NCERT Solutions for Class 11 Maths Chapter 4 - Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths Chapter 4 - Complex Numbers and Quadratic Equations

Komal MiglaniUpdated on 19 Aug 2025, 03:21 PM IST

Have you ever faced a situation where solving an equation led to the square root of a negative number? Do you know how engineers, physicists, and mathematicians deal with this situation? Yes, this is where Complex Numbers come in! This chapter discusses the idea of Complex numbers and quadratic equations. A complex number has both real and imaginary parts. The imaginary parts are represented by i(iota). Students will carry out basic algebraic operations on complex numbers in the same way as on real numbers, such as addition, subtraction, multiplication, and division. The quadratic equations, or second-degree order equations, teach one how to determine the roots of these equations through the discriminant and quadratic formulas. This article on NCERT solutions for class 11 maths will help students to understand all the concepts more thoroughly.

This Story also Contains

  1. NCERT Solution for Class 11 Maths Chapter 4 Solutions: Download PDF
  2. NCERT Solutions for Class 11 Maths Chapter 4: Exercise Questions
  3. Class 11 Maths NCERT Chapter 4: Extra Question
  4. Complex Numbers and Quadratic Equations Class 11 Chapter 4: Topics
  5. Complex Numbers and Quadratic Equations Class 11 Solutions: Important Formulae
  6. Approach to Solve Questions of Complex Numbers and Quadratic Equations Class 11
  7. What Extra Should Students Study Beyond NCERT for JEE?
  8. NCERT Solutions for Class 11 Maths: Chapter Wise
NCERT Solutions for Class 11 Maths Chapter 4 - Complex Numbers and Quadratic Equations
NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

The NCERT solutions for Class 11 complex numbers and quadratic equations deliver step-by-step processes to solve exercises and simplify complicated problems. In addition to textbook exercises, students are recommended to solve Class 11 Maths Chapter 4 Question Answers to reinforce their learning and ensure they are well-prepared for exams. Through the use of resources such as the NCERT Exemplar solutions for Class 11 Maths, students can solve more complex problems and enhance their knowledge of both complex numbers and quadratic equations. Students need to refer to the NCERT Class 10 Maths books to gain more knowledge. With consistent practice and a clear understanding of these concepts, students can approach both theoretical and applied mathematics confidently. For syllabus, notes, and PDF, refer to NCERT.

NCERT Solution for Class 11 Maths Chapter 4 Solutions: Download PDF

Careers360 experts have prepared these NCERT Solutions for Class 11 Maths Chapter 4 to make learning maths easier and help you do better in class and exams. NCERT Class 11 Maths Chapter 4 Solutions PDF are also available.

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NCERT Solutions for Class 11 Maths Chapter 4: Exercise Questions

Class 10 Maths chapter 4 solutions Exercise: 4.1
Page number: 82-83
Total questions: 14

Question 1: Express the given complex number in the form a+ib:
$(5i)(-\frac35 i)$?

Answer:

Given $a+i b:(5 i) \times\left(-\frac{3}{5} i\right)$

$(5 i) \times\left(-\frac{3}{5} i\right)=-\left(5 \times \frac{3}{5}\right)\left(i^2\right)$

$\begin{aligned} & =-3\left(i^2\right) \\ & =3 \end{aligned}$

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

$i^9+i^{19}$?

Answer:

We know that $i^4 = 1$
Now, we will reduce $i^9+i^{19}$ into

$i^9+i^{19}=\left(i^4\right)^2 \cdot i+\left(i^4\right)^3 \cdot i^3$
$= (1)^2.i+(1)^3.(-i)$ $(\because i^4 = 1 , i^3 = -i\ and \ i^2 = -1)$
$=i-i $
$= 0$
Now, in the form of $a+ib$, we can write it as
$0+i0$
Therefore, the answer is $0+i0$.

Question 3: Express each of the complex number in the form a+ib.
$i^{-39}$

Answer:

We know that $i^4 = 1$
Now, we will reduce $i^{-39}$ into

$i^{-39}$ $= (i^{4})^{-9}.i^{-3}$
$= (1)^{-9}.(-i)^{-1}$ $(\because i^4 = 1 , i^3 = -i)$
$= \frac{1}{-i}$
$= \frac{1}{-i} \times \frac{i}{i}$
$= \frac{i}{-i^2}$ $(\because i^2 = -1)$
$= \frac{i}{-(-1)}$
$=i$
Now, in the form of $a+ib$ we can write it as
$0+i1$
Therefore, the answer is $0+i1$.

