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Class 11 NCERT Solutions – Chapter 5 Complex Numbers And Quadratic Equations – Miscellaneous Exercise on Chapter 5 | Set 2
  • Last Updated : 01 Apr, 2021

Question 11. If a + ib =  \frac{(x+i)^2}{2x^2+1}, prove that a2 + b2\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 z1 = 2 – i, z2 = -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 + 5i) is the conjugate of – 6 – 24i.

Solution:

Let us assume z = (x – iy) (3 + 5i)

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 -24i

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 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 × 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 = 2x

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 (a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = A2 + B2.

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

(a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = 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).

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