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NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations

Edited By Ramraj Saini | Updated on Sep 21, 2023 11:22 PM IST

Complex Numbers and Quadratic Equations Class 11 Questions And Answers

NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations are discussed here. These NCERT solutions are created by expert team at Careers360 keeping in mind of latest syllabus of CBSE 2023-24. in the earlier classes you have studied the quadratic equations. You must have come across some equations like x2 + 2 = 0, x2 = -2, for which there is no real solution. How to solve these quadratic equations? In this class 11 maths chapter 5 question answer, you will learn to solve equations like x2 + 2 = 0.

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This Story also Contains
  1. Complex Numbers and Quadratic Equations Class 11 Questions And Answers
  2. Complex Numbers and Quadratic Equations Class 11 Questions And Answers PDF Free Download
  3. Complex Numbers and Quadratic Equations Class 11 Solutions - Important Formulae
  4. Complex Numbers and Quadratic Equations Class 11 NCERT Solutions (Intext Questions and Exercise)
  5. Highlights Of NCERT Class 11 Chapter 5 Complex Numbers And Quadratic Equations
  6. Complex Numbers And Quadratic Equations Exercise Wise Solutions
  7. NCERT Solutions For Class 11 Mathematics - Chapter Wise
  8. Benefits of NCERT Class 11 Maths ch 5 Question Answer
  9. NCERT Solutions For Class 11 - Subject wise
  10. NCERT Books and NCERT Syllabus
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations
NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations

The CBSE Syllabus for 2023 includes the chapter Complex Numbers and Quadratic Equations, which comprises significant mathematical theorems and formulae. The NCERT textbook provides ample practice problems to cover all these concepts, facilitating students' comprehension of advanced concepts in the future. To aid in this process, Careers360 offers detailed solutions for all complex numbers class 11 problems in the textbook. These complex numbers and quadratic equations class 11 solutions are particularly beneficial for students who aim to pass their exams with last-minute preparations. However, the primary focus of NCERT solutions for class 11 is to master concepts and develop a deeper understanding of the subject.

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Complex Numbers and Quadratic Equations Class 11 Questions And Answers PDF Free Download

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Complex Numbers and Quadratic Equations Class 11 Solutions - Important Formulae

Complex Numbers:

A complex number is expressed as a + bi, where 'a' and 'b' are real numbers, and 'i' is the imaginary unit.

Imaginary Numbers: The square root of a negative real number is called an imaginary number, represented as √-1 = i.

Equality of Complex Number: Two complex numbers z1 = x1 + iy1 and z2 = x2 + iy2 are equal if x1 = x2 and y1 = y2.

Algebra of Complex Numbers:

Addition: (z1 + z2) = (x1 + x2) + i(y1 + y2)

Subtraction: (z1 - z2) = (x1 - x2) + i(y1 - y2)

Multiplication: (z1 * z2) = (x1x2 - y1y2) + i(x1y2 + x2y1)

Division: (z1 / z2) = [(x1x2 + y1y2) + i(x2y1 - x1y2)] / (x22 + y22), where z2 ≠ 0.

Conjugate of Complex Number: The conjugate of a complex number z = x + iy is represented as z¯ = x - iy.

Modulus of a Complex Number: |z| = √(x2 + y2)

Argument of a Complex Number: The angle made by the line joining the point z to the origin, with the positive X-axis in an anti-clockwise sense is called the argument (arg) of the complex number.

Principal Value of Argument:

  • When x > 0 and y > 0 ⇒ arg(z) = θ

  • When x < 0 and y > 0 ⇒ arg(z) = π - θ

  • When x < 0 and y < 0 ⇒ arg(z) = -(π - θ)

  • When x > 0 and y < 0 ⇒ arg(z) = -θ

Polar Form of a Complex Number: z = |z| (cosθ + isinθ), where θ = arg(z).

The general polar form of z is z = |z| [cos(2nπ + θ) + isin(2nπ + θ)], where n is an integer.

Free download NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations for CBSE Exam.

Complex Numbers and Quadratic Equations Class 11 NCERT Solutions (Intext Questions and Exercise)

Class 11 maths chapter 5 NCERT solutions - Exercise: 5.1


Question:1 Express each of the complex number in the form a+ib .

(5i)(35i)

Answer:

On solving

(5i)(35i)
we will get

(5i)(35i)=5×(35)×i×i
=3×i2 (i2=1)
=3×1
=3

Now, in the form of a+ib we can write it as
=3+0i

Question:2 Express each of the complex number in the form a+ib .

Double exponent: use braces to clarify

Answer:

We know that i4=1
Now, we will reduce i9+i19 into

Double exponent: use braces to clarify =(i4)2.i+(i4)3.i3
=(1)2.i+(1)3.(i) (i4=1,i3=i and i2=1)
=ii=0
Now, in the form of a+ib we can write it as
o+io
Therefore, the answer is o+io

Question:3 Express each of the complex number in the form a+ib.

i39

Answer:

We know that i4=1
Now, we will reduce i39 into

i39 =(i4)9.i3
=(1)9.(i)1 (i4=1,i3=i)
=1i
=1i×ii
=ii2 (i2=1)
=i(1)
=i
Now, in the form of a+ib we can write it as
o+i1
Therefore, the answer is o+i1

Question:4 Express each of the complex number in the form a+ib.

