Application

To

The Chief MinisterHimachal PradeshShimla

Sub :-  Request for transfer as Principal (School cadre)

 Respected Sir   I want to draw your kind attention toward these facts that:-        I have been serving as Principal at Govt Sen Sec School TIKKARI NEWAL Distt SHIMLA since Feb 15th 2021.    I had also served in hard area as a Lecturer in Mathematics in Govt Senior Sec School Shillai  Distt. Sirmour for more than 3 years.    My wife is not medically fit and she is taking medical treatment with Neurosurgeon at Tanda Medical College.    This station is approximately at the distance of 500 km from my home station and of 2 days  journey by bus.    So in the light of above facts I request you to transfer me from Govt Sen Sec School TIKKARI NEWAL Distt SHIMLA against vacant post at Govt Sen Sec School BHARARPAT  Distt MANDI being caused on 31st March 2023 in relaxation of ban on transfer and in condonation of short stay.    With ThanksYours Sincerely



Rajesh KumarPrincipalGSS TIKKARI NEWALDistt Shimla

Home AddressVPO MAHAKALDistt KANGRA   H.P.

Contact No 94181-21725

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HOW TO UNDERSTAND RELATIONS AND FUNCTIONS ,INVERSE OF A FUNCTION

we are going to  discuss Relations and Functions , "How to understand  Relations and  Functions, Inverse of a Function" under the topic  Relations and Functions.

Ordered-Pair Numbers :-

Ordered-pair number is written within a set of parentheses and separated by a comma.

For example, (5, 6) is an ordered-pair number; the order is designated by the first element 5 and the second element 6. The pair (3, 6) is not the same as (6,3) because they have different order. Sets of ordered-pair numbers can represent relations or functions.

Example of ordered pair :

(3,8),(2,1),(7,6)

Relation

A relation is a  set of ordered-pair numbers.

consider the following table

________________________________________________________________________

Numbers of students      1             2          3         4           5          6

_______________________________________________________________________

Marks Obtained             96          98       97         78        77         86

_______________________________________________________________________

In the above table the numbers of students and marks obtained by them  is a relation and can be written as a set of ordered-pair numbers.

A= {(1, 96), (2, 98), (3, 97), (4, 88),(5,77),(6,86)}

When we collect all the elements written in 1st column of the ordered pairs and placed in a set then the set so formed is called  Domain of the relation.
The domain of A= {1, 2, 3, 4,5,6}

As all the elements written in 2nd column of the ordered pairs and placed in a set then the set so formed is called  Range of the relation.

The range of A = {  96,98,97,88,77,86}

we can better understand this concept with the help of this video

Function

A function is a relation in which every first element in ordered pairs have unique second element associated with them. Second  elements may or may not be same.

---------------------------------------------------------

Example

 {(1, 2), (2, 3), (3, 4), (4, 5),(5,6)}  is an example of function 

 { (1, 2), (2, 3), (3, 4), (4, 5),(5,6) } is a function because all the  first elements are different.

Example

{(1, 3), (3, 3), (2, 1), (4, 2)}  is an example of function 

 {(1, 3), (2, 3), (2, 1), (4, 2)}  is a function because all the first elements are different.

Example

{ (1, 6), (2, 5), (1, 9), (4, 3) }  is not an  example of function 

As in  {(1, 6), (2, 5), (1, 9), (4, 3)}  the element "1 "   appeared twice .

Example

{(2, 15), (3, 15), (4, 15), (5, 13),(6,18)}  is  an  example of function 

As in  {(2, 15), (3, 15), (4, 15), (5, 15)}   all the first elements are different.

Example

{(1, 1), (-1, 1),(2,4),(-2,4), (3, 9), (-3, 9),(4,16),(-4,16)}  is an  example of function although   the element "1" and "-1" ,"2" and "-2" , "3" and "-3" ,"4","-4" have same images. This is an example of many one function.

Question:-   Find x and y if: 

(i) (5x + 3, y) = (4x + 5,  2)

(ii) (x – y, x + y) = (8, 12)
(iii) ( 2x-y , y+5 ) = ( -2,3 )

Solution

(1)  Given  (5x + 3 , y) = (4x + 5, 2)
So By the equality of ordered pair elements,
1st element of the ordered number written on the left hand side will be equal to the 1st element of the ordered pair number written on the  right hand side . Therefore 

5x + 3 = 4x + 5   and y =  2 
5x-4x = 5 -3   and y = 2 
x = 2 and y = 2

(ii) So By the equality of ordered pair elements
x – y = 8 and  x + y = 12

Solving these two equations for x and y 

 2x =20  and    10+ y =12 

x=10   y = 2

(iii) So By the equality of ordered pair elements
2x-y  =-2  , y+5 = 3 
2x = -2+y  , y = 3-5
2x = -2+y  , y = -2
Putting the value of y in 1st Equation ,we get
2x = -2 - 2
2x = -4
x = -2
so x= -2 and y =-2

Types of Relations

A relation R in a set A is called

(i) reflexive, if (a, a) ∈ R, for every a ∈ A,

(ii) symmetric, if (a, b) ∈ R implies that (a, b) ∈ R, for all a,b ∈ A.

