Ncert Solutions Maths class 11th

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.

27. Kindly go through the solution

26. Kindly go through the solution

25. Kindly go through the solution

24. Kindly go through the solution

23. x² – x + 2 = 0

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

We have, a = 1, b = –1 andc = 2

Hence, discriminant of the equation is

b² – 4ac = (-1)² – 4 * 1 * 2 = 1 – 8 = –7

Therefore, the solution of the quadratic equation is

21. –x² + x – 2 = 0

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

We have, a = –1, b = 1 and c = –2

Hence, discriminant of the equation is

b² – 4ac = 1² – 4 * (-1)* (-2) = 1 – 8 = –7

Therefore, the solution of the quadratic equation is

20. x² + 3x + 9 = 0

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

We have, a = 1, b = 3 and c = 9

Hence, discriminant of the equation is

b² – 4ac = 3² – 4 * 1* 9 = 9 – 36 = –27

Therefore, the solution of the quadratic equation is

18. x² + 3 = 0

=>x² = –3

=>x = $\pm \surd 3$

=>x= ± √3$i$ [since, √-1 = i]

=>x = 0 ± √3$i$