Maths

40.

So, the only solution of the given equation is 0.

Hence, there is no non – zero integral solution of the given equation.

39. (x + iy)³ = u + iv

=>x³ + (iy)³ + 3.x.iy (x + iy) = u + iv   [since, (a + b)³ = a³ + b³ + 3ab (a + b)]

=>x³ – iy³ + 3x²yi + 3xy²i² = u + iv

=>x³ – iy³ + 3x²yi – 3xy² = u + iv                [since, i² = -1]

=> (x³ – 3xy²) + i (3x²y – y³) = u + iv

Equating real and imaginary part we get,

u = x³ – 3xy² and v = 3x²y – y³

Now,  $\frac{u}{x}$ + $\frac{v}{y}$

= $\frac{x^3-3x{y}^2}{x}$ + $\frac{3x^{2}y- y^3}{y}$

= $\frac{x\left(x^2- 3y^2\right)}{x}$ + $\frac{y\left(3x^2- y^2\right)}{y}$

= x² – 3y² + 3x² – y²

= 4x² – 4y²

= 4 (x² – y²)

Hence proved.

38. $\frac{1+i}{1-i}$ – $\frac{1-i}{1+i}$

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

= $\frac{1^2+i^2+2.1.i –\left(1^2+i^2- 2.1.i\right)}{1^2- i^2}$ [Since, (a + b)² = a² + b² + 2ab

(a – b)² = a² + b² – 2ab

a² – b² = (a + b) (a – b)]

= $\frac{1-1+2i-1+1+2i}{1+1}$ [Since, i² = –1]

= $\frac{4i}{2}$

= 2i

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