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

$3(7+7i)+i(7+7i)$ $= 21+21i+7i+7i^2$
$= 21+21i+7i+7(-1)$ $(\because i^2 = -1)$
$= 21+21i+7i-7$
$=14+28i$

Therefore, the answer is $14+28i$

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

$(1-i)-(-1+6i)$

Answer:

Given problem is
$(1-i)-(-1+6i)$
Now, we will reduce it to

$(1-i)-(-1+6i)$$=1-i+1-6i$
$= 2-7i$

Therefore, the answer is $2-7i$

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

$\left ( \frac{1}{5}+i\frac{2}{5} \right )-\left ( 4+i\frac{5}{2} \right )$

Answer:

Given problem is
$\left ( \frac{1}{5}+i\frac{2}{5} \right )-\left ( 4+i\frac{5}{2} \right )$
Now, we will reduce it to

$\left ( \frac{1}{5}+i\frac{2}{5} \right )-\left ( 4+i\frac{5}{2} \right ) = \frac{1}{5}+i\frac{2}{5}-4-i\frac{5}{2}$
$= \frac{1-20}{5}+i\frac{(4-25)}{10}$
$= -\frac{19}{5}-i\frac{21}{10}$

Therefore, the answer is $-\frac{19}{5}-i\frac{21}{10}$

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

$\left [ \left ( \frac{1}{3}+i\frac{7}{3} \right )+\left ( 4+i\frac{1}{3} \right ) \right ]-\left ( -\frac{4}{3}+i \right )$

Answer:

Given problem is
$\left [ \left ( \frac{1}{3}+i\frac{7}{3} \right )+\left ( 4+i\frac{1}{3} \right ) \right ]-\left ( -\frac{4}{3}+i \right )$
Now, we will reduce it into

$\left [ \left ( \frac{1}{3}+i\frac{7}{3} \right )+\left ( 4+i\frac{1}{3} \right ) \right ]-\left ( -\frac{4}{3}+i \right ) = \frac{1}{3}+i\frac{7}{3} + 4+i\frac{1}{3} + \frac{4}{3}-i$
$=\frac{1+4+12}{3}+i\frac{(7+1-3)}{3}$
$=\frac{17}{3}+i\frac{5}{3}$

Therefore, the answer is $\frac{17}{3}+i\frac{5}{3}$

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

$(1-i)^4$

Answer:

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

$(1-i)^4 = ((1-i)^2)^2$
$= (1^2+i^2-2.1.i)^2$ $(using \ (a-b)^2= a^2+b^2-2ab)$

$=(1-1-2i)^2$ $(\because i^2 = -1)$
$= (-2i)^2$
$= 4i^2$
$= -4$

Therefore, the answer is $-4+i0$

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

$\left ( \frac{1}{3}+3i \right )^3$

Answer:

Given problem is
$\left ( \frac{1}{3}+3i \right )^3$
Now, we will reduce it into

$\left ( \frac{1}{3}+3i \right )^3=\left ( \frac{1}{3} \right )^3+(3i)^3+3.\left ( \frac{1}{3} \right )^2.3i+3.\frac{1}{3}.(3i)^2$ $(using \ (a+b)^3=a^3+b^3+3a^2b+3ab^2)$
$= \frac{1}{27}+27i^3+i + 9i^2$

$= \frac{1}{27}+27(-i)+i + 9(-1)$ $(\because i^3=-i \ and \ i^2 = -1)$
$=\frac{1}{27}-27i+i-9$
$=\frac{1-243}{27}-26i$
$=-\frac{242}{27}-26i$

Therefore, the answer is

$-\frac{242}{27}-26i$

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

$\left ( -2-\frac{1}{3}i \right )^3$

Answer:

Given problem is
$\left ( -2-\frac{1}{3}i \right )^3$
Now, we will reduce it into

$\left ( -2-\frac{1}{3}i \right )^3=-\left ( (2)^3+\left ( \frac{1}{3}i \right )^3 +3.(2)^2\frac{1}{3}i+3.\left ( \frac{1}{3}i \right )^2.2 \right )$ $(using \ (a+b)^3=a^3+b^3+3a^2b+3ab^2)$
$=-\left ( 8+\frac{1}{27}i^3+3.4.\frac{1}{3}i+3.\frac{1}{9}i^2.2 \right )$

$=-\left ( 8+\frac{1}{27}(-i)+4i+\frac{2}{3}(-1) \right )$ $(\because i^3=-i \ and \ i^2 = -1)$
$=-\left ( 8-\frac{1}{27}i+4i-\frac{2}{3} \right )$
$=-\left ( \frac{(-1+108)}{27}i+\frac{24-2}{3} \right )$
$=-\frac{22}{3}-i\frac{107}{27}$