3(7+7i)+i(7+7i)

Answer:

Given problem is
3(7+7i)+i(7+7i)
Now, we will reduce it into

3(7+7i)+i(7+7i) =21+21i+7i+7i2
=21+21i+7i+7(1) (i2=1)
=21+21i+7i7
=14+28i

Therefore, the answer is 14+i28

Question:5 Express each of the complex number in the form a+ib .

(1i)(1+6i)

Answer:

Given problem is
(1i)(1+6i)
Now, we will reduce it into

(1i)(1+6i)=1i+16i
=27i

Therefore, the answer is 27i

Question:6 Express each of the complex number in the form a+ib .

(15+i25)(4+i52)

Answer:

Given problem is
(15+i25)(4+i52)
Now, we will reduce it into

(15+i25)(4+i52)=15+i254i52
=1205+i(425)10
=195i2110

Therefore, the answer is 195i2110

Question:7 Express each of the complex number in the form a+ib .

[(13+i73)+(4+i13)](43+i)

Answer:

Given problem is
[(13+i73)+(4+i13)](43+i)
Now, we will reduce it into

[(13+i73)+(4+i13)](43+i)=13+i73+4+i13+43i
=1+4+123+i(7+13)3
=173+i53

Therefore, the answer is 173+i53

Question:8 Express each of the complex number in the form a+ib .

(1i)4

Answer:

The given problem is
(1i)4
Now, we will reduce it into

(1i)4=((1i)2)2
=(12+i22.1.i)2 (using (ab)2=a2+b22ab)

=(112i)2 (i2=1)
=(2i)2
=4i2
=4

Therefore, the answer is 4+i0

Question:9 Express each of the complex number in the form a+ib .

(13+3i)3

Answer:

Given problem is
(13+3i)3
Now, we will reduce it into

(13+3i)3=(13)3+(3i)3+3.(13)2.3i+3.13.(3i)2 (using (a+b)3=a3+b3+3a2b+3ab2)
=127+27i3+i+9i2

=127+27(i)+i+9(1) (i3=i and i2=1)
=12727i+i9
=12432726i
=2422726i

Therefore, the answer is

2422726i

Question:10 Express each of the complex number in the form a+ib .

(213i)3

Answer:

Given problem is
(213i)3
Now, we will reduce it into

(213i)3=((2)3+(13i)3+3.(2)213i+3.(13i)2.2) (using (a+b)3=a3+b3+3a2b+3ab2)
=(8+127i3+3.4.13i+3.19i2.2)

=(8+127(i)+4i+23(1)) (i3=i and i2=1)
=(8127i+4i23)
=((1+108)27i+2423)
=223i10727

Therefore, the answer is 223i10727

Question:11 Find the multiplicative inverse of each of the complex numbers.

43i

Answer:

Let z=43i
Then,
z¯=4+3i
And
|z|2=42+(3)2=16+9=25
Now, the multiplicative inverse is given by
z1=z¯|z|2=4+3i25=425+i325

Therefore, the multiplicative inverse is

425+i325

Question:12 Find the multiplicative inverse of each of the complex numbers.

5+3i

Answer:

Let z=5+3i
Then,
z¯=53i
And
|z|2=(5)2+(3)2=5+9=14
Now, the multiplicative inverse is given by
z1=z¯|z|2=53i14=514i314

Therefore, the multiplicative inverse is 514i314

Question:13 Find the multiplicative inverse of each of the complex numbers.

i

Answer:

Let z=i
Then,
z¯=i
And
|z|2=(0)2+(1)2=0+1=1
Now, the multiplicative inverse is given by
z1=z¯|z|2=i1=0+i

Therefore, the multiplicative inverse is 0+i1

Question:14 Express the following expression in the form of a+ib:

(3+i5)(3i5)(3+2i)(3i2)

Answer:

Given problem is
(3+i5)(3i5)(3+2i)(3i2)
Now, we will reduce it into

(3+i5)(3i5)(3+2i)(3i2)=32(5i)2(3+2i)(3i2) (using (ab)(a+b)=a2b2)
=95i23+2i3+2i
=95(1)22i (i2=1)
=1422i×2i2i
=72i2i2
=72i2
Therefore, answer is 0i722


Class 11 maths chapter 5 ncert solutions - Exercise: 5.2

Question:1 Find the modulus and the arguments of each of the complex numbers.

z=1i3

Answer:

Given the problem is
z=1i3
Now, let
rcosθ=1   and   rsinθ=3
Square and add both the sides
r2(cos2θ+sin2θ)=(1)2+(3)2 (cos2θ+sin2θ=1)
r2=1+3
r2=4
r=2 (r>0)
Therefore, the modulus is 2
Now,
2cosθ=1   and   2sinθ=3
cosθ=12   and   sinθ=32
Since, both the values of cosθ and sinθ is negative and we know that they are negative in III quadrant
Therefore,
Argument = (ππ3)=2π3
Therefore, the argument is

