# Important Questions Class 11 Maths Chapter 5

## Important Questions Class 11 Mathematics Chapter 5

### Important Questions for CBSE Class 11 Mathematics Chapter 5 – Complex Numbers and Quadratic Equations

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### Study Important Questions for Class 11 Mathematics Chapter 5 – Complex Numbers and Quadratic Equations

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Very Short Answer Questions: (1 Mark)

Q.1 Evaluate the value of i−39.

Ans: Let us solve the given expression further –

i−39=i−38−1

⇒i−39=(i2)−19.1i

⇒i−39=(−1)−19.1i

⇒i−39=−11ix ii

Therefore,

i−39=i

Q.2 Evaluate the expression (1+i)4

Ans: Let us simplify the given expression –

(1+i)4=[(1+i)2]2

⇒(1+i2+i2)2

⇒(1+i2−1)2

⇒4i2

⇒−4

Hence, (1+i)4=−4+0i

Q.3 Find the conjugate of −3i−5.

Ans: Let us consider a complex number

z=−3i−5

Therefore, the conjugate will be

z¯=3i−5

Q.4 Express in the form of a+ib:(3i−7)+(7−4i)−(6+3i)+i23.

Ans: Let us simplify the given expression –

(3i−7)+(7−4i)−(6+3i)+i23=3i−7+7−4i−6−3i+i23

⇒−4i−6+i22+1

⇒−4i−6+(i2)11.i

⇒−4i−6+(−1)11.i

⇒−4i−6−i

Therefore, (3i−7)+(7−4i)−(6+3i)+i23=−6−5i.

Q.5 Solve for x and y, 3x+(2x−y)i=6−3i.

Ans: We will equate the real part of the right-hand side with the real part of the left-hand side. Similarly, we will equate their imaginary parts as well.

Therefore,

3x=6 and 2x−y=−3

⇒x=2 and ⇒2(2)−y=−3

⇒4−y=−3

⇒y=7

Hence, x=2 and y=7

Q.6 Find the value of 1+i2+i4+i6+i8+……+i20.

Ans: Let us simplify the given expression –

1+i2+i4+i6+i8+……+i20=1−1+(i2)2+(i2)3+(i2)4+……+(i2)10

=(−1)2+(−1)3+(−1)4+……+(−1)10

=1−1+1−1+……+1

=1

Therefore, 1+i2+i4+i6+i8+……+i20=1

Q.7 Multiply 3−2i by its conjugate.

Ans: Let there be a complex number z=3−2i

Hence, its conjugate will be z¯=3+2i

Therefore, the product of the complex number with its conjugate will be –

zz¯=(3−2i)(3+2i)

⇒9+6i−6i−4i2

⇒9+4

Hence, (3−2i)(3+2i)=13

Q.8 Write the real and imaginary parts 1−2i2.

Ans: Let us simplify the given expression –

1−2i2=1−2(−1)

=1+2

=3

Hence, the real and imaginary parts of 1−2i2 are 3 and 0.

Q.1 For what real value of x and y are numbers equal (1+i)y2+(6+i) and (2+i)x

Ans: Let us equate both the numbers as –

(1+i)y2+(6+i)=(2+i)x

⇒(y2+6)+i(y2+1)=2x+ix

⇒y2+6=2x

and

y2+1=x

After solving both the equations obtained, we obtain–

x=5 and y=±2

Q.2 Find the real numbers x and y if (x−iy)(3+5i) is the conjugate of −6−24i.

Ans: We know that the conjugate of the given complex number z=−6−24i will be z¯=−6+24i.

Now, let us simplify the expression (x−iy)(3+5i)-

Hence, (x−iy)(3+5i)=3x+5xi−3yi+5y

⇒(3x+5y)+i(5x−3y)

Now, we will compare the values of the expression with the conjugate of the complex number.

Therefore,

−6+24i=(3x+5y)+i(5x−3y)

Thus, we have –

−6=3x+5y and 24=5x−3y.

Now, we will solve both equations for x and y.

Hence, we have 51=−17y

⇒y=−3

After substituting the value in the first equation we get –

⇒x=3.

Therefore, the real numbers x and y are 3 and −3.

Q.1 If z=x+iy and w= 1-izz-i. Show that | w|=1⇒z is purely real.

Ans: Given we have z=x+iy

Hence, we have

w= 1-i (x+iy)x+iy-i

w = 1+y−ixx+i(y−1)

As, |w|=1

Hence,

1+y−ixx+i(y−1) = 1

|1+y−ix||x+i(y−1)| =1

(1+y)2 + x2x2 + (y-1)2 =1

⇒(1+y)2+x2=x2+(y−1)2

⇒1+y2+2y=y2+1−2y

⇒4y=0

Hence, z=x+0i implying that z is purely real.

Q.2  Find two numbers such that their sum is 6

and the product is 14

Ans:

Let us consider x and y as the two numbers.

Hence, we have –

x+y=6

and

xy=14.

Therefore,

x(6−x)=14

⇒6x−x2=14

⇒x2−6x+14=0

x= 6 ± 36-562

x= 6 ± 20i2

x= 3 ± 5i

Hence y= 6- (3 ± 5i)

y = 3 5i

### Class 11 Mathematics Chapter 5 Important Question

Extramarks provide Chapter 5 Class 11 Mathematics Important Questions. Complex numbers and Quadratic Equations in Class 11 explains how to solve sums using complex numbers.

Topics Covered in the Chapter Complex Numbers and Quadratic Equations

Before going through Mathematics Class 11 Chapter 5 Important Questions, students can review the topics covered in this chapter given below.