Therefore, the answer is $-\frac{22}{3}-i\frac{107}{27}$

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

$4-3i$

Answer:

Let $z = 4-3i$
Then,
$\bar z = 4+ 3i$
And
$|z|^2 = 4^2+(-3)^2 = 16+9 =25$
Now, the multiplicative inverse is given by
$z^{-1}= \frac{\bar z}{|z|^2}= \frac{4+3i}{25}= \frac{4}{25}+i\frac{3}{25}$

Therefore, the multiplicative inverse is

$\frac{4}{25}+i\frac{3}{25}$

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

$\sqrt{5}+3i$

Answer:

Let $z = \sqrt{5}+3i$
Then,
$\bar z = \sqrt{5}-3i$
And
$|z|^2 = (\sqrt5)^2+(3)^2 = 5+9 =14$
Now, the multiplicative inverse is given by
$z^{-1}= \frac{\bar z}{|z|^2}= \frac{\sqrt5-3i}{14}= \frac{\sqrt5}{14}-i\frac{3}{14}$

Therefore, the multiplicative inverse is $\frac{\sqrt5}{14}-i\frac{3}{14}$

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

$-i$

Answer:

Let $z = -i$
Then,
$\bar z = i$
And
$|z|^2 = (0)^2+(1)^2 = 0+1 =1$
Now, the multiplicative inverse is given by
$z^{-1}= \frac{\bar z}{|z|^2}= \frac{i}{1}= 0+i$

Therefore, the multiplicative inverse is $0+i1$

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

$\frac{(3+i\sqrt{5})(3-i\sqrt{5})}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-i\sqrt{2})}$

Answer:

Given problem is
$\frac{(3+i\sqrt{5})(3-i\sqrt{5})}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-i\sqrt{2})}$
Now, we will reduce it to

$\frac{(3+i\sqrt{5})(3-i\sqrt{5})}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-i\sqrt{2})} = \frac{3^2- (\sqrt5i)^2}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-i\sqrt{2})}$ $(using \ (a-b)(a+b)=a^2-b^2)$
$=\frac{9-5i^2}{\sqrt3+\sqrt2i-\sqrt3+\sqrt2i}$
$=\frac{9-5(-1)}{2\sqrt2i}$ $(\because i^2 = -1)$
$=\frac{14}{2\sqrt2i}\times \frac{\sqrt2i}{\sqrt2i}$
$=\frac{7\sqrt2i}{2i^2}$
$=-\frac{7\sqrt2i}{2}$
Therefore, the answer is $(0-i\frac{7\sqrt2}{2})$

Class 10 Maths chapter 4 solutions Miscellaneous Exercise
Page number: 85-86
Total questions: 14

Question 1: Evaluate $\left[i^{18}+\left(\frac{1}{i}\right)^{25}\right]^3$ .

Answer:

The given problem is
$\left[i^{18}+\left(\frac{1}{i}\right)^{25}\right]^3$
Now, we will reduce it to
$\left[i^{18}+\left(\frac{1}{i}\right)^{25}\right]^3=\left[\left(i^4\right)^4 \cdot i^2+\frac{1}{\left(i^4\right)^6 \cdot i}\right]^3$
$=\left [ 1^4.(-1)+\frac{1}{1^6.i} \right ]^3$ $(\because i^4 = 1, i^2 = -1 )$
$= \left [ -1+\frac{1}{i} \right ]^3$
$= \left [ -1+\frac{1}{i} \times \frac{i}{i}\right ]^3$
$= \left [ -1+\frac{i}{i^2} \right ]^3$
$= \left [ -1+\frac{i}{-1} \right ]^3 = \left [ -1-i \right ]^3$
Now,
$-(1+i)^3=-(1^3+i^3+3.1^2.i+3.1.i^2)$ $(using \ (a+b)^3=a^3+b^3+3.a^2.b+3.a.b^2)$
$= -(1 - i +3i+3(-1))$ $(\because i^3=-i , i^2 = -1)$
$= -(1 - i +3i-3)= -(-2+2i)$
$=2-2i$
Therefore, answer is $2-2i$

Question 2: For any two complex numbers $\small z_1$ and $\small z_2$ , prove that $\small Re (z_1z_2)=Re\hspace {1mm}z_1\hspace {1mm}Re\hspace {1mm}z_2-Imz_1\hspace {1mm}Imz_2$

Answer:

Let two complex numbers are
$z_1=x_1+iy_1$
$z_2=x_2+iy_2$
Now,
$z_1.z_2=(x_1+iy_1).(x_2+iy_2)$
$=x_1x_2+ix_1y_2+iy_1x_2+i^2y_1y_2$
$=x_1x_2+ix_1y_2+iy_1x_2-y_1y_2$ $(\because i^2 = -1)$
$=x_1x_2-y_1y_2+i(x_1y_2+y_1x_2)$
$Re(z_1z_2)= x_1x_2-y_1y_2$
$=Re(z_1z_2)-Im(z_1z_2)$

Hence proved

Question 3: Reduce $\small \left ( \frac{1}{1-4i}-\frac{2}{1+i} \right )\left ( \frac{3-4i}{5+i} \right )$ to the standard form.

Answer:

Given problem is
$\small \left ( \frac{1}{1-4i}-\frac{2}{1+i} \right )\left ( \frac{3-4i}{5+i} \right )$
Now, we will reduce it to

$\small \left ( \frac{1}{1-4i}-\frac{2}{1+i} \right )\left ( \frac{3-4i}{5+i} \right ) = \left ( \frac{(1+i)-2(1-4i)}{(1+i)(1-4i)} \right )\left ( \frac{3-4i}{5+i} \right )$
$=\left ( \frac{1+i-2+8i}{1-4i+i-4i^2} \right )\left ( \frac{3-4i}{5+i} \right )$
$=\left ( \frac{-1+9i}{1-3i-4(-1)} \right )\left ( \frac{3-4i}{5+i} \right )$
$=\left ( \frac{-1+9i}{5-3i} \right )\left ( \frac{3-4i}{5+i} \right )$
$=\left ( \frac{-3+4i+27i-36i^2}{25+5i-15i-3i^2} \right )= \left ( \frac{-3+31i+36}{25-10i+3} \right )= \frac{33+31i}{28-10i}= \frac{33+31i}{2(14-5i)}$

Now, multiply numerator and denominator by $(14+5i)$
$\Rightarrow \frac{33+31i}{2(14-5i)}\times \frac{14+5i}{14+5i}$
$\Rightarrow \frac{462+165i+434i+155i^2}{2(14^2-(5i)^2)}$ $(using \ (a-b)(a+b)=a^2-b^2)$
$\Rightarrow \frac{462+599i-155}{2(196-25i^2)}$
$\Rightarrow \frac{307+599i}{2(196+25)}= \frac{307+599i}{2\times 221}= \frac{307+599i}{442}= \frac{307}{442}+i\frac{599}{442}$

Therefore, answer is $\frac{307}{442}+i\frac{599}{442}$

Question 4: If $\small x-iy=\sqrt{\frac{a-ib}{c-id}}$ , prove that $\small (x^2+y^2)^2=\frac{a^2+b^2}{c^2+d^2}.$

Answer:

The given problem is

$\small x-iy=\sqrt{\frac{a-ib}{c-id}}$
Now, multiply the numerator and denominator by

$\sqrt{c+id}$
$x-iy = \sqrt{\frac{a-ib}{c-id}\times \frac{c+id}{c+id}}$
$= \sqrt{\frac{(ac+bd)+i(ad-bc)}{c^2-i^2d^2}}= \sqrt{\frac{(ac+bd)+i(ad-bc)}{c^2+d^2}}$
Now, square both sides
$(x-iy)^2=\left ( \sqrt{\frac{(ac+bd)+i(ad-bc)}{c^2+d^2}} \right )^2$
$=\frac{(ac+bd)+i(ad-bc)}{c^2+d^2}$
$x^2-y^2-2ixy=\frac{(ac+bd)+i(ad-bc)}{c^2+d^2}$
On comparing the real and imaginary parts, we obtain

$x^2-y^2 = \frac{ac+bd}{c^2+d^2} \ \ and \ \ -2xy = \frac{ad-bc}{c^2+d^2} \ \ \ -(i)$

Now,
$(x^2+y^2)^2= (x^2-y^2)^2+4x^2y^2$
$= \left ( \frac{ac+bd}{c^2+d^2} \right )^2+\left ( \frac{ad-bc}{c^2+d^2} \right )^2 \ \ \ \ (using \ (i))$
$=\frac{a^2c^2+b^2d^2+2acbd+a^2d^2+b^2c^2-2adbc}{(c^2+d^2)^2}$
$=\frac{a^2c^2+b^2d^2+a^2d^2+b^2c^2}{(c^2+d^2)^2}$
$=\frac{a^2(c^2+d^2)+b^2(c^2+d^2)}{(c^2+d^2)^2}$
$=\frac{(a^2+b^2)(c^2+d^2)}{(c^2+d^2)^2}$
$=\frac{(a^2+b^2)}{(c^2+d^2)}$

Hence proved.