2π3

Question:2 Find the modulus and the arguments of each of the complex numbers.

z=3+i

Answer:

Given the problem is
z=3+i
Now, let
rcosθ=3   and   rsinθ=1
Square and add both the sides
r2(cos2θ+sin2θ)=(3)2+(1)2 (cos2θ+sin2θ=1)
r2=1+3
r2=4
r=2 (r>0)
Therefore, the modulus is 2
Now,
2cosθ=3   and   2sinθ=1
cosθ=32   and   sinθ=12
Since values of cosθ is negative and value sinθ is positive and we know that this is the case in II quadrant
Therefore,
Argument = (ππ6)=5π6
Therefore, the argument is

5π6

Question:3 Convert each of the complex numbers in the polar form:

1i

Answer:

Given problem is
z=1i
Now, let
rcosθ=1   and   rsinθ=1
Square and add both the sides
r2(cos2θ+sin2θ)=(1)2+(1)2 (cos2θ+sin2θ=1)
r2=1+1
r2=2
r=2 (r>0)
Therefore, the modulus is 2
Now,
2cosθ=1   and   2sinθ=1
cosθ=12   and   sinθ=12
Since values of sinθ is negative and value cosθ is positive and we know that this is the case in the IV quadrant
Therefore,
θ=π4            (lies in IV quadrant)
Therefore,
1i=rcosθ+irsinθ
=2cos(π4)+i2sin(π4)
=2(cos(π4)+isin(π4))

Therefore, the required polar form is 2(cos(π4)+isin(π4))

Question:4 Convert each of the complex numbers in the polar form:

1+i

Answer:

Given the problem is
z=1+i
Now, let
rcosθ=1   and   rsinθ=1
Square and add both the sides
r2(cos2θ+sin2θ)=(1)2+(1)2 (cos2θ+sin2θ=1)
r2=1+1
r2=2
r=2 (r>0)
Therefore, the modulus is 2
Now,
2cosθ=1   and   2sinθ=1
cosθ=12   and   sinθ=12
Since values of cosθ is negative and value sinθ is positive and we know that this is the case in II quadrant
Therefore,
θ=(ππ4)=3π4            (lies in II quadrant)
Therefore,
1+i=rcosθ+irsinθ
=2cos(3π4)+i2sin(3π4)
=2(cos(3π4)+isin(3π4))

Therefore, the required polar form is 2(cos(3π4)+isin(3π4))

Question:5 Convert each of the complex numbers in the polar form:

1i

Answer:

Given problem is
z=1i
Now, let
rcosθ=1   and   rsinθ=1
Square and add both the sides
r2(cos2θ+sin2θ)=(1)2+(1)2 (cos2θ+sin2θ=1)
r2=1+1
r2=2
r=2 &nbsnbsp; (r>0)
Therefore, the modulus is 2
Now,
2cosθ=1   and   2sinθ=1
cosθ=12   and   sinθ=12
Since values of both cosθ and sinθ is negative and we know that this is the case in III quadrant
Therefore,
θ=(ππ4)=3π4            (lies in III quadrant)
Therefore,
1i=rcosθ+irsinθ
=2cos(3π4)+i2sin(3π4)
=2(cos(3π4)+isin(3π4))

Therefore, the required polar form is 2(cos(3π4)+isin(3π4))

Question:6 Convert each of the complex numbers in the polar form:

3

Answer:

Given problem is
z=3
Now, let
rcosθ=3   and   rsinθ=0
Square and add both the sides
r2(cos2θ+sin2θ)=(3)2+(0)2 (cos2θ+sin2θ=1)
r2=9+0
r2=9
r=3 (r>0)
Therefore, the modulus is 3
Now,
3cosθ=3   and   3sinθ=0
cosθ=1   and   sinθ=0
Since values of cosθ is negative and sinθ is Positive and we know that this is the case in II quadrant
Therefore,
θ=π            (lies in II quadrant)
Therefore,
3=rcosθ+irsinθ
=3cos(π)+i3sin(π)
=3(cosπ+isinπ)

Therefore, the required polar form is 3(cosπ+isinπ)

Question:7 Convert each of the complex numbers in the polar form:

3+i

Answer:

Given problem is
z=3+i
Now, let
rcosθ=3   and   rsinθ=1
Square and add both the sides
r2(cos2θ+sin2θ)=(3)2+(1)2 (cos2θ+sin2θ=1)
r2=3+1
r2=4
r=2 (r>0)
Therefore, the modulus is 2
Now,
2cosθ=3   and   2sinθ=1
cosθ=32   and   sinθ=12
Since values of Both cosθ and sinθ is Positive and we know that this is the case in I quadrant
Therefore,
θ=π6            (lies in I quadrant)
Therefore,
3+i=rcosθ+irsinθ
=2cos(π6)+i2sin(π6)
=2(cosπ6+isinπ6)

