x^2+1)^2}\\ =\frac{(x^2+1)^2}{(2x^2+1)^2}
Hence, proved,
a2 + b2 = \frac{(x^2+1)^2}{(2x^2+1)^2}
Question 12. Let z 1 = 2 – **i**, z** 2 = -2 + **i****. Find**
****(i)** Re\left(\frac{z_1z_2}{\overline{z_1}}\right)
****(ii)** Im\left(\frac{1}{z_1\overline{z_2}}\right)
**Solution:
****(i)** Given:
z1 = 2 - i, z2 = -2 + i
(i) z1z2 = (2 - i)(-2 + i) = -4 + 2i + 2i - i2 = -4 + 4i - (-1) = -3 + 4i
\overline{z_1} = 2 + i
Therefore,
\frac{z_1z_2}{\overline{z_1}}=\frac{-3+4i}{2+i}
On multiplying numerator and denominator by (2 - i), we get
\frac{z_1z_2}{\overline{z_1}}=\frac{(-3+4i)(2-i)}{(2+i)(2-i)}=\frac{-6+3i+8i-4i^2}{2^2+1^2}=\frac{-6+11i-4(-1)}{2^2+1^2}\\ =\frac{-2+11i}{5}=\frac{-2}{5}+\frac{11}{5}i
On comparing the real parts, we have
Re\left(\frac{z_1z_2}{\overline{z_1}}\right)=\frac{-2}{5}
****(ii)** \frac{1}{z_1\overline{z_2}}=\frac{1}{(2-1)(2+i)}=\frac{1}{(2)^2+(1)^2}=\frac{1}{5}\\
On comparing the imaginary part, we get
Im\left(\frac{1}{z_1\overline{z_2}}\right) = 0
**Question 13. Find the modulus and argument of the complex number \frac{1+2i}{1-3i}
**Solution:
Let, z = \frac{1+2i}{1-3i} , then
z = \frac{1+2i}{1-3i}\times\frac{1+3i}{1+3i}=\frac{1+3i+2i+6i^2}{1^2+3^6}=\frac{1+5i+6(-1)}{1+9}\\ =\frac{-5+5i}{10}=\frac{-5}{10}+\frac{5i}{10}=\frac{-1}{2}+\frac{1}{2}i
Let z = r cosθ + ir sinθ
So,
r cosθ = \frac{-1}{2} and r sinθ = \frac{1}{2}
On squaring and adding, we get
r2(cos2θ + sin2θ) = \left(\frac{-1}{2}\right)^2+\left(\frac{1}{2}\right)^2
r2 = \frac{1}{4}+\frac{1}{4}=\frac{1}{2}\ \ \ \ \ \ [Conventionally,\ r>0]
r = \frac{1}{\sqrt2}
Now,
\frac{1}{\sqrt2} cosθ = \frac{-1}{2} and \frac{1}{\sqrt2} sinθ = \frac{1}{2}
= cosθ = \frac{-1}{\sqrt2} and sinθ = \frac{1}{\sqrt2}
Therefore,
θ = \pi-\frac{\pi}{4}=\frac{3\pi}{4} [As θ lies in the II quadrant]
Question 14. Find the real numbers **x and **y if ( x – **iy****) (3 + 5** i) is the conjugate of – 6 – 24 i****.**
**Solution:
Let us assume z = (_x – iy) (3 + 5__i)
z = 3x + xi - 3yi - 5yi2 = 3x + 5xi - 3yi + 5y = (3x + 5y) + i(5x - 3y)
Therefore,
\overline{z} =(3x + 5y) - i(5x - 3y)
Also given, \overline{z} = -6 - 24i
And,
(3x + 5y) – i(5x – 3y) = -6 -24__i
After equating real and imaginary parts, we get
3x + 5y = -6 …… (i)
5x – 3y = 24 …… (ii)
After doing (i) x 3 + (ii) x 5, we have
(9x + 15y) + (25x – 15y) = -18 + 120
34x = 102
x = 102/34 = 3
Putting the value of _x in equation (i), we get
3(3) + 5y = -6
5y = -6 – 9 = -15
y = -3
Therefore, the values of _x and _y are 3 and –3 respectively.