Question 5: If $\small z_1=2-i, z_2=1+i$ , find $\small \left |\frac{z_1+z_2+1}{z_1-z_2+1} \right |$ .

Answer:

It is given that
$\small z_1=2-i, z_2=1+i$
Then,
$\small \left |\frac{z_1+z_2+1}{z_1-z_2+1} \right | =\left | \frac{2-i+1+i+1}{2-i-1-i+1} \right | = \left | \frac{4}{2(1-i)} \right |= \left | \frac{2}{(1-i)} \right |$
Now, multiply the numerator and denominator by $1+i$
$\Rightarrow \left | \frac{2}{(1-i)} \times \frac{1+i}{1+i} \right |=\left |\frac{2(1+i)}{1^2-i^2} \right |=\left | \frac{2(1+i)}{1+1} \right |= \left| 1+i \right |$
Now,
$|1+i| = \sqrt{1^2+1^2}=\sqrt{1+1}=\sqrt{2}$
Therefore, the value of

$\small \left |\frac{z_1+z_2+1}{z_1-z_2+1} \right |$ is $\sqrt{2}$

Question 6: If $\small a+ib=\frac{(x+i)^2}{2x^2+1}$ , prove that $\small a^2+b^2=\frac{(x^2+1)^2}{(2x^2+1)^2}$ .

Answer:

It is given that
$\small a+ib=\frac{(x+i)^2}{2x^2+1}$
Now, we will reduce it to

$\small a+ib=\frac{(x+i)^2}{2x^2+1} = \frac{x^2+i^2+2xi}{2x^2+1}=\frac{x^2-1+2xi}{2x^2+1}=\frac{x^2-1}{2x^2+1}+i\frac{2x}{2x^2+1}$
On comparing the real and imaginary parts, we will get
$a=\frac{x^2-1}{2x^2+1}\ and \ b=\frac{2x}{2x^2+1}$
Now,
$a^2+b^2=\left ( \frac{x^2-1}{2x^2+1} \right )^2+\left ( \frac{2x}{2x^2+1} \right )^2$
$= \frac{x^4+1-2x^2+4x^2}{(2x^2+1)^2}$
$= \frac{x^4+1+2x^2}{(2x^2+1)^2}$
$= \frac{(x^2+1)^2}{(2x^2+1)^2}$
Hence proved

Question 7 (i): Let $\small z_1=2-i,z_2=-2+i.$ Find

$\small Re\left ( \frac{z_1z_2}{\bar{z_1}} \right )$

Answer:

It is given that
$\small z_1=2-i \ and \ z_2=-2+i$
Now,
$z_1z_2= (2-i)(-2+i)= -4+2i+2i-i^2=-4+4i+1= -3+4i$
And
$\bar z_1 = 2+i$
Now,
$\frac{z_1z_2}{\bar z_1}= \frac{-3+4i}{2+i}= \frac{-3+4i}{2+i}\times \frac{2-i}{2-i}= \frac{-6+3i+8i-4i^2}{2^2-i^2}= \frac{-6+11i+4}{4+1}$ $= \frac{-2+11i}{5}= -\frac{2}{5}+i\frac{11}{5}$
Now,
$Re\left ( \frac{z_1z_2}{z_1} \right )= -\frac{2}{5}$
Therefore, the answer is

$-\frac{2}{5}$

Question 7 (ii): Let $\small z_1=2-i,z_2=-2+i.$ Find

$\small Im\left ( \frac{1}{z_1\bar{z_1}} \right )$

Answer:

It is given that
$z_1= 2-i$
Therefore,
$\bar z_1= 2+i$
NOw,
$z_1\bar z_1= (2-i)(2+i)= 2^2-i^2=4+1=5$ $(using \ (a-b)(a+b)= a^2-b^2)$
Now,
$\frac{1}{z_1\bar z_1}= \frac{1}{5}$
Therefore,
$Im\left ( \frac{1}{z_1\bar z_1} \right )= 0$
Therefore, the answer is 0.

Question 8: Find the real numbers x and y if $\small (x-iy)(3+5i)$ is the conjugate of $\small -6-24i$.