Therefore, the required polar form is 2(cosπ6+isinπ6)

Question:8 Convert each of the complex numbers in the polar form:

i

Answer:

Given problem is
z=i
Now, let
rcosθ=0   and   rsinθ=1
Square and add both the sides
r2(cos2θ+sin2θ)=(0)2+(1)2 (cos2θ+sin2θ=1)
r2=0+1
r2=1
r=1 (r>0)
Therefore, the modulus is 1
Now,
1cosθ=0   and   1sinθ=1
cosθ=0   and   sinθ=1
Since values of Both cosθ and sinθ is Positive and we know that this is the case in I quadrant
Therefore,
θ=π2            (lies in I quadrant)
Therefore,
i=rcosθ+irsinθ
=1cos(π2)+i1sin(π2)
=cosπ2+isinπ2

Therefore, the required polar form is cosπ2+isinπ2


Class 11 maths chapter 5 ncert solutions - Exercise: 5.3

Question:1 Solve each of the following equations: x2+3=0

Answer:

Given equation is
x2+3=0
Now, we know that the roots of the quadratic equation is given by the formula
b±b24ac2a
In this case value of a = 1 , b = 0 and c = 3
Therefore,
0±024.1.(3)2.1=±122=±23i2=±3i
Therefore, the solutions of requires equation are ±3i

Question:2 Solve each of the following equations: 2x2+x+1=0

Answer:

Given equation is
2x2+x+1=0
Now, we know that the roots of the quadratic equation are given by the formula
b±b24ac2a
In this case value of a = 2 , b = 1 and c = 1
Therefore,
1±124.2.12.2=1±184=1±74=1±7i4
Therefore, the solutions of requires equation are

1±7i4

Question:3 Solve each of the following equations: x2+3x+9=0

Answer:

Given equation is
x2+3x+9=0
Now, we know that the roots of the quadratic equation are given by the formula
b±b24ac2a
In this case value of a = 1 , b = 3 and c = 9
Therefore,
3±324.1.92.1=3±9362=3±272=3±33i2
Therefore, the solutions of requires equation are

3±33i2

Question:4 Solve each of the following equations: x2+x2=0

Answer:

Given equation is
x2+x2=0
Now, we know that the roots of the quadratic equation is given by the formula
b±b24ac2a
In this case value of a = -1 , b = 1 and c = -2
Therefore,
1±124.(1).(2)2.(1)=1±182=1±72=1±7i2
Therefore, the solutions of equation are

1±7i2

Question:5 Solve each of the following equations: x2+3x+5=0

Answer:

Given equation is
x2+3x+5=0
Now, we know that the roots of the quadratic equation are given by the formula
b±b24ac2a
In this case value of a = 1 , b = 3 and c = 5
Therefore,
3±324.1.52.1=3±9202=3±112=3±11i2
Therefore, the solutions of the equation are 3±11i2

Question:6 Solve each of the following equations: x2x+2=0

Answer:

Given equation is
x2x+2=0
Now, we know that the roots of the quadratic equation are given by the formula
b±b24ac2a
In this case value of a = 1 , b = -1 and c = 2
Therefore,
(1)±(1)24.1.22.1=1±182=1±72=1±7i2
Therefore, the solutions of equation are 1±7i2

Question:7 Solve each of the following equations: 2x2+x+2=0

Answer:

Given equation is
2x2+x+2=0
Now, we know that the roots of the quadratic equation is given by the formula
b±b24ac2a
In this case the value of a=2,b=1 and c=2
Therefore,
1±124.2.22.2=1±1822=1±722=1±7i22
Therefore, the solutions of the equation are 1±7i22

Question:8 Solve each of the following equations: 3x22x+33=0

Answer:

Given equation is
3x22x+33=0
Now, we know that the roots of the quadratic equation are given by the formula
b±b24ac2a
In this case the value of a=3,b=2 and c=33
Therefore,
(2)±(2)24.3.332.3=2±23623=2±3423 =2±34i23
Therefore, the solutions of the equation are 2±34i23

Question:9 Solve each of the following equations: x2+x+12=0

Answer:

Given equation is
x2+x+12=0
Now, we know that the roots of the quadratic equation is given by the formula
b±b24ac2a
In this case the value of a=1,b=1 and c=12
Therefore,
1±124.1.122.1=1±1222=1±(221)2 =1±(221)i2
Therefore, the solutions of the equation are

1±(221)i2

Question:10 Solve each of the following equations:

x2+x2+1=0

Answer:

Given equation is
x2+x2+1=0
Now, we know that the roots of the quadratic equation are given by the formula
b±b24ac2a
In this case the value of a=1,b=12 and c=1
Therefore,
12±(12)24.1.12.1=12±1242=12±722 =1±7i22
Therefore, the solutions of the equation are

1±7i22

Complex numbers and quadratic equations class 11 solutions - Miscellaneous Exercise

Question:1 Evaluate Double exponent: use braces to clarify .

