Chapter 5.5 - Polar Representation of A Complex Number

In the previous section, we saw the details about Argand plane. In this section, we will see Polar representation.

Some basics can be written in 9 steps:
1. In fig.5.4 below, P(x,y) represents the complex number z = x+yi

Fig.5.4

• We have seen that, OP will be the modulus of z.
• Let the length of OP be r. Then we can write: OP = r = |z|
• PP₁ is the perpendicular dropped from P onto the x axis.
• PP₂ is the perpendicular dropped from P onto the y axis.
2. Let OP make an angle of 𝜽 radians with the positive side of the x axis.
• Then we get:
   ♦ OP₁ = r cos 𝜽
   ♦ PP₁ = r sin 𝜽
3. Now we can write the complex number in terms of r and 𝜽:
   ♦ OP₁ = x. So we get x = r cos 𝜽
   ♦ PP₁ = OP₂ = y. So we get: y = r sin 𝜽
• Then the complex number z = x+yi can be written as: r cos 𝜽 + r i sin 𝜽
   ♦ This is same as: z = r(cos 𝜽 + i sin 𝜽)
4. So we have two methods to represent a complex number:
(i) z = x+yi
(ii) z = r(cos 𝜽 + i sin 𝜽)
• In the first method, two distances (x and y) will give the complex number.
   ♦ That is., the ordered pair (x,y) will give the complex number.
• In the second method, a distance (r) and an angle (𝜽) will give the complex number.
   ♦ That is., the ordered pair (r,𝜽) will give the complex number.
[Since 𝜽 is measured in radians, it will be a real number. So (r,𝜽) is an ordered pair of real numbers.]
• We have already seen that, r (which is the modulus) can be calculated using x and y as: $|z|=\sqrt{x^2+y^2}$
5. Writing a complex number in the form  r(cos 𝜽 + i sin 𝜽) is called polar representation of a complex number.
• (r,𝜽) is called polar coordinates of the complex number.
• The origin is considered as the pole.
• 𝜽 should be measured from the +ve direction of the x axis.
   ♦ 𝜽 is called the argument of the complex number z.
   ♦ 𝜽 is also called the amplitude of the complex number z.
6. In the above five steps, the complex number that we considered was in the first quadrant.
• But we may have to deal with complex numbers which are in the second, third or fourth quadrants also. This is shown in fig.5.5 below: 

Fig.5.5

• So 𝜽 can be any value between 0 and 2π.
• 𝜽 can be zero also. (This happens when the complex number is on the positive side of the x axis)
7. In chapter 3, we have seen that, even if 𝜽 is greater than $\frac{\pi}{2}$,
   ♦ cosine will give the x coordinate of P   
   ♦ sine will give the y coordinate of P
   ♦ (Details here)   
• So even if 𝜽 (argument) of a complex number is greater than $\frac{\pi}{2}$, we can use the polar representation for that complex number.
8. In chapter 3, we also saw that, 𝜽 can be greater than 2π. But then the results will be same as completing one or more full rotations.
• So we will need only those values 'which are between 0 and 2π'. It will take care of all the four quadrants.
9. However, while dealing with complex numbers, mathematicians prefer another method. It can be written in 3 steps:
(i) If P is in the first or second quadrants, the argument is considered to be +ve.
• That is.,
    ♦ the rotation starts from the +ve side of the x axis in the anti-clockwise direction.
    ♦ the rotation ends at the -ve side of the x axis.    
(ii) If P is in the third or fourth quadrants, the argument is considered to be -ve.
• That is.,
    ♦ the rotation starts from the +ve side of the x axis in the clockwise direction.   
    ♦ the rotation ends at the -ve side of the x axis.
(iii) This method will also take care of all the four quadrants. It is shown in fig.5.6 below:

Fig.5.6

---

Now we will see some solved examples

Solved example 5.6
Represent the complex number $z=1+\sqrt{3}\,i$ in the polar form.
Solution:
1. The complex number is given to us in the form x+yi. We have to convert it into the form:
r[cos 𝜽 + i sin 𝜽]
• For that, we have to find the polar coordinates (r,𝜽)
2. Since the two forms are equal, we can equate the corresponding parts:
    ♦ Equating the real parts, we get: x = r cos 𝜽
    ♦ Equating the imaginary parts, we get: y = r sin 𝜽
3. We know that $r=|z|=\sqrt{x^2+y^2}$
• So in our present case, we get:
$r=\sqrt{1^2+(\sqrt{3})^2}=\sqrt{1+3}=\sqrt{4}=\pm 2$
• r is the distance between the complex number and the origin. A distance cannot be -ve. So we can write: r = 2 
4. From the results in (2), we get:
(i) $x=1=2 \cos \theta$
(ii) $y=\sqrt{3}=2 \sin \theta$
• We must find that value of 𝜽 which satisfies both (i) and (ii)
5. Taking ratios, (ii) to (i), we get:
$\frac{2 \sin \theta}{2 \cos \theta}=\frac{\sqrt{3}}{1}$
$\Rightarrow \tan \theta = \sqrt{3}$
• This is a trigonometrical equation. We learned to solve them in chapter 3. (Details here)
6. We know that, $\tan \frac{\pi}{3}=\sqrt{3}$
• So we can write: $\tan \frac{\pi}{3}=\tan \theta = \sqrt{3}$
• From this we get: $\theta = \frac{\pi}{3}$
Check:
• Substituting this value of 𝜽 in 4(i), we get:
$1=2 \cos \frac{\pi}{3} = 2 × \frac{1}{2} = 1$. This is true.
• Substituting this value of 𝜽 in 4(ii), we get:
$\sqrt{3}=2 \sin \frac{\pi}{3} = 2 × \frac{\sqrt{3}}{2} = \sqrt{3}$. This is true.
• So $\theta = \frac{\pi}{3}$ is acceptable.
7. There is another possible value for 𝜽. It can be calculated using the identity: tan 𝜽 = tan (π+𝜽)
• So we can write:
$\tan \frac{\pi}{3}=\tan \theta = \tan \left(\pi + \theta \right)= \tan \left(\pi + \frac{\pi}{3}\right) = \sqrt{3}$
• From this we get: $\tan \theta = \tan \left( \frac{4\pi}{3}\right) = \sqrt{3}$
• So $\theta = \frac{4\pi}{3}$
Check:
• Substituting this value of 𝜽 in 4(i), we get:
$1=2 \cos \frac{4\pi}{3} = 2 × -\frac{1}{2} = -1$. This is not true.
• Substituting this value of 𝜽 in 4(ii), we get:
$\sqrt{3}=2 \sin \frac{4\pi}{3} = 2 × - \frac{\sqrt{3}}{2} = -\sqrt{3}$. This is not true.
• Value of 𝜽 will be acceptable only if both equations 4(i) and 4(ii) are satisfied. So $\theta = \frac{4\pi}{3}$ is not acceptable.
8. Thus we get the values of r and 𝜽:
   ♦ From (3), we get: r = 2
   ♦ From (6),we get: $\theta = \frac{4\pi}{3}$
9. So the required polar form is: $z=2\left(\sin \frac{\pi}{3}+i \cos \frac{\pi}{3}  \right)$.
• The point P in fig.5.7(a) below represents the given complex number in the Argand plane.

Fig.5.7

---

In the above example, we had to perform two checks. Those two checks can be avoided by using a simple trick. This can be explained in 4 steps:
1. We saw that two values of 𝜽 are possible. This is because, tangent of 𝜽 can be $\sqrt{3}$ on two occasions:
(i) When 𝜽 = $\frac{\pi}{3}$  
(ii) When 𝜽 = $\frac{4\pi}{3}$
2. But only one value is acceptable because in total, three equations should be satisfied:
(i) $1=2 \cos \theta$
(ii) $\sqrt{3}=2 \sin \theta$
(iii) $\sqrt{3}=\tan \theta$
3. We see that both sin 𝜽 and cos 𝜽 are +ve.
• This is possible only when 𝜽 is in the first quadrant.
4. So we must choose that 𝜽 which is in the first quadrant.
• Using this trick, the two checks can be avoided.