Answer:

Let
$z = \small (x-iy)(3+5i) = 3x+5xi-3yi-5yi^2= 3x+5y+i(5x-3y)$
Therefore,
$\bar z = (3x+5y)-i(5x-3y) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)$
Now, it is given that
$\bar z = -6-24i \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)$
Comparing (i) and (ii), we will get
$(3x+5y)-i(5x-3y) = -6-24i$
On comparing the real and imaginary parts, we will get
$3x+5y=-6 \ \ \ and \ \ \ 5x-3y = 24$
On solving these, we will get
$x = 3 \ \ \ and \ \ \ y =- 3$

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

Question 9: Find the modulus of $\small \frac{1+i}{1-i}-\frac{1-i}{1+i}$ .

Answer:

Let
$z =\small \frac{1+i}{1-i}-\frac{1-i}{1+i}$
Now, we will reduce it to
$z =\small \frac{1+i}{1-i}-\frac{1-i}{1+i} = \frac{(1+i)^2-(1-i)^2}{(1+i)(1-i)}= \frac{1^2+i^2+2i-1^2-i^2+2i}{1^2-i^2}$ $= \frac{4i}{1+1}= \frac{4i}{2}=2i$
Now,
$r\cos\theta = 0 \ \ and \ \ r\sin \theta = 2$
Square and add both sides, we will get,
$r^2\cos^2\theta+r^2\sin^2 \theta = 0^2+2^2$
$r^2(\cos^2\theta+\sin^2 \theta) = 4$
$r^2 = 4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because \cos^2\theta+\sin^2 \theta = 1)$
$r = 2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because r > 0)$

Therefore, the modulus of

$\small \frac{1+i}{1-i}-\frac{1-i}{1+i}$ is 2

Question 10: If $\small (x+iy)^3=u+iv$ , then show that $\small \frac{u}{x}+\frac{v}{y}=4 (x^2-y^2).$

Answer:

it is given that
$\small (x+iy)^3=u+iv$
Now, expand the Left-hand side
$x^3+(iy)^3+3.(x)^2.iy+3.x.(iy)^2= u + iv$
$x^3+i^3y^3+3x^2iy+3xi^2y^2= u + iv$
$x^3-iy^3+3x^2iy-3xy^2= u + iv$ $(\because i^3 = -i \ \ and \ \ i^2 = -1)$
$x^3-3xy^2+i(3x^2y-y^3)= u + iv$
On comparing real and imaginary part. we will get,
$u = x^3-3xy^2 \ \ \ and \ \ \ v = 3x^2y-y^3$
Now,
$\frac{u}{x}+\frac{v}{y}= \frac{x(x^2-3y^2)}{x}+\frac{y(3x^2-y^2)}{y}$
$= x^2-3y^2+3x^2-y^2$
$= 4x^2-4y^2$
$= 4(x^2-y^2)$
Hence proved

Question 11: If $\small \alpha$ and $\small \beta$ are different complex numbers with $\small |\beta|=1$ , then find $\small \left | \frac{\beta -\alpha }{1-\bar{\alpha }\beta } \right |$ .

Answer:

Let
$\alpha = a+ib$ and $\beta = x+iy$
It is given that
$\small |\beta|=1\Rightarrow \sqrt{x^2+y^2} = 1\Rightarrow x^2+y^2 = 1$
and
$\small \bar \alpha = a-ib$
Now,
$\small \left | \frac{\beta -\alpha }{1-\bar{\alpha }\beta } \right | = \left | \frac{(x+iy)-(a+ib)}{1-(a-ib)(x+iy)} \right | = \left | \frac{(x-a)+i(y-b)}{1-(ax+iay-ibx-i^2yb)} \right |$
$\small = \left | \frac{(x-a)+i(y-b)}{(1-ax-yb)-i(bx-ay)} \right |$
$\small = \frac{\sqrt{(x-a)^2+(y-b)^2}}{\sqrt{(1-ax-yb)^2+(bx-ay)^2}}$
$\small = \frac{\sqrt{x^2+a^2-2xa+y^2+b^2-yb}}{\sqrt{1+a^2x^2+b^2y^2-2ax+2abxy-by+b^2x^2+a^2y^2-2abxy}}$
$\small = \frac{\sqrt{(x^2+y^2)+a^2-2xa+b^2-yb}}{\sqrt{1+a^2(x^2+y^2)+b^2(x^2+y^2)-2ax+2abxy-by-2abxy}}$
$\small = \frac{\sqrt{1+a^2-2xa+b^2-yb}}{\sqrt{1+a^2+b^2-2ax-by}}$ $\small (\because x^2+y^2 = 1 \ given)$
$\small =1$

Therefore, value of $\small \left | \frac{\beta -\alpha }{1-\bar{\alpha }\beta } \right |$ is 1

Question 12: Find the number of non-zero integral solutions of the equation $\small |1-i|^x=2^x$.