Answer:

The given problem is
Double exponent: use braces to clarify
Now, we will reduce it into
Double exponent: use braces to clarify
=[14.(1)+116.i]3 (i4=1,i2=1)
=[1+1i]3
=[1+1i×ii]3
=[1+ii2]3
=[1+i1]3=[1i]3
Now,
(1+i)3=(13+i3+3.12.i+3.1.i2) (using (a+b)3=a3+b3+3.a2.b+3.a.b2)
=(1i+3i+3(1)) (i3=i,i2=1)
=(1i+3i3)=(2+2i)
=22i
Therefore, answer is 22i

Question:2 For any two complex numbers z1 and z2 , prove that Re(z1z2)=Rez1Rez2Imz1Imz2

Answer:

Let two complex numbers are
z1=x1+iy1
z2=x2+iy2
Now,
z1.z2=(x1+iy1).(x2+iy2)
=x1x2+ix1y2+iy1x2+i2y1y2
=x1x2+ix1y2+iy1x2y1y2 (i2=1)
=x1x2y1y2+i(x1y2+y1x2)
Re(z1z2)=x1x2y1y2
=Re(z1z2)Im(z1z2)

Hence proved

Question:3 Reduce (114i21+i)(34i5+i) to the standard form.

Answer:

Given problem is
(114i21+i)(34i5+i)
Now, we will reduce it into

(114i21+i)(34i5+i)=((1+i)2(14i)(1+i)(14i))(34i5+i)
=(1+i2+8i14i+i4i2)(34i5+i)
=(1+9i13i4(1))(34i5+i)
=(1+9i53i)(34i5+i)
=(3+4i+27i36i225+5i15i3i2)=(3+31i+362510i+3)=33+31i2810i=33+31i2(145i)

Now, multiply numerator an denominator by (14+5i)
33+31i2(145i)×14+5i14+5i
462+165i+434i+155i22(142(5i)2) (using (ab)(a+b)=a2b2)
462+599i1552(19625i2)
307+599i2(196+25)=307+599i2×221=307+599i442=307442+i599442

Therefore, answer is 307442+i599442

Question:4 If xiy=aibcid , prove that (x2+y2)2=a2+b2c2+d2.

Answer:

the given problem is

xiy=aibcid
Now, multiply the numerator and denominator by

c+id
xiy=aibcid×c+idc+id
=(ac+bd)+i(adbc)c2i2d2=(ac+bd)+i(adbc)c2+d2
Now, square both the sides
(xiy)2=((ac+bd)+i(adbc)c2+d2)2
=(ac+bd)+i(adbc)c2+d2
x2y22ixy=(ac+bd)+i(adbc)c2+d2
On comparing the real and imaginary part, we obtain

x2y2=ac+bdc2+d2  and  2xy=adbcc2+d2   (i)

Now,
(x2+y2)2=(x2y2)2+4x2y2
=(ac+bdc2+d2)2+(adbcc2+d2)2    (using (i))
=a2c2+b2d2+2acbd+a2d2+b2c22adbc(c2+d2)2
=a2c2+b2d2+a2d2+b2c2(c2+d2)2
=a2(c2+d2)+b2(c2+d2)(c2+d2)2
=(a2+b2)(c2+d2)(c2+d2)2
=(a2+b2)(c2+d2)

Hence proved

Question:5(i) Convert the following in the polar form:

1+7i(2i)2

Answer:

Let
z=1+7i(2i)2=1+7i4+i24i=1+7i414i=1+7i34i

Now, multiply the numerator and denominator by 3+4i
z=1+7i34i×3+4i3+4i=3+4i+21i+28i232+42=25+25i25=1+i
Now,
let
rcosθ=1  and  rsinθ=1
On squaring both and then add
r2(cos2θ+sin2θ)=(1)2+12
r2=2
r=2                    (r>0)
Now,
2cosθ=1  and  2sinθ=1
cosθ=12  and  sinθ=12
Since the value of cosθ is negative and sinθ is positive this is the case in II quadrant
Therefore,
θ=ππ4=3π4            (lies in II quadrant)
z=rcosθ+irsinθ
=2cos3π4+i2sin3π4
=2(cos3π4+isin3π4)
Therefore, the required polar form is

2(cos3π4+isin3π4)

Question:5(ii) Convert the following in the polar form:

1+3i12i

Answer:

Let
z=1+3i12i

Now, multiply the numerator and denominator by 1+2i
z=1+3i12i×1+2ii+2i=1+2i+3i61+4=5+5i5=1+i
Now,
let
rcosθ=1  and  rsinθ=1
On squaring both and then add
r2(cos2θ+sin2θ)=(1)2+12
r2=2
r=2                    (r>0)
Now,
2cosθ=1  and  2sinθ=1
cosθ=12  and  sinθ=12
Since the value of cosθ is negative and sinθ is positive this is the case in II quadrant
Therefore,
θ=ππ4=3π4            (lies in II quadrant)
z=rcosθ+irsinθ
=2cos3π4+i2sin3π4
=2(cos3π4+isin3π4)
Therefore, the required polar form is