**Question 15. Find the modulus of \frac{1+i}{1-i}-\frac{1-i}{1+i}
**Solution:
\frac{1+i}{1-i}-\frac{1-i}{1+i}=\frac{(1+i)^2-(1-i)^2}{(1-i)(1+i)}\\ =\frac{1+i^2+2i-1-i^2+2i}{1^2+1^2}\\ =\frac{4i}{2}=2i\\ \therefore\left|\frac{1+i}{1-i}-\frac{1-i}{1+i}\right|=|2i|=\sqrt{2^2}=2
Question 16. If ( x + **iy**)** 3 = **u + **iv****, then show that** \frac{u}{y}+\frac{v}{y} = 4(x2 - y2)
**Solution:
(x + iy)3 = u + iv
x3 + (iy)3 + 3 × x × iy(x + iy) = u + iv
x3 + i3y3 + 3x2yi + 3xy2 = u + iv
x3 - iy3 + 3x2yi - 3xy2 = u + iv
(x3 - 3xy2) + i(3x2y - y3) = u + iv
On equating real and imaginary parts, we get
u = x3 - 3xy2, v = 3x2y - y3
\frac{u}{x}+\frac{v}{y}=\frac{x^3-3xy^2}{x}+\frac{3x^2y-y^3}{y}\\ =\frac{x(x^2-3y^2)}{x}+\frac{y(3x^2y-y)}{y}
= x2 - 3y2 + 3x2 - y2
= 4x2 - 4y2
= 4(x2 - y2)
\therefore\frac{u}{x}+\frac{v}{y}=4(x^2-y^2)
Hence proved
**Question 17. If α and β are different complex numbers with |β| = 1, then find \left|\frac{\beta-\alpha}{1-\overline{\alpha}\beta}\right|
**Solution:
Assume α = a + ib and β = x + iy
Given: |β| = 1
So, \sqrt{x^2+y^2}=1
= x2 + y2 = 1 ....(1)
\left|\frac{\beta-\alpha}{1-\overline{\alpha}\beta}\right|=\left|\frac{(x+iy)(a+ib)}{1-(a-ib)(x+iy)}\right|\\ =\left|\frac{(x-a)+i(y-b)}{1-(ax+aiy-ibx+by)}\right|\\ =\left|\frac{(x-a)+i(y-b)}{(1-ax-by)+i(bx-ay)}\right|\\ =\left|\frac{(x-a)+i(y-b)}{(1-ax-by)+i(bx-ay)}\right|\ \ \ \ \ \left[\left|\frac{z_1}{z_2}\right|=\left|\frac{z_1}{z_2}\right|\right]\\ =\frac{\sqrt{(x-a)^2+(y-b)^2}}{\sqrt{(1-ax-by)^2+(bx-ay)^2}}\\ =\frac{\sqrt{x^2+a^2-2ax+y^2+b^2-2by}}{\sqrt{1+a^2x^2+b^2y^2-2ax+2abxy-2by+b^2x^2+a^2y^2-2abxy}}\\ =\frac{\sqrt{(x^2+y^2)+a^2+b^2-2ax-2by}}{\sqrt{1+a^2(x^2+y^2)+b^2(y^2+x^2)-2ax-2by}}\\ =\frac{\sqrt{1+a^2+b^2-2ax-2by}}{\sqrt{1+a^2+b^2-2ax-2by}}\ \ \ \ \ \ \ \ [Using\ (1)]
= 1
\therefore\left|\frac{\beta-\alpha}{1-\overline{\alpha}\beta}\right|=1
**Question 18. Find the number of non-zero integral solutions of the equation |1 – i| x = 2 x
**Solution:
|1 - i|x = 2t
(\sqrt{1^2+(-1)^2})^x=2^x\\ (\sqrt2)^x=2^x\\ 2^{\frac{x}{2}}=2^x\\ \frac{x}{2}=x
x = 2x
2x - 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.
**Question 19. If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that (a 2 + b 2 ) (c 2 + d 2 ) (e 2 + f 2 ) (g 2 + h 2 ) = A 2 + B 2 .
**Solution:
Given:
(a + ib)(c + id)(e + if)(g + ih) = A + iB
Therefore,
|(a + ib)(c + id)(e + if)(g + ih)| = |A + iB|
= |(a + ib)| × |(c + id)| × |(e + if)| × |(g + ih)| = |A + iB|
\sqrt{a^2+b^2}\times\sqrt{c^2+d^2}\times\sqrt{e^2+f^2}\times\sqrt{g^2+h^2}=\sqrt{A^2+B^2}
On squaring both sides, we get
(__a_2 + __b_2) (__c_2 + __d_2) (__e_2 + __f_2) (__g_2 + __h_2) = A2 + B2
Hence, proved.
**Question 20. If, then find the least positive integral value of m.\left(\frac{1+i}{1-i}\right)^m=1
**Solution:
\left(\frac{1+i}{1-i}\right)^m=1
\left(\frac{1+i}{1-i}\times\frac{1+i}{1+i}\right)^m=1\\ \left(\frac{(1+i)^2}{1^2+1^2}\right)^m=1\\ \left(\frac{1-1+2i}{2}\right)^m=1\\ \left(\frac{2i}{2}\right)^m=1
im = 1
Hence, m = 4k, where k is some integer.
Hence, the least positive integer is 1.
Thus, the least positive integral value of _m is 4 (= 4 × 1).
Conclusion
The Complex Numbers and Quadratic Equations form the foundation for the understanding higher mathematics particularly in fields involving the electrical engineering, quantum physics and applied mathematics. Mastering these concepts allows students to the solve quadratic equations with the non-real solutions and work comfortably with the complex numbers in the various mathematical contexts. This chapter is essential for the building a strong mathematical base enabling students to the approach complex problems with the confidence.