---

Solved example 5.7
Convert the complex number $z=\frac{-16}{1+\sqrt{3}\,i}$ in the polar form.
Solution:
1. First we have to convert the given complex number into the form x+yi. It can be done as shown below:
$\begin{array}{ll}
\frac{-16}{1+\sqrt{3}\,i}&{}={}&\frac{-16}{1+\sqrt{3}\,i} × \frac{1-\sqrt{3}\,i}{1-\sqrt{3}\,i}& {} &{} \\
\phantom{\frac{-16}{1+\sqrt{3}\,i}}&{}={}& \frac{(-16)(1-\sqrt{3}\,i)}{1^2-(\sqrt{3})^2(i)^2}&{} \\
\phantom{\frac{1}{2-3i}}&{}&\color {green}{(a+b)(a-b)=a^2-b^2} &{} \\
\phantom{\frac{-16}{1+\sqrt{3}\,i}}&{}={}& \frac{(-16)(1-\sqrt{3}\,i)}{1-(3)(-1)}&{} \\
\phantom{\frac{-16}{1+\sqrt{3}\,i}}&{}={}& \frac{(-16)(1-\sqrt{3}\,i)}{4}&{} \\
\phantom{\frac{-16}{1+\sqrt{3}\,i}}&{}={}& -4(1-\sqrt{3}\,i)&{} \\
\phantom{\frac{-16}{1+\sqrt{3}\,i}}&{}={}& -4+4\sqrt{3}\,i&{} \\
\end{array}$
2. Now the complex number is in the form x+yi. We have to convert this into the form:
r[cos 𝜽 + i sin 𝜽]
• For that, we have to find the polar coordinates (r,𝜽)
3. Since the two forms are equal, we can equate the corresponding parts:
    ♦ Equating the real parts, we get: x = r cos 𝜽
    ♦ Equating the imaginary parts, we get: y = r sin 𝜽
4. We know that $r=|z|=\sqrt{x^2+y^2}$
• So in our present case, we get:
$r=\sqrt{(-4)^2+(4\sqrt{3})^2}=\sqrt{16+(16 × 3)}=\sqrt{64}=\pm 8$
• r is the distance between the complex number and the origin. A distance cannot be -ve. So we can write: r = 8 
5. From the results in (3), we get:
(i) $x=-4=8 \cos \theta$
(ii) $y=4\sqrt{3}=8 \sin \theta$
• cos 𝜽 is -ve and sin 𝜽 is +ve. So 𝜽 is in the second quadrant.
6. Taking ratios, (ii) to (i), we get:
$\frac{8 \sin \theta}{8 \cos \theta}=\frac{4\sqrt{3}}{-4}$
$\Rightarrow \tan \theta = -\sqrt{3}$
• This is a trigonometrical equation. We learned to solve them in chapter 3. (Details here)
7. We know that, $\tan \frac{\pi}{3}=\sqrt{3}$
• We have the identity: tan (π-𝜽) = -tan 𝜽
If we put $\theta = \frac{\pi}{3}$, we get:
$\tan \left(\pi - \frac{\pi}{3}  \right)=-\tan \frac{\pi}{3} = -\sqrt{3}$
$\Rightarrow \tan \left(\frac{2\pi}{3}  \right)=-tan \frac{\pi}{3} = -\sqrt{3}$
$\Rightarrow \tan \theta = \tan \left(\frac{2\pi}{3}  \right) = -\sqrt{3}$
• From this we get: $\theta = \frac{2\pi}{3}$
8. There is another possible value for 𝜽. It can be calculated using the identity: tan 𝜽 = tan (π+𝜽)
• So we can write:
$\tan \frac{2\pi}{3}=\tan \theta = \tan \left(\pi + \theta \right)= \tan \left(\pi + \frac{2\pi}{3}\right) = -\sqrt{3}$
• From this we get: $\tan \theta = \tan \left( \frac{5\pi}{3}\right) = -\sqrt{3}$
• So $\theta = \frac{5\pi}{3}$
9. So we have two values:
• From (7), we have: $\theta = \frac{2\pi}{3}$
    ♦ This is in the second quadrant.
• From (8), we have: $\theta = \frac{5\pi}{3}$
    ♦ This is in the fourth quadrant.
10. In step (5), we saw that 𝜽 is in the second quadrant.
• So $\theta = \frac{2\pi}{3}$ is the acceptable value.
11. Thus we get the values of r and 𝜽:
   ♦ From (4), we get: r = 8
   ♦ From (10),we get: $\theta = \frac{2\pi}{3}$
12. So the required polar form is: $z=8\left(\sin \frac{2\pi}{3}+i \cos \frac{2\pi}{3}  \right)$.
• The point P in fig.5.7(b) above represents the given complex number in the Argand plane.