2(cos3π4+isin3π4)

Question:6 Solve each of the equation: 3x24x+203=0

Answer:

Given equation is
3x24x+203=0
Now, we know that the roots of the quadratic equation are given by the formula
b±b24ac2a
In this case the value of

a=3,b=4 and c=203
Therefore,
(4)±(4)24.3.2032.3=4±16806=4±646 =4±8i6=23±i43
Therefore, the solutions of requires equation are

23±i43

Question:7 Solve each of the equation: x22x+32=0

Answer:

Given equation is
x22x+32=0
Now, we know that the roots of the quadratic equation are given by the formula
b±b24ac2a
In this case the value of a=1,b=2 and c=32
Therefore,
(2)±(2)24.1.322.1=2±462=2±22 =2±i22=1±i22
Therefore, the solutions of requires equation are

1±i22

Question:8 Solve each of the equation: 27x210x+1=0 .

Answer:

Given equation is
27x210x+1=0
Now, we know that the roots of the quadratic equation are given by the formula
b±b24ac2a
In this case the value of a=27,b=10 and c=1
Therefore,
(10)±(10)24.27.12.27=10±10010854=10±854 =10±i2254=527±i227
Therefore, the solutions of requires equation are 527±i227

Question:9 Solve each of the equation: 21x228x+10=0

Answer:

Given equation is
21x228x+10=0
Now, we know that the roots of the quadratic equation are given by the formula
b±b24ac2a
In this case the value of a=21,b=28 and c=10
Therefore,
(28)±(28)24.21.102.21=28±78484042=28±5642 =28±i21442=23±i1421
Therefore, the solutions of requires equation are

23±i1421

Question:10 If z1=2i,z2=1+i , find |z1+z2+1z1z2+1| .

Answer:

It is given that
z1=2i,z2=1+i
Then,
|z1+z2+1z1z2+1|=|2i+1+i+12i1i+1|=|42(1i)|=|2(1i)|
Now, multiply the numerator and denominator by 1+i
|2(1i)×1+i1+i|=|2(1+i)12i2|=|2(1+i)1+1|=|1+i|
Now,
|1+i|=12+12=1+1=2
Therefore, the value of

|z1+z2+1z1z2+1| is 2

Question:11 If a+ib=(x+i)22x2+1 , prove that a2+b2=(x2+1)2(2x2+1)2 .

Answer:

It is given that
a+ib=(x+i)22x2+1
Now, we will reduce it into

a+ib=(x+i)22x2+1=x2+i2+2xi2x2+1=x21+2xi2x2+1=x212x2+1+i2x2x2+1
On comparing real and imaginary part. we will get
a=x212x2+1 and b=2x2x2+1
Now,
a2+b2=(x212x2+1)2+(2x2x2+1)2
=x4+12x2+4x2(2x2+1)2
=x4+1+2x2(2x2+1)2
=(x2+1)2(2x2+1)2
Hence proved

Question:12(i) Let z1=2i,z2=2+i. Find

Re(z1z2z1¯)

Answer:

It is given that
z1=2i and z2=2+i
Now,
z1z2=(2i)(2+i)=4+2i+2ii2=4+4i+1=3+4i
And
z¯1=2+i
Now,
z1z2z¯1=3+4i2+i=3+4i2+i×2i2i=6+3i+8i4i222i2=6+11i+44+1 =2+11i5=25+i115
Now,
Re(z1z2z1)=25
Therefore, the answer is

25

Question:12(ii) Let z1=2i,z2=2+i. Find

Im(1z1z1¯)

Answer:

It is given that
z1=2i
Therefore,
z¯1=2+i
NOw,
z1z¯1=(2i)(2+i)=22i2=4+1=5 (using (ab)(a+b)=a2b2)
Now,
1z1z¯1=15
Therefore,
Im(1z1z¯1)=0
Therefore, the answer is 0

Question:13 Find the modulus and argument of the complex number 1+2i13i .

Answer:

Let
z=1+2i13i
Now, multiply the numerator and denominator by (1+3i)

z=1+2i13i×1+3i1+3i=1+3i+2i+6i212(3i)2=1+5i619i2=5+5i10 =12+i12
Therefore,
rcosθ=12  and  rsinθ=12
Square and add both the sides
r2cos2θ+r2sin2θ=(12)2+(12)2
r2(cos2θ+sin2θ)=(14)+(14)
r2=12                    (sin2θ+cos2θ=1)
r=12                 (r>0)
Therefore, the modulus is 12
Now,
12cosθ=12  and  12sinθ=12
cosθ=12  and  sinθ=12
Since the value of cosθ is negative and the value of sinθ is positive and we know that it is the case in II quadrant
Therefore,
Argument =(ππ4)=3π4

Therefore, Argument and modulus are 3π4  and  12 respectively

Question:14 Find the real numbers x andy if (xiy)(3+5i) is the conjugate of 624i .