Answer:

Given problem is
$\small |1-i|^x=2^x$
Now,
$( \sqrt{1^2+(-1)^2 })^x=2^x$
$( \sqrt{1+1 })^x=2^x$
$\left ( \sqrt{2 }\right )^x=2^x$
$2^{\frac{x}{2}}= 2^x$
$\frac{x}{2}=x$
$\frac{x}{2}=0$
x = 0 is the only possible solution to the given problem

Therefore, there is 0 number of non-zero integral solution of the equation $\small |1-i|^x=2^x$

Question 13: If $\small (a+ib)(c+id)(e+if)(g+ih)=A+iB,$ then show that $\small (a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2)=A^2+B^2$

Answer:

It is given that
$\small (a+ib)(c+id)(e+if)(g+ih)=A+iB,$
Now, take the modulus on both sides
$\left | (a+ib)(c+id)(e+if)(g+ih) \right |= \left | A+iB \right |$
$|(a+ib)||(c+id)||(e+if)||(g+ih)|= \left | A+iB \right |$ $(\because |z_1z_2|=|z_1||z_2|)$
$(\sqrt{a^2+b^2})(\sqrt{c^2+d^2})(\sqrt{e^2+f^2})(\sqrt{g^2+h^2})= (\sqrt{A^2+B^2})$
Square both sides, we will get

$({a^2+b^2})({c^2+d^2})({e^2+f^2})({g^2+h^2})= (A^2+B^2)$

Hence proved

Question 14: If $\small \left ( \frac{1+i}{1-i} \right )^m=1,$ then find the least positive integral value of $\small m$ .

Answer:

Let
$z = \left ( \frac{1+i}{1-i} \right )^m$
Now, multiply both numerator and denominator by $(1+i)$
We will get,
$z = \left ( \frac{1+i}{1-i}\times \frac{1+i}{1+i} \right )^m$
$= \left ( \frac{(1+i)^2}{1^2-i^2} \right )^m$
$= \left ( \frac{1^2+i^2+2i}{1+1} \right )^m$
$= \left ( \frac{1-1+2i}{2} \right )^m$ $(\because i^2 = -1)$
$= \left ( \frac{2i}{2} \right )^m$
$= i^m$
We know that $i^4 = 1$
Therefore, the least positive integral value of $\small m$ is 4.

Also, read,

Class 11 Maths NCERT Chapter 4: Extra Question

Question:
If $\alpha$ is a root of the equation $x^2+x+1=0$ and $\sum_{\mathrm{k}=1}^{\mathrm{n}}\left(\alpha^{\mathrm{k}}+\frac{1}{\alpha^{\mathrm{k}}}\right)^2=20$, then n is equal to ______

Solution:
We have,
$\begin{aligned}
& \alpha=\omega \\
& \therefore\left(\omega^{\mathrm{k}}+\frac{1}{\omega^{\mathrm{k}}}\right)^2=\omega^{2 \mathrm{k}}+\frac{1}{\omega^{2 \mathrm{k}}}+2 \\
& =\omega^{2 \mathrm{k}}+\omega^{\mathrm{k}}+2 \quad \because \omega^{3 \mathrm{k}}=1
\end{aligned}$

$\begin{aligned}
& \therefore \sum_{k=1}^n\left(\omega^{2 k}+\omega^k+2\right)=20 \\
& \Rightarrow\left(\omega^2+\omega^4+\omega^6+\ldots+\omega^{2 n}\right)+\left(\omega+\omega^2+\omega^3+\ldots+\right. \\
& \left.\omega^n\right)+2 n=20
\end{aligned}$
Now if $n=3 m, \quad m \in I$
Then $0+0+2 \mathrm{n}=20 \Rightarrow \mathrm{n}=10$ (not satisfy) if $n=3 m+1$, then

$\begin{aligned}
& \omega^2+\omega+2 n=20 \\
& -1+2 n=20 \Rightarrow n=\frac{21}{2} \text { (not possible) } \\
& \text { if } n=3 m+2, \\
& \left(\omega^8+\omega^{10}\right)+\left(\omega^4+\omega^5\right)+2 n=20 \\
& \Rightarrow\left(\omega^2+\omega\right)+\left(\omega+\omega^2\right)+2 n=20 \\
& 2 n=22
\end{aligned}$

$\begin{aligned} & \mathrm{n}=11 \text { satisfy } \mathrm{n}=3 \mathrm{~m}+2 \\ & \therefore \mathrm{n}=11\end{aligned}$

Hence, the correct answer is 11.

