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For any two complex numbersz1 and z2, prove that
Re (z1z2)= Re z1 Re z2 – Im z1 Im z2
Solve the equation
The given quadratic equation is
This equation can also be written as 
On comparing this equation with ax2 + bx + c = 0, we obtain
a = 2, b = –4, and c = 3
Therefore, the discriminant of the given equation is
D = b2 – 4ac = (–4)2 – 4 × 2 × 3 = 16 – 24 = –8
Therefore, the required solutions are
Solve the equation 27x2 – 10x + 1 = 0
The given quadratic equation is 27x2 – 10x + 1 = 0
On comparing the given equation with ax2 + bx + c = 0, we obtain
a = 27, b = –10, and c = 1
Therefore, the discriminant of the given equation is
D = b2 – 4ac = (–10)2 – 4 × 27 × 1 = 100 – 108 = –8
Therefore, the required solutions are
Solve the equation 21x2 – 28x + 10 = 0
The given quadratic equation is 21x2 – 28x + 10 = 0
On comparing the given equation with ax2 + bx + c = 0, we obtain
a = 21, b = –28, and c = 10
Therefore, the discriminant of the given equation is
D = b2 – 4ac = (–28)2 – 4 × 21 × 10 = 784 – 840 = –56
Therefore, the required solutions are
If a + ib =
, prove that a2 + b2 = 
a+ib=.......................(1)
taking conjugate on both sides ,
a-ib= ........................(2)
multiplying (1) and (2)
(a+ib)(a-ib)=*
a²+b²=[(x+i)(x-i)]²/[2x²+1]²
=
Let 
. Find
(i) 
, (ii) 
(i) 
On multiplying numerator and denominator by (2 – i), we obtain
On comparing real parts, we obtain
(ii) 
On comparing imaginary parts, we obtain
Find the modulus and argument of the complex number
.
Let
, then
On squaring and adding, we obtain
Therefore, the modulus and argument of the given complex number are 
respectively.
Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of –6 – 24i.
Let 
It is given that, 
Equating real and imaginary parts, we obtain
Multiplying equation (i) by 3 and equation (ii) by 5 and then adding them, we obtain
Putting the value of x in equation (i), we obtain
Thus, the values of x and y are 3 and –3 respectively.
If (x + iy)3 = u + iv, then show that
.
On equating real and imaginary parts, we obtain
Hence, proved.
If α and β are different complex numbers with
= 1, then find
.
Let α = a + ib and β = x + iy
It is given that, 
Find the number of non-zero integral solutions of the equation
.
Thus, 0 is the only integral solution of the given equation. Therefore, the number of non-zero integral solutions of the given equation is 0.
If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that
(a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = A2 + B2.
On squaring both sides, we obtain
(a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = A2 + B2
Hence, proved.
If x – iy =
prove that
.
x-iy=......................(1)
taking conjugate on both sides,
......................(2)
multiply equations (1) and (2)
(x+iy)(x-iy)=*
squaring on both sides
(x²+y²)²==
hence proved
Solve the equation
.
The given quadratic equation is
This equation can also be written as 
On comparing this equation with ax2 + bx + c = 0, we obtain
a = 9, b = –12, and c = 20
Therefore, the discriminant of the given equation is
D = b2 – 4ac = (–12)2 – 4 × 9 × 20 = 144 – 720 = –576
Therefore, the required solutions are
Convert the following in the polar form:
(i) 
, (ii) 
(i) Here, 
Let r cos θ = –1 and r sin θ = 1
On squaring and adding, we obtain
r2 (cos2θ + sin2θ) = 1 + 1
⇒ r2 (cos2θ + sin2θ) = 2
⇒ r2 = 2 [cos2θ + sin2θ = 1]
∴z = r cos θ + i r sin θ
This is the required polar form.
(ii) Here, 
Let r cos θ = –1 and r sin θ = 1
On squaring and adding, we obtain
r2 (cos2θ + sin2θ) = 1 + 1
⇒r2 (cos2θ + sin2θ) = 2
⇒ r2 = 2 [cos2θ + sin2θ = 1]
∴z = r cos θ + i r sin θ
This is the required polar form.
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to the standard form. 
find 
. 
. 
, then find the least positive integral value of m.