Answer:

Let
z=(xiy)(3+5i)=3x+5xi3yi5yi2=3x+5y+i(5x3y)
Therefore,
z¯=(3x+5y)i(5x3y)               (i)
Now, it is given that
z¯=624i                    (ii)
Compare (i) and (ii) we will get
(3x+5y)i(5x3y)=624i
On comparing real and imaginary part. we will get
3x+5y=6   and   5x3y=24
On solving these we will get
x=3   and   y=3

Therefore, the value of x and y are 3 and -3 respectively

Question:15 Find the modulus of 1+i1i1i1+i .

Answer:

Let
z=1+i1i1i1+i
Now, we will reduce it into
z=1+i1i1i1+i=(1+i)2(1i)2(1+i)(1i)=12+i2+2i12i2+2i12i2 =4i1+1=4i2=2i
Now,
rcosθ=0  and  rsinθ=2
square and add both the sides. we will get,
r2cos2θ+r2sin2θ=02+22
r2(cos2θ+sin2θ)=4
r2=4                  (cos2θ+sin2θ=1)
r=2                  (r>0)

Therefore, modulus of

1+i1i1i1+i is 2

Question:16 If (x+iy)3=u+iv , then show that ux+vy=4(x2y2).

Answer:

it is given that
(x+iy)3=u+iv
Now, expand the Left-hand side
x3+(iy)3+3.(x)2.iy+3.x.(iy)2=u+iv
x3+i3y3+3x2iy+3xi2y2=u+iv
x3iy3+3x2iy3xy2=u+iv (i3=i  and  i2=1)
x33xy2+i(3x2yy3)=u+iv
On comparing real and imaginary part. we will get,
u=x33xy2   and   v=3x2yy3
Now,
ux+vy=x(x23y2)x+y(3x2y2)y
=x23y2+3x2y2
=4x24y2
=4(x2y2)
Hence proved

Question:17 If α and β are different complex numbers with |β|=1 , then find |βα1α¯β| .

Answer:

Let
α=a+ib and β=x+iy
It is given that
|β|=1x2+y2=1x2+y2=1
and
α¯=aib
Now,
|βα1α¯β|=|(x+iy)(a+ib)1(aib)(x+iy)|=|(xa)+i(yb)1(ax+iayibxi2yb)|
=|(xa)+i(yb)(1axyb)i(bxay)|
=(xa)2+(yb)2(1axyb)2+(bxay)2
=x2+a22xa+y2+b2yb1+a2x2+b2y22ax+2abxyby+b2x2+a2y22abxy
=(x2+y2)+a22xa+b2yb1+a2(x2+y2)+b2(x2+y2)2ax+2abxyby2abxy
=1+a22xa+b2yb1+a2+b22axby (x2+y2=1 given)
=1

Therefore, value of |βα1α¯β| is 1

Question:18 Find the number of non-zero integral solutions of the equation |1i|x=2x .

Answer:

Given problem is
|1i|x=2x
Now,
(12+(1)2)x=2x
(1+1)x=2x
(2)x=2x
2x2=2x
x2=x
x2=0
x = 0 is the only possible solution to the given problem

Therefore, there are 0 number of non-zero integral solutions of the equation |1i|x=2x

Question:19 If (a+ib)(c+id)(e+if)(g+ih)=A+iB, then show that (a2+b2)(c2+d2)(e2+f2)(g2+h2)=A2+B2

Answer:

It is given that
(a+ib)(c+id)(e+if)(g+ih)=A+iB,
Now, take mod on both sides
|(a+ib)(c+id)(e+if)(g+ih)|=|A+iB|
|(a+ib)||(c+id)||(e+if)||(g+ih)|=|A+iB| (|z1z2|=|z1||z2|)
(a2+b2)(c2+d2)(e2+f2)(g2+h2)=(A2+B2)
Square both the sides. we will get

(a2+b2)(c2+d2)(e2+f2)(g2+h2)=(A2+B2)

Hence proved

Question:20 If (1+i1i)m=1, then find the least positive integral value of m .

Answer:

Let
z=(1+i1i)m
Now, multiply both numerator and denominator by (1+i)
We will get,
z=(1+i1i×1+i1+i)m
=((1+i)212i2)m
=(12+i2+2i1+1)m
=(11+2i2)m (i2=1)
=(2i2)m
=im
We know that i4=1
Therefore, the least positive integral value of m is 4

Highlights Of NCERT Class 11 Chapter 5 Complex Numbers And Quadratic Equations

The NCERT Solutions for Chapter 5 - Complex Numbers and Quadratic Equations in Class 11 Maths consists of 3 exercises and a miscellaneous exercise to provide enough practice problems for the students to comprehend all the concepts. The PDF of NCERT Solutions for Class 11 Maths includes detailed explanations of the following topics and sub-topics:

  • 5.1 Introduction
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This section of ch 5 maths class 11 covers quadratic equations with no real solutions and their solutions using complex numbers. The solution to the quadratic equation ax^2 + bx + c = 0 has been derived when the discriminant D = b^2 - 4ac is less than 0.