Complex Numbers and Quadratic Equations Class 11 Chapter 4: Topics

The topics discussed in the NCERT Solutions for class 11, chapter 4, Complex Numbers and Quadratic Equations are:

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

1. Addition: $\quad \mathrm{z}_1+\mathrm{z}_2=\left(\mathrm{x}_1+\mathrm{x}_2\right)+\mathrm{i}\left(\mathrm{y}_1+\mathrm{y}_2\right)$
2. Subtraction: $\quad \mathrm{z}_1-\mathrm{z}_2=\left(\mathrm{x}_1-\mathrm{x}_2\right)+\mathrm{i}\left(\mathrm{y}_1-\mathrm{y}_2\right)$
3. Multiplication: $\quad \mathrm{z}_1 \times \mathrm{z}_2=\left(\mathrm{x}_1 \mathrm{x}_2-\mathrm{y}_1 \mathrm{y}_2\right)+\mathrm{i}\left(\mathrm{x}_1 \mathrm{y}_2+\mathrm{x}_2 \mathrm{y}_1\right)$
4. Division: $\left(z_2 \neq 0\right) \quad \frac{z_1}{z_2}=\frac{\left(\mathbf{x}_1 \mathbf{x}_2+\mathbf{y}_1 \mathbf{y}_2\right)+i\left(\mathbf{y}_1 \mathbf{x}_2-\mathbf{x}_1 \mathbf{y}_2\right)}{\mathbf{x}_2^2+\mathbf{y}_2^2}$

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

Modulus of a Complex Number: |z| = $\sqrt{x_2+y_2}$

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:
$z=|z|[\cos (\theta+2 n \pi)+i \sin (\theta+2 n \pi)]$, where n is an integer.

Approach to Solve Questions of Complex Numbers and Quadratic Equations Class 11

Here are some approaches that students can use to solve complex numbers and quadratic equation problems.

  • Familiarise yourself with the standard form of complex numbers, i.e., a + bi, where ‘a’ and ‘b’ are real numbers and ‘i’ is the imaginary unit.
  • For quadratic equations, the standard form is $ax^2+bx+c$. The discriminant, D = $b^2 -4ac$ can be used to determine the nature of the roots.
  • Practice formulas like addition, subtraction, multiplication, division, conjugate, and modulus of a complex number.
    Grasp the concepts of roots and their properties for finding the sum and product of roots. This would be extremely helpful while attempting MCQ questions.
    - Sum $=\alpha+\beta=-\frac{b}{a}$
    - Product $=\alpha \beta=\frac{c}{a}$
  • Develop a strong understanding of plotting complex numbers as points and vectors.
    Also, realise that a quadratic graph is a parabola, and you must analyse the roots depending on where it crosses the x-axis.
  • Also, realise the Euler and polar form of complex numbers employed in multiplication and division. De Moivre's theorem is also of great use.

NCERT Solutions for Class 11 Maths: Chapter Wise

The links below allow students to access all the Maths solutions from the NCERT book.

Also, read,

NCERT Solutions For Class 11 - Subject-wise

Given below are the subject-wise NCERT solutions of class 11 NCERT:

NCERT Books and NCERT Syllabus

Here are some useful links for NCERT books and the NCERT syllabus for class 11:

Frequently Asked Questions (FAQs)

Q: What is the definition of a complex number?
A:

A complex number is of the form z = a + ib, where a is the real part, b is the imaginary part, and the value of the square of i is –1.

Q: What is the polar form of a complex number?
A:

It is z = |z|(cos θ+i sin θ), where |z| is the modulus and θ is the argument.

Q: What are important topics of the chapter Complex Numbers and Quadratic Equations ?
A:

This chapter covers topics such as:

  • Complex Numbers
  • Algebra of Complex Numbers
  • The Modulus and the Conjugate of a Complex Number
  • Argand Plane and Polar Representation
Q: How does the NCERT solutions are helpful ?
A:

NCERT solutions are very helpful for students because they give clear and comprehensive answers to all textbook questions. Made by experienced subject experts, these solutions explain all important topics and concepts in a simple way.

Q: Can I download these NCERT solutions in PDF format?
A:

Many trusted educational websites, such as Careers360, offer free downloadable PDFS of the NCERT Solutions for class 11 Maths. Careers360 provides step-by-step explanations for all textbook exercises, helping students understand concepts clearly and prepare effectively for exams. 

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