Complex Numbers and Quadratic Equations

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]

= $\frac{614+1198i}{884}$

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

= $\frac{307+599i}{442}$

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

28. To proof, Re (z1z2) = Re z1 Re z2 – Imz1 Imz2

Let z­­₁ = x₁ + iy₁ and z₂ = x₂ + iy₂ be two complex number.

Then, z₁.z₂ = (x₁ + iy₁) (x₂ + iy₂)

=x₁x₂ + ix₁y₂ + ix₂y₁ + i²y₁y₂

= x₁x₂ + ix₁y₂ + ix₂y₁ – y₁y₂            [since, i² = -1]

= (x₁x₂ – y₁y₂) + i (x₁y₂ + x₂y₁)

As, Re (z₁z₂) = (x₁) (x₂) – (y₁) (y₂)

Now, RHS = Re z₁ Re z₂ – Imz₁Imz₂ = x₁x₂ – y₁y₂

Therefore, Re (z₁z₂) = Rez₁Rez₂ – Imz₁Imz₂

Hence proved.