  • 5.2 Complex Numbers It defines complex numbers and explains the real and imaginary parts of complex numbers with examples.
  • 5.3 Algebra of Complex Numbers This section covers the basic BODMAS operations on complex numbers i.e:

5.3.1 Addition of two complex numbers

5.3.2 Difference of two complex numbers

5.3.3 Multiplication of two complex numbers

5.3.4 Division of two complex number

5.3.5 Power of i

5.3.6 The square roots of a negative real number

5.3.7 Identities

After completing these exercises of complex no class 11 , students will have a better understanding of the fundamental BODMAS operations on complex numbers, as well as their properties, the power of i, square roots of negative real numbers, and complex number identitie

  • 5.4 The Modulus and the Conjugate of a Complex Number This section provides detailed explanations of the modulus and conjugate of a complex number with solved examples.
  • 5.5 Argand Plane and Polar Representation This section explains the complex plane or Argand plane and polar representation of complex numbers, including how to write ordered pairs for complex numbers.
  1. A complex number is a number expressed in the form of a + ib where a and b are real numbers. The real part of the complex number is denoted by "a" while the imaginary part is denoted by "b".
  2. If z1 = a + ib and z2 = c + id are two complex numbers, then their addition is (a + c) + i(b + d) and their product is (ac - bd) + i(ad + bc).
  3. For any non-zero complex number z = a + ib (where a and b are not equal to zero), there exists a unique complex number called the multiplicative inverse of z, denoted by 1/z or z(-1).
  4. For any integer k, i(4k) equals 1, i(4k+1) equals i, i(4k+2) equals -1, and i(4k+3) equals -i.
  5. The polar form of the complex number z = x + iy is represented as r(cosθ + i sinθ), where r is the modulus or absolute value of the complex number, and θ is the argument or angle made by the complex number with the positive x-axis.
  6. A polynomial equation of degree n has exactly n roots, which may or may not be complex.

Complex Numbers And Quadratic Equations Exercise Wise Solutions

NCERT Solutions For Class 11 Mathematics - Chapter Wise

Benefits of NCERT Class 11 Maths ch 5 Question Answer

Clarity of concepts: The NCERT Solutions for maths chapter 5 class 11 provide clear explanations and examples that help students to understand the concepts easily. The solutions are designed to make the learning process easier and more enjoyable.

Practice: The class 11 complex numbers and quadratic equations NCERT solutions come with a wide range of practice problems that help students to practice and master the concepts covered in the chapter. The more problems they solve, the better they become at the topic.

Exam preparation: NCERT Solutions for ch 5 maths class 11 are designed to help students prepare for their exams. The solutions provide a comprehensive overview of the chapter, including all the important topics and subtopics.

NCERT Solutions For Class 11 - Subject wise

Complex Numbers And Quadratic Equation Class 11 Chapter-Some Important Point To Remember

As mentioned in the first paragraph i=1 and

i2=1i3=1×i=ii4=1×1=1

and any number can be represented as a complex number of the form a+ ib where a is the real part and b is the imaginary part, for example, 1=1+0i.

So a complex number of the form a+ ib can be represented as r(cosθ+isinθ)

and

\\r=\sqrt{a^2+b^2}\\a=rcos\theta\\b=rsin\theta

and the above representation is known as the polar form of a complex number. T he polar form of the complex number makes the problem very easy to solve. There are many problems in the NCERT solutions for class 11 maths chapter 5 complex numbers and quadratic equations which are explained using the polar form of the complex number and some are solved using 2-D geometry. So, NCERT solutions for class 11 maths chapter 5 complex numbers and quadratic equations can make learning easier for you so that you can score well.

NCERT Books and NCERT Syllabus

Happy Reading !!!

Frequently Asked Questions (FAQs)

1. What are important topics of the chapter Complex Numbers and Quadratic Equations ?

Class 11 maths NCERT solutions chapter 5 includes topics such as Complex numbers, algebra of complex numbers, modulus and the conjugate of a complex number, argand plane and polar representation, and quadratic equations are the important topics in this chapter.  to get command in these topics students can practice problems from complex numbers class 11 pdf. 

2. How does the NCERT solutions are helpful ?

NCERT solutions are highly beneficial for students as they provide well-structured and comprehensive solutions to the textbook questions. These solutions are designed by subject matter experts and cover all the important topics and concepts in a clear and concise manner. Students can practice complex numbers class 11 solutions to get good hold on the concepts of chapter 5 class 11.

3. What are the most difficult chapters in the class 11 maths ?

Most of the students consider permutation and combination, trigonometry as the most difficult chapters in class 11 maths but with the rigorous practice students can get command on them also.

4. How many chapters are there in CBSE class 11 maths ?

There are 16 chapters starting from set to probability in the CBSE class 11 maths.

5. Does CBSE provides the solutions of NCERT class 11 maths ?

No, CBSE doesn’t provided NCERT solutions for any class or subject.

6. Where can I find the complete solutions of NCERT for class 11 maths ?

Here you will get the detailed NCERT solutions for class 11 maths  by clicking on the link.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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