Ncert Solutions Maths class 11th

37. Let z = (x – iy) (3 + 5i)

= 3x + 5xi – 3yi – 5yi²

= (3x + 5y) + (5x – 3y)i

Given,  $\overline{z}$ = –6 – 24i

=> (3x + 5y) – (5x – 3y)i = –6 – 24i

Equating real and imaginary part,

3x + 5y = –6 - (1)

5x – 3y = 24 - (2)

Multiplying (1) by 3 and (2) by 5 and adding them, we get

9x + 15y + 25x – 15y = –18 + 120

=> 34x = 102

=>x = 102/34 = 3

Putting x = 3 in (1) we get,

3 * 3 + 5y = –6

=> 9 + 5y = –6

=> 5y = –6 – 9

=> 5y = –15

=>y = –15/5 = –3

Hence, the values of x and y are 3 and –3 respectively.

36.  Z₁ = 2 – i, z₂ = –2 + i

Z₁z₂ = (2 – i)(–2 + i)

= –4 + 2i + 2i – i2

= –4 + 4i + 1[since, i² = –1]

= –3 + 4i

$\overline{z_1}$ = 2 + i

i. $\frac{z_1{z}_2}{\overline{z_1}}$ = $\frac{- 3+4i}{2+i}$

= $\frac{-3+4i}{2+i}$ * $\frac{2-i}{2-i}$ [multiply denominator and numerator by (2 – i)]

= $\frac{-6+3i+8i-4i^2}{2^2- i^2}$

= $\frac{-6+11i+4}{4-\left(-1\right)}$ [since, i2 = –1]

= $\frac{-6+4+11i}{4+1 }$

= $\frac{-2+11i}{5}$

= $\frac{-2}{5}$ + $\frac{11}{5}i$

So, Re( $\frac{z_1{z}_2}{\overline{z_1}}$ ) = $\frac{-2}{5}$

ii. $\frac{1}{z_1\overline{z_1}}$ = $\frac{1}{\left(2-i\right)\left(2+i\right)}$

= $\frac{1}{2^2- i^2}$

= $\frac{1}{4+1}$ [since, i2 = –1]

= $\frac{1}{5}$ + 0i

Therefore, Im $\left(\frac{1}{z_1\overline{z_1}}\right)$ = 0

35. Let, z = a + ib

= $\frac{{\left(x+i\right)}^2}{2x^2+ 1}$

= $\frac{x^2+ i^2+2xi}{2x^2+ 1}$ [since, (a + b)² = a² + b² + 2ab]

= $\frac{x^2- 1+2xi}{2x^2+ 1}$ [since, i2 = –1]

= $\frac{x^2- 1}{2x^2+ 1}$ + $\frac{i2x}{2x^2+ 1}$

So, |z|² = a² + b²

= $\frac{{\left(x^2- 1\right)}^2}{{\left(2x^2+ 1\right)}^2}$ + $\frac{{(2x)}^2}{{\left(2x^2+ 1\right)}^2}$

= $\frac{{\left(x^2\right)}^2+ 1^2- 2.x^2.1+ 4x^2}{{\left(2x^2+ 1 \right)}^2}$ [since, (a + b)² = a² + b² + 2ab]

= $\frac{x^4+ 1-2x^2+ 4x^2}{{\left(2x^2+ 1\right)}^2}$

= $\frac{x^4+ 2x^2+ 1}{{\left(2x^2+ 1\right)}^2}$

= $\frac{{\left(x^2+ 1\right)}^2}{{\left(2x^2+ 1\right)}^2}$ [as, (a + b)² = a² + b² + 2ab]

Hence proved.

34. z₁ = 2 – i ,z₂ = 1 + i

$\left|\frac{z_1+ z_2+ 1}{z_1- z_2+ 1}\right|$

= $\left|\frac{2-i+1+i+1}{2-i-1-i+1}\right|$

= $\left|\frac{4}{2-2i}\right|$

= $\left|\frac{4}{2\left(1-i\right)}\right|$

= $\left|\frac{2}{1i}\right|$

= $|\frac{2}{1-i}$ * $\frac{1+i}{1+i}|$ [multiply numerator and denominator by (1 + i)]

= $\left|\frac{2+2i}{1^2-i^2}\right|$

= $\left|\frac{2+2i}{1-\left(-1\right)}\right|$ [since, i2 = –1]

= $\left|\frac{2+2i}{1+1}\right|$

= $\left|\frac{2\left(1+i\right)}{2}\right|$

= |1 + i|

32. 27x² – 10x + 1 = 0

Comparing the given equation with ax² + bx + c = 0

We have, a = 27, b = –10  andc = 1

Hence, discriminant of the equation is

b² – 4ac = ( 10)² – 4 * 27 * 1 = 100 – 108 =  –8

Therefore, the solution of the quadratic equation is

31.

Multiplying the above equation by 2, we get

and Comparing with ax² + bx + c = 0

We have, a = 2, b = –4 and c = 3

Hence, discriminant of the equation is

b² – 4ac = (-4)² – 4 * 2* 3 = 16 – 24 = –8

Therefore, the solution of the quadratic equation is

29. $\left(\frac{1}{1-4i}-\frac{2}{1+i}\right)$ $\left(\frac{3-4i}{5+i}\right)$

= $\left[\frac{1+i-2\left(1-4i\right)}{\left(1-4i\right)\left(1+i\right)}\right]$ $\left[ \frac{3-4i}{5+i}\right]$

= $\left(\frac{1+i-2+8i}{1+i-4i-4i^2}\right)$ $\left( \frac{3-4i}{5+i}\right)$

= $\left(\frac{9i-1}{5-3i}\right)$ $\left( \frac{3-4i}{5+i}\right)$

= $\frac{\left(9i-1\right)\left(3-4i\right)}{\left(5-3i\right)\left(5+i\right)}$

= $\frac{27i-36i^2-3+4i}{25+5i-15i-3i^2}$

= . [since, i² = –1]

= $\frac{33+31i}{28-10i}$

= $\frac{33+31i}{28-10i}$ * $\frac{28+10i}{28+10i}$ [multiplying denominator and numerator by 28 + 10i]

= $\frac{924+330i+868i+310i^2}{784-100i^2}$

= $\frac{924+1198i-310}{784+100}$ [since, i² = –1]

= $\frac{614+1198i}{884}$

= $\frac{2\left(307+599i\right)}{884}$

= $\frac{307+599i}{442}$

= $\frac{307}{442}$ + $i\frac{599}{442}$