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Learn Miscellaneous Exercise 5 with Free Lessons & Tips

Evaluate:

Comments

For any two complex numbersz1 and z2, prove that

Re (z1z2)= Re z1 Re z2 – Im z1 Im z2

Comments

Reduce to the standard form.

Comments

If xiy =prove that.

Comments

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

Comments

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

Comments

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

Comments

If find .

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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

=

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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

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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.

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Find the real numbers x and y if (xiy) (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.

Comments

Find the modulus of .

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If (x + iy)3 = u + iv, then show that.

On equating real and imaginary parts, we obtain

Hence, proved.

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If α and β are different complex numbers with = 1, then find.

Let α = a + ib and β = x + iy

It is given that,

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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.

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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.

Comments

If xiy =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

Comments

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

Comments

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.

Comments

If, then find the least positive integral value of m.

Comments

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