Complex Numbers And Quadratic Equations Class 11th Notes - Free NCERT Class 11 Maths Chapter 5 notes - Download PDF

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NCERT Class 11 Maths Chapter 4 Notes Complex Numbers and Quadratic Equations

Complex Numbers and Quadratic Equations help us solve equations that have no real solutions by introducing the imaginary number $i$. This chapter explores how complex numbers work and how they are used to find the roots of quadratic equations.

Complex Number

A complex number is defined as the combination of real and imaginary numbers.

Example: $3+4i$

Here, $3$ is real and $4i$ is the imaginary part.

Here is $i =\sqrt{(-1)}$, then we can say that $i^2 = -1$

Algebra Of Complex Numbers

Addition Of Two Complex Numbers

Let $x=a+ib$ and $y=c+id$

Therefore, $x+y= (a+c) +i(b+d)$

Properties

• The closure law: The sum of two complex numbers is also a complex number

• The commutative law $x+y=y+x$

• The associative law $(x+y)+z=x+(y+z)$

Difference Of Two Complex Numbers
Let $x=a+ib$ and $y=c+id$

Therefore $x-y= x+(-y)$

Multiplication of Two Complex Numbers
Let $x=a+ib$ and $y=c+id$

Therefore, $xy=ac-bd+i(ad+bc)$

Properties

• The closure law. The multiplication of two complex numbers is also a complex number

• The commutative law $xy=yx$

• The associative law $(xy)z=x(yz)$

• The distributive law $x(y+z) = xy+xz$ and $(x+y)z=xz+yz$

Division Of Two Complex Numbers

Let $x=a+ib$ and $y=c+id$

Therefore $\frac{x}{y}$ = $x.(\frac{1}{y})$

Power of $i$

In general, any integer k

$i^{4 k}=1, i^{4 k+1}=i, i^{4 k+2}=-1, i^{4 k+3}=-i$

The square roots of a negative real number

$i^2=-1, \text { and }(-i)^2=i^2=-1$

The Modulus And The Conjugate Of A Complex Number

If $x=a+i b$, then $|x|=\sqrt{a^2+b^2}$

Properties

1. $|x y|=|x||y|$

2. $|\frac{x}{y}|=\frac{|x|}{|y|}$

3. $(xy)'=x'y'$

4. $(x±y)’=x’±y’$

5. $(\frac{x}{y})’=\frac{x’}{y’}$

Argand Plane And Polar Representation

The complex number $a+ib$ corresponding to ordered pairs $(a, b)$ can also be represented geometrically in the XY plane.

In the Argand plane, the modulus of a complex number $a+ib =\sqrt{(a^2+b^2)}$ is the distance from the origin to the point.

Polar Representation Of A Complex Number

We note that the point P is uniquely determined by the ordered pair of real numbers $(r, θ)$ called the polar coordinates of the point P.

We get $a=r\cosθ, b=r\sinθ,$ thus $x=r(\cosθ +i \sin θ)$

Example:

Express the following in the form $\mathrm{a}+\mathrm{ib}: i^{-35}$
Solution : $i^{-35}=\frac{1}{i^{35}}=\frac{1}{\left(i^2\right)^{17} i}=\frac{1}{-i} × \frac{i}{i}=i$

Quadratic Equation

The equation that consists of the highest power in the polynomial is 2.

Let $ax^2+bx+c=0$ be a quadratic equation, then its discriminant is

$D=b^2-4ac$

Let us assume that $D<0$

Then $x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$

Example:

Solve $x^2+x+1=0$

Solution:

Here, $b^2-4ac=1^2-4×1×1=-3<0$

$\begin{aligned} & \text { Therefore } x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\ &= \frac{-1 \pm \sqrt{1^2-4 × 1 × 1}}{2 × 1}=\frac{-1 \pm \sqrt{3} i}{2}\end{aligned}$

This concludes the Complex Numbers and Quadratic Equations class 11 notes.

Complex Numbers And Quadratic Equations: Previous Year Question and Answer

Given below are some previous year question answers of various examinations from the NCERT class 11 chapter 4 Complex Numbers And Quadratic Equations:

Question 1: 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.

Question 2: Consider the equation $\mathrm{x}^2+4 \mathrm{x}-\mathrm{n}=0$, where $\mathrm{n} \in[20,100]$ is a natural number. Then the number of all distinct values of n, for which the given equation has integral roots, is equal to

Solution:
Let us simplify the equation and solve for $x$,
$\begin{aligned}
& x^2+4 x+4=n+4 \\
& (x+2)^2=n+4 \\
& x=-2 \pm \sqrt{n+4} \\
& \because 20 \leq n \leq 100 \\
& \sqrt{24} \leq \sqrt{n+4} \leq \sqrt{104} \\
& \Rightarrow \sqrt{n+4} \in\{5,6,7,8,9,10\}
\end{aligned}$
$\therefore$ $6$ integral values of $n$ are possible.
Hence, the correct answer is $6$.

Question 3: Let $\mathrm{z} \in \mathrm{C}$ be such that $\frac{\mathrm{z}^2+3 \mathrm{i}}{\mathrm{z}-2+\mathrm{i}}=2+3 \mathrm{i}$. Then the sum of all possible values of $z^2$ is:

Solution:
Given:
$\frac{\mathrm{z}^2+3 \mathrm{i}}{\mathrm{z}-2+\mathrm{i}}=2+3 \mathrm{i}$
Apply cross multiplication,
$z^2+3 i=z(2+3 i)-7-4 i$
$z^2-z(2+3 i)+7+7 i=0$
Since it is quadratic, say it has two roots, $z_1,z_2$
$\begin{aligned} & z_1^2+z_2^2=\left(z_1+z_2\right)^2-2 z_1 z_2 \\ & =4-9+12 i-14-14 i \\ & =-19-2 i\end{aligned}$
Hence, the correct answer is $-19-2 \mathrm{i}$.

NCERT Class 11 Maths Notes – Chapter-Wise Links

For the convenience of students, Careers360 provides complete NCERT Class 11 Maths notes together in one location. Simply click the links below to access.

Subject-Wise NCERT Exemplar Solutions

After finishing the textbook exercises, students can use the following links to check the NCERT exemplar solutions for a better understanding of the concepts.

• NCERT Exemplar Class 11 Solutions
• NCERT Exemplar Class 11 Maths
• NCERT Exemplar Class 11 Physics
• NCERT Exemplar Class 11 Chemistry
• NCERT Exemplar Class 11 Biology

Subject-Wise NCERT Solutions

Students can also check these well-structured, subject-wise solutions.

• NCERT Solutions for Class 11 Mathematics
• NCERT Solutions for Class 11 Chemistry
• NCERT Solutions for Class 11 Physics
• NCERT Solutions for Class 11 Biology

NCERT Books and NCERT Syllabus

Before the start of a new academic year, students should refer to the latest syllabus to determine the chapters they’ll be studying. Below are the updated syllabus links, along with some recommended reference books.

• NCERT Books Class 11 Maths
• NCERT Syllabus Class 11 Maths
• NCERT Books Class 11
• NCERT Syllabus Class 11