• The real number and an imaginary number
• Complex number definition
• Integral powers of iota
• Purely real and purely imaginary complex numbers.
• Complex number equality
• Algebra in complex numbers – addition, subtraction, multiplication, and division
• Conjugate of a complex number.
• Modulus in a complex number
• Argand plane
• Polar form

Before understanding the concept of the complex number, one must first understand the meaning of the real number and the imaginary number.

Important Definitions

Real Number: A number on the number line that is in the form of a positive, negative, rational, irrational, zero, fraction, integer, etc., is called a real number.

For example: 34, -3, 7, 22/67, 0

Imaginary Number: The numbers except the real numbers is called imaginary number. It is the root of a negative number.

For example: -67, -23/92 etc

Complex Number: A complex number is defined as a number that can be expressed in the form of a + ib.

Here, a and b are real numbers and i is iota which will be discussed.

The value of iota is R-1.

Therefore, z (complex number) = a + ib where a is the real part, and ib is the imaginary part.

a = Re (z)

b= Im (z)

The Integral Powers of Lota

i=-1

i² = -1

i³ = i².i = (-1). i = -i

i⁴ = i².i² = (-1) (-1) = 1

Now we can generalise it, therefore;

i4n+1  = i

i4n+2= = -1

i4n+3 = -i

i4n+4= 1

It can also be generalised in this form,

If n= even integer then, in = (-1)n/2

And, if n= odd integer then, in  = (-1)(n−1)/2

What is a Purely Real Complex Number and a Purely Imaginary Complex Number?

When the imaginary part of a complex number is zero, it is referred to as a purely real complex number.

Im(z) = 0

When the real part of a complex number is zero, it is referred to as a purely imaginary complex number.

Re(z) = 0

Complex Number Equality

Two complex numbers are taken, z1 and z2

z1 = a1+ ib1

z2 = a2 + ib2

If z1 = z2

i.e. a1 + ib1 = a2 + ib2

then, a1= a2 and b1 = b2

Therefore, Re(z1) = Re(z2)

And Im(z1) = Im(z2)

Algebra in Complex Number

Let, z1 = a1 + ib1 and, z2 = a2+ ib2

Adding both the complex numbers we get,

z1 + z2 = (a1 + ib1) + (a2 + ib2)

=  (a1+ a2) + i(b1 + b2)

Properties:

1. Commutative: z1+ z2= z2+ z1
2. Associative: (z1+ z) + z3= z1+ (z2+ z3)
3. Additive identity: 0+z = z+0 = z

Solved Examples of Complex Numbers

Q1. Express (5 – 3i)³ in the form a + ib.

Ans:  We have, (5 – 3i)³

= 5³ – 3 × 5²× (3i) + 3 × 5 (3i)² – (3i)³

= 125 – 225i – 135 + 27i

= – 10 – 198i

Q2. Simplify

1. a) 16i + 10i(3-i)
2. b) (7i)(5i)
3. c) 11i + 13i – 2i

Ans:

1. a) 16i + 10i(3-i)

= 16i + 10i(3) + 10i (-i)

= 16i +30i – 10 i2

= 46 i – 10 (-1)

= 46i + 10

1. b) (7i)(5i) = 35  i2 = 35 (-1) = -35
1. c) 11i + 13i – 2i = 22i

Conjugate of a Complex Number

It is denoted by z¯

z¯ = a – ib

Modulus of a Complex Number

If z= a+ib

Then, ।z। = a2 + b2

Argand Plane and Polar Plane

A plane just like the XY plane where the complex number a+ib has the coordinates, a and b is called the Argand plane. It is also known as the Gaussian plane.

The argument of a complex number, z is shown by arg(z)= θ = tan-1(a/b)

arg(z) can also be written as amp(z).

z is 2nπ + θ is the general value of arg(z) and the length of OP = a2+ b2

The principal values of the argument lie in the interval (- π, π].

(i) Given x> 0 and y > 0 then, arg (z) = 0

(ii) Given x < 0 and y> 0 then, arg (z) = π -0

(iii) Given x < 0 and y < 0 then, arg (z) = – (π – θ)

(iv) Given x> 0 and y < 0 then, arg (z) = -θz).

Polar Form

If z = a + ib is a complex number, then z in polar form can be written as,

z = |z| (cos θ + i sin θ) where, θ = arg (z)

If the general value of the argument is 0, then the polar form of z is

z = |z|cos(2nπ+θ)+isin(2nπ+θ), where n is an integer.

Q.1 If the complex numbers z1, z2 and z3 represent the vertices of an equilateral triangle and |z1| = |z2| = |z3|, then

Marks:1

Ans

z1+z2+z3 = 0.

Q.2

Marks:1

Ans

Q.3
If z1=2+i,z2=23i and z3=4+5i, evaluate (i)?Re?(z1?z2z3) (ii)?Im?(z1?z2z3)
Marks:4

Ans
z1‹…z2 z3=(2+i) (23i ) 4+5i =(2+i) (2+3i) (45i)16+25 =4441+2741iRez1‹…z2 z3 =4441iImz1‹…z2 z3 =2741
Q.4
Find the modulus and the argument of the following: (i)?(2?i)2 (ii)1+i1?i
Marks:4

Ans
i2i2=414i=34i4i|=9+16=5modulus = 5Since (3,4i) lies in third quadrant, therefore its argument is given by¸

=+tan143ii1+i1i =1+i1i—1+i1+i=11+2i1+1 =i=0+0+i =1=1„modulus =1and argument  is given by=tan110
Q.5
Convert the complex number ?3+i in the polar form and represent it in Argand Plane.
Marks:6

Ans
Let z=3+i=r(cos¸+isin¸)Comparing real part and imaginary part, we getrcos¸=3 and rsin¸=1On squaring and adding, we getr2cos2¸+sin2¸=32+12r2=4r=2cos¸=32 and sin¸=12The value of ¸ satistying both the equation is by, ¸=56Hence Z=2cos56+isin56