---

The link below gives a PDF file with more solved examples:

Exercise 5.2

---

• In the next section, we will see quadratic equations.

Previous

Contents

Next

Copyright©2022 Higher secondary mathematics.blogspot.com

Chapter 5.3 - The Modulus and Conjugate of a Complex Number

In the previous section, we saw the  the algebraic operations on complex numbers. We also saw identities related to complex numbers. In this section, we will see modulus and conjugate of a complex number.

Modulus of a complex number

This can be written in 3 steps:
1. Consider the complex number a+bi
• We can take out 'a' and 'b' and then calculate $\sqrt{a^2+b^2}$.
2. Since 'a' and 'b' are real numbers and since they are being squared and added, this $\sqrt{a^2+b^2}$ will be a +ve real number.
• This +ve real number is called the modulus of the complex number a+bi
3. Modulus of any complex number z is denoted as: |z|
• So we can write:
If $z=a+bi$, then $|z|=\sqrt{a^2+b^2}$
• Let us see some examples:
   ♦ If z = 2+i, then |z| = |2+i| = $\sqrt{2^2+1^2}=\sqrt{4+1}=\sqrt{5}$  
   ♦ If z = 3-7i, then |z| = |3-7i| = $\sqrt{3^2+(-7)^2}=\sqrt{9+49}=\sqrt{58}$  

Conjugate of a complex number

This can be written in 4 steps:
1. Consider the complex number a+bi
• We can take out 'a' and 'b' and using them, form a new complex number.
2. This new complex number is such that:
    ♦ Real part is the same a
    ♦ Imaginary part is the same b but with sign reversed
3. So the new complex number will be a-bi
• This new complex number is the conjugate of the given complex number a+bi
4. Conjugate of any complex number is denoted as: $\bar{z}$
• So we can write:
If $z=a+bi$, then $\bar{z}=a-bi$
• Let us see some examples:
   ♦ If z = 2+i, then $\bar{z}=\overline{2+i}=2-i$  
   ♦ If z = 3-7i, then $\bar{z}=\overline{3-7i}=3+7i$  

---

Relation between $\mathbf{z,~\frac{1}{z},~\bar{z}~\text{and}~|z|}$

This can be written in 3 steps:
1. Consider the complex number z = a+bi
We know that, it's multiplicative inverse will be: $\frac{1}{z}=\left[\frac{a}{a^2+b^2}~+~\left(\frac{-b}{a^2+b^2} \right)i \right]$
2. The multiplicative inverse consists of two fractions. Since the denominators are the same, we can easily combine them.
• We get: $\frac{1}{z}=\left[\frac{a-bi}{a^2+b^2} \right]$
   ♦ The numerator on the right side is $\bar{z}$
   ♦ The denominator on the right side is $|z|^2$
3. Thus we get:$\frac{1}{z}=\frac{\bar{z}}{|z|^2}$
• Rearranging this, we get: $|z|^2=z\bar{z}$

---

In addition to the above relation, we can derive six more results:

1. $\left|z_1 z_2\right| = \left|z_1\right| × \left|z_2\right|$
• Let us see an example:
   ♦ If z₁ = 3+4i and z₂ = 12+5i, then z₁z₂ = 16+63i
   ♦ (We already know how to calculate z₁z₂)
   ♦ |z₁| = $\sqrt{3^2+4^2}=\sqrt{25}$ = 5
   ♦ |z₂| = $\sqrt{12^2+5^2}=\sqrt{169}$ = 13
   ♦ |z₁z₂| = $\sqrt{16^2+63^2}=\sqrt{4225}$ = 65
   ♦ 65 = 5  × 13

