# Chapter – 5(Miscellaneous Exercise)

## Complex Numbers and Quadratic Equations

Chapter 5 Miscellaneous ncert math solution class 11

Question 1: Evaluate:

Solution : Given

Question 2: For any two complex numbers and , prove that:

Solution: Let and

Therefore,

Now

Question 3: Reduce to the standard form.

Solution :Given that

Question 4: If prove that

Solution : Given that

Squaring both side

Taking conjugate of eq(i) using formula

Multiply (i) and (ii)

Question 5: Convert into polar form:

(i)

(ii)

Solution : (i)

Let

Comparing the real and imaginary parts, we have

Squaring and adding equation (1) and (2), we have

Therefore, modulus

Now, dividing equation (2) by (1), we have

From the equations (1), (2) and (3), it is clear that and are negative but is positive. So, lies in II quadrant. Therefore,

Therefore, the polar form of is given by

(ii)

Let

Comparing the real and imaginary parts, we have

Squaring and adding equation (1) and (2), we have

Therefore, modulus

Now, dividing equation (2) by (1), we have

From the equations (1), (2) and (3), it is clear that and are negative but is positive. So, lies in Il quadrant. Therefore,

Therefore, the polar form of is given by

Solve each of the equation in Exercises 6 to 9.

Question 6:

Solution: Given that

.

On comparing the given equation with , we obtain , and

Therefore, the discriminant of the given equation is given by

Therefore, the required solutions are

Question 7: :

Solution : Given that

.

On comparing the given equation with , we obtain , and

Therefore, the discriminant of the given equation is given by

Therefore, the required solutions are

Question 8:

Solution : Given that

The given quadratic equation is .

On comparing the given equation with , we obtain , and

Therefore, the discriminant of the given equation is given by

Therefore, the required solutions are

Question 9:

Solution : The given quadratic equation is .

On comparing the given equation with , we obtain , and

Therefore, the discriminant of the given equation is given by

Therefore, the required solutions are

Question 10: If , find .

Solution : Since,

Therefore,

Question 11: If , prove that

Solution: Here,

and

Now,

Question 12: Let . Find

(i)

(ii)

Solution : (i)

Therfore,

(ii)

Therefore,

Question 13: Find the modulus and argument of the complex number .

Solution :

Let

Comparing the real and imaginary parts, we have

Squaring and adding equation (1) and (2), we have

Therefore, modulus

Now, dividing equation (2) by (1), we have

From the equations (1), (2) and (3), it is clear that and are negative but is positive. So, lies in II quadrant. Therefore,

Argument

Question 14: Find the real numbers and if is the conjugate of .

Solution : Given that: is the conjugate of

Comparing the real and imaginary parts, we have

and

Form the equation (1), we have

Putting the value of in equation (2), we get

Putting the value of in equation (1), we have

Question 15: Find the modulus of

Solution : Given that

Therefore,

Question 16: If , then show that .

Solution : Given that:

Comparing the real and imaginary parts, we have

and

and

Question 17: If and are different complex number with , then find .

Solution : We know that

Question 18: Find the number of non-zero integral solutions of the equation .

Solution : Given that:

Thus, 0 is the only integral solution of the given equation.

Therefore, the number of nonzero integral solutions of the given equation is 0 .

Question 19: If , then show that .

Solution : Given that:

Squaring both sides, we have

Question 20: If , then find the least integral value of .

Solution : Given that:

Therefore, the least integral value of .

Exercise 5.1 complex no. ncert math solution class 11

Exercise 5.2 complex no. ncert math solution class 11

Exercise 5.3 complex no. ncert math solution class 11

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