Important Questions: Complex Numbers & Quadratic Equations | Mathematics (Maths) Class 11 - Commerce PDF Download

Q1: Express the given complex number (-3) in the polar form.
Ans: Given, complex number is -3.
Let r cos θ = -3 …(1)
and r sin θ = 0 …(2)
Squaring and adding (1) and (2), we get
r²cos²θ + r²sin²θ = (-3)²
Take r² outside from L.H.S, we get
r²(cos²θ + sin²θ) = 9
We know that, cos²θ + sin²θ = 1, then the above equation becomes,
r² = 9
r = 3 (Conventionally, r > 0)
Now, subsbtitute the value of r in (1) and (2)
3 cos θ = -3 and 3 sin θ = 0
cos θ = -1 and sin θ = 0
Therefore, θ = π
Hence, the polar representation is,
-3 = r cos θ + i r sin θ
3 cos π + 3 sin π = 3(cos π + i sin π)
Thus, the required polar form is 3 cos π+ 3i sin π = 3(cos π+i sin π)

Q2: For any two complex numbers z₁ and z₂, show that Re(z₁z₂) = Rez₁ Rez₂– Imz₁Imz₂
Ans: Given: z₁ and z₂ are the two complex numbers
To prove: Re(z₁z₂) = Rez₁ Rez₂– Imz₁Imz₂
Let z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂
Now, z₁ z₂ =(x₁ + iy₁)(x₂ + iy₂)
Now, split the real part and the imaginary part from the above equation:
⇒ x₁(x₂ + iy₂) + iy1(x₂ + iy₂)
Now, multiply the terms:
= x₁x₂ + ix₁y₂ + ix₂y₁ + i₂y₁y₂
We know that, i₂ = -1, then we get
= x₁x₂ + ix₁y₂ + ix₂y₁ - y₁y₂
Now, again seperate the real and the imaginary part:
= (x₁x₂ - y₁y₂) + i (x₁y₂ + x₂y₁)
From the above equation, take only the real part:
⇒ Re (z₁z₂) = (x₁x₂ - y₁y₂)
It means that,
⇒ Re(z₁z₂) = Rez₁ Rez₂– Imz₁Imz₂
Hence, the given statement is proved.

Q3: Write the given complex number (1 – i) – ( –1 + i6) in the form a + ib
Ans: Given Complex number: (1 – i) – ( –1 + i6)
Multiply (-) by the term inside the second bracket ( –1 + i6)
= 1 – i +1 – i6
= 2 – 7i, which is of the form a + ib.

Q4: Solve the given quadratic equation 2x² + x + 1 = 0.
Ans: Given quadratic equation: 2x² + x + 1 = 0
Now, compare the given quadratic equation with the general form ax² + bx + c = 0
On comparing, we get
a = 2, b = 1 and c = 1
Therefore, the discriminant of the equation is:
D = b²– 4ac
Now, substitute the values in the above formula
D = (1)² – 4(2)(1)
D = 1- 8
D = -7
Therefore, the required solution for the given quadratic equation is
x =[-b ± √D]/2a
x = [-1 ± √-7]/2(2)
We know that, √-1 = i
x = [-1 ± √7i] / 4
Hence, the solution for the given quadratic equation is (-1 ± √7i) / 4.

Q5: Find the modulus of [(1 + i)/(1 - i)] – [(1 - i)/(1 + i)]
Ans: Given: [(1 + i)/(1 - i)] – [(1 - i)/(1 + i)]
Simplify the given expression, we get:
[(1+i)/(1-i)] – [(1-i)/(1+i)] = [(1+i)²– (1-i)²]/ [(1+i)(1-i)]
= (1 + i² + 2i - 1- i² + 2i)) / (1² + 1²)
Now, cancel out the terms,
= 4i/2
= 2i
Now, take the modulus,
| [(1 + i)/(1 - i)] – [(1- i)/(1 + i)]| =|2i| = √2² = 2
Therefore, the modulus of [(1+i)/(1-i)] – [(1-i)/(1+i)] is 2.