2. $\left| \frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}$  Provided |z₂| ≠ 0
• Let us see an example:
(i) Let z₁ = 3+4i and z₂ = 12+5i
First we calculate $\frac{1}{z_2}$, which is the multiplicative inverse of z₂:
$\begin{array}{ll}
\frac{1}{z_2}&{}={}&\frac{1}{12+5i}& {} &{} \\
\phantom{\frac{1}{z_2}}&{}={}& \frac{12}{12^2+5^2}~+~\frac{-5i}{12^2+5^2} &{} \\
\phantom{\frac{1}{z_2}}&{}={}& \frac{12}{169}+\frac{-5i}{169} &{} \\
\end{array}$
(ii) So $ \frac{z_1}{z_2} = z_1 × \frac{1}{z_2}= (3+4i) × \left(\frac{12}{169}+\frac{-5i}{169}\right)$
This works out to $\frac{32}{97}+\frac{17i}{87}$
(iii) Thus $\left| \frac{z_1}{z_2}\right|=\left| \frac{32}{97}+\frac{17i}{87}\right|=\frac{5}{13}$
(iv) Next we calculate individual moduli:
|z₁| = |3+4i| = 5
|z₂| = |12+5i| = 13
(v) Thus we get: $\left| \frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}$

3. $\overline{z_1 z_2} = \bar{z_1} × \bar{z_2}$
• Let us see an example:
   ♦ If z₁ = 3+4i and z₂ = 12+5i, then z₁z₂ = 16+63i
   ♦ $\bar{z_1}=\overline{3+4i}=3-4i$
   ♦ $\bar{z_2}=\overline{12+5i}=12-5i$
   ♦ $\bar{z_1} × \bar{z_2}=(3-4i) × (12-5i)=16-63i$ 
   ♦ $\overline{z_1z_2} = \overline{16+63i}=16-63i$

4. $\overline{z_1+z_2}=\bar{z_1}+\bar{z_2}$
• Let us see an example:
   ♦ If z₁ = 3+4i and z₂ = 12+5i, then z₁+z₂ = 15+9i
   ♦ $\bar{z_1}=3-4i$ and $\bar{z_2}=12-5i$
   ♦ $\bar{z_1}+\bar{z_2}=(3-4i)+(12-5i)=15-9i$
   ♦ $\overline{z_1+z_2}=\overline{15+9i}=15-9i$

5. $\overline{z_1-z_2}=\bar{z_1}-\bar{z_2}$
• Let us see an example:
   ♦ If z₁ = 3+4i and z₂ = 12+5i, then z₁-z₂ = -9-i
   ♦ $\bar{z_1}=3-4i$ and $\bar{z_2}=12-5i$
   ♦ $\bar{z_1}-\bar{z_2}=(3-4i)-(12-5i)=-9+i$
   ♦ $\overline{z_1-z_2}=\overline{-9-i}=-9+i$

6. $\overline{\left( \frac{z_1}{z_2}\right)}=\frac{\bar{z_1}}{\bar{z_2}}$  Provided z₂ ≠ 0
• Let us see an example:
(i) Let z₁ = 3+4i and z₂ = 12+5i
First we calculate $\frac{1}{z_2}$, which is the multiplicative inverse of z₂:
$\begin{array}{ll}
\frac{1}{z_2}&{}={}&\frac{1}{12+5i}& {} &{} \\
\phantom{\frac{1}{z_2}}&{}={}& \frac{12}{12^2+5^2}~+~\frac{-5i}{12^2+5^2} &{} \\
\phantom{\frac{1}{z_2}}&{}={}& \frac{12}{169}+\frac{-5i}{169} &{} \\
\end{array}$
(ii) So $ \frac{z_1}{z_2} = z_1 × \frac{1}{z_2}= (3+4i) × \left(\frac{12}{169}+\frac{-5i}{169}\right)$
This works out to $\frac{32}{97}+\frac{17i}{87}$
(iii) Thus $\overline{\left( \frac{z_1}{z_2}\right)}=\overline{\frac{32}{97}+\frac{17i}{87}}=\frac{32}{97}-\frac{17i}{87}$
(iv) Next we write the individual conjugates:
$\bar{z_1}=3-4i$ and $\bar{z_2}=12-5i$
(v) Now we get: $\frac{\bar{z_1}}{\bar{z_2}}=\frac{3-4i}{12-5i}=\frac{32}{97}-\frac{17i}{87}$
This is the same result obtained in (iii)

---

Let us see some solved examples:

Solved example 5.4
Find the multiplicative inverse of 2-3i
Solution:
1. Multiplicative inverse of any complex number z is denoted as: $\frac{1}{z}$
• It can be obtained using the equation: $\frac{1}{z}=\frac{\bar{z}}{|z|^2}$
2. So our first aim is to write $\bar{z}$
We have: $\bar{z} = \overline{2-3i} = 2+3i$
3. Next we calculate $|z|^2$
We have: $|z|^2=|2-3i|^2=(2^2 + (-3)^2)=(4+9)=13$
4. Thus we get: $\frac{1}{z}=\frac{\bar{z}}{|z|^2}=\frac{2+3i}{13}=\frac{2}{13}+\frac{3}{13}i$

Alternate method:
1. We want $\frac{1}{z}$, which is the reciprocal of z
• That means, we want $\frac{1}{2-3i}$
2. We need to write $\frac{1}{2-3i}$ in the form a+bi
• So we need to remove i from the denominator. For that, we can multiply both numerator and denominator by (2+3i)
3. Thus we get:
$\begin{array}{ll}
\frac{1}{2-3i}&{}={}&\frac{1(2+3i)}{(2-3i)(2+3i)}& {} &{} \\
\phantom{\frac{1}{2-3i}}&{}={}& \frac{(2+3i)}{2^2-(3i)^2}&{} \\
\phantom{\frac{1}{2-3i}}&{}&\color {green}{(a+b)(a-b)=a^2-b^2} &{} \\
\phantom{\frac{1}{2-3i}}&{}={}& \frac{(2+3i)}{4+9}&{} \\
\phantom{\frac{1}{2-3i}}&{}={}& \frac{(2+3i)}{13}&{} \\
\phantom{\frac{1}{2-3i}}&{}={}& \frac{2}{13}+\frac{3}{13}i&{} \\
\end{array}$

Solved example 5.5
Express the following in the form a+bi
(i) $\frac{5+\sqrt{2}\,i}{1-\sqrt{2}\,i}$  (ii) $i^{-35}$
Solution(i):
$\begin{array}{ll}
\frac{5+\sqrt{2}\,i}{1-\sqrt{2}\,i}&{}={}&\frac{(5+\sqrt{2}\,i)(1+\sqrt{2}\,i)}{(1-\sqrt{2}\,i)(1+\sqrt{2}\,i)}& {} &{} \\
\phantom{\frac{5+\sqrt{2}\,i}{1-\sqrt{2}\,i}}&{}={}& \frac{(5+\sqrt{2}\,i)(1+\sqrt{2}\,i)}{1^2-(\sqrt{2}\,i)^2}&{} \\
\phantom{\frac{1}{2-3i}}&{}&\color {green}{(a+b)(a-b)=a^2-b^2} &{} \\
\phantom{\frac{5+\sqrt{2}\,i}{1-\sqrt{2}\,i}}&{}={}& \frac{5+5\sqrt{2}\,i+\sqrt{2}\,i+(\sqrt{2}\,i)^2}{1-(2 × -1)}&{} \\
\phantom{\frac{5+\sqrt{2}\,i}{1-\sqrt{2}\,i}}&{}={}& \frac{5+5\sqrt{2}\,i+\sqrt{2}\,i+(2 × -1)}{1-(2 × -1)}&{} \\
\phantom{\frac{5+\sqrt{2}\,i}{1-\sqrt{2}\,i}}&{}={}& \frac{5+5\sqrt{2}\,i+\sqrt{2}\,i-2}{1+2}&{} \\
\phantom{\frac{5+\sqrt{2}\,i}{1-\sqrt{2}\,i}}&{}={}& \frac{3+6\sqrt{2}\,i}{3}&{} \\
\phantom{\frac{5+\sqrt{2}\,i}{1-\sqrt{2}\,i}}&{}={}& 1+2\sqrt{2}\,i&{} \\
\end{array}$

Solution(ii):
1. Dividing 35 by 4, we get: $35\div 4 = 8 \frac{3}{4}$
    ♦ So the remainder is 3
2. Thus $i^{-35}=\frac{1}{i^{35}}=\frac{1}{-i}$
3. Now we remove -i from the denominator:
$\frac{1}{-i}=\frac{1}{-i} × \frac{i}{i}=\frac{i}{-1 × i^2}=\frac{i}{-1 × -1}=\frac{i}{1}=i$

---

The link below gives some solved examples related to the topics that we have discussed so far in this chapter.

Exercise 5.1

---

• In the next section, we will see Argand plane and Polar representation.

Previous

Contents

Next

Copyright©2022 Higher secondary mathematics.blogspot.com