(iii) transitive, if (a, b) ∈ R and (b, c) ∈ R implies that (a, c) ∈ R, for all a, b,c ∈ A.

Equivalence Relation

A relation R in a set A is said to be an equivalence  relation if R is reflexive, symmetric and transitive.

1 ) Let B be the set of all triangles in a plane with R a relation in B given by

R = {(T1, T2) : T1 is congruent to T2}. Then R is an equivalence relation.

2 ) Let R be the relation defined in the set A = {1, 2, 3, 4, 5, 6, 7}  by

R = {(a, b) : both a and b are either odd or even}. Then R is an equivalence

one-one Function

A function f : X → Y is defined to be one-one (or injection ), if the images of distinct elements of X under f are distinct, i.e., for every x, y ∈ X, f (x) = f (y) implies x = y. Otherwise, f is called many-one.

Onto Function

A function f : X → Y is said to be onto (or surjective), if every element of Y is the image of some element of X under f, i.e., for every y ∈ Y, there exists an

element x in X such that f (x) = y.

Example

1   Function f : R → R, given by f (x) = 2x, is one-one and Onto As all the elements  have only one and uniqe image under f.

2  Function f : N → N, given by f (x) = 2x, is one-one but not onto.Because  the elements  have only one and unique image under f Therefore it is one one function .But not all elements of N have image under f 

e. g .  1,3,5,7... are not the image of any elements of N under f so it is not onto function

Example

The function f : N → N, given by f (1) = f (2) = 1 and f (x) = x – 1,

for every x > 2, is onto but not one-one.

Solution

Since f is Not one-one, as f (1) = f (2) = 1. 

But f is Onto, as given any y ∈ N, y ≠ 1,

Choose x = y + 1 s.t.

 f (y + 1) = y + 1 – 1

f (y + 1)  = y. 

Also for 1 ∈ N, 

we are given  f (1) = 1

Inverse of a Function

A function f : X → Y is defined to be invertible, if there exists a function g : Y → X such that gof = IX and fog = IY. The function g is called the inverse of f and is denoted by f –1

Example

Let S = {1, 2, 3}. Determine whether the functions f : S → S defined as below have inverses. Find f , if it exists.

(a) f = {(1, 1), (2, 2), (3, 3)}

(b) f = {(2, 2), (3, 1), (4, 1)}

(c) f = {(1, 5), (3, 4), (2, 1)}

Solution

(a) It is to  proved that  f is one-one and onto Hence f is invertible with the inverse f –1 of  f given by f –1 = {(1, 1), (2, 2), (3, 3)} = f.

(b) Since f (3) = f (4) = 1, f is not one-one, so that f is not invertible.

(c) Here  f   is one-one and onto, so that f is invertible with

 f –1 = {(5, 1), (4, 3), (1, 2)}.

Composition of Functions

Let f : A → B and g : B → C be two functions. Then the composition of f and g, denoted by gof, is defined as the function gof : A → C given by

gof (x) = g(f (x)), ∀ x ∈ A

Example

fof(x) = (16x + 12 + 18x -12 ) / ( 24x + 18 - 24x +16)

fof(x) = (34 x ) / ( 34)

fof(x) =  x  =  I(x)

Example

Let f : {2, 3, 4, 5} → {3, 4, 5, 9} and g : {3, 4, 5, 9} → {7, 11, 15} be functions defined as f (2) = 3, f (3) = 4, f (4) = f (5) = 5 and g(3) = g(4) = 7 and g(5) = g(9) = 11. Find gof = ?

Solution

We are given

 gof (2) = g (f (2)) 

               = g(3) 

               = 7

 gof (3) = g(f (3)

             = g(4)

              = 7,

gof (4) = g(f (4)) 

           = g(5) 

             = 11 

and  gof (5) = g(f (5))

                   = g (5)                     

                    = 11

So gof ={(2,7),(3,7),(4,11),(5,11)} 

Example

Conclusion

Thanks for devoting your precious time to my post on "How to understand  Relations and  Functions, Inverse of a Function " of this blog . If you liked this post please follow me on my blog for more post. 

If you want to Read  HOW  TO  UNDERSTAND BINARY  OPERATIONS , RELATIONS  AND  FUNCTIONS || COMMUTATIVE || ASSOCIATIVE  click here 

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COMPLEX NUMBERS || MULTIPLICATION || MODULUS || INVERSE || SQUARE ROOTS

Let us discuss concept of number, complex numbers, rational,numbers, real numbers, prime numbers ,complex imaginary numbers, factors of a number, complex number , introduction to complex numbers , operations with complex numbers such as addition of complex numbers , subtraction, multiplying complex numbers, conjugate, modulus  polar form and their Square roots of the complex numbers and complex numbers questions and answers .

Complex Numbers

Introduction to complex numbers

Any number of the type a+ib where a,b are Real numbers and i =√ (-1) is called complex number. Where "a" is called real part and "b" is called imaginary  part . The set of complex number is denoted by  C . These numbers are also called  non real numbers .

Complex Numbers Examples  

3,7 + 3i , -2 + 9i , 3i - 2 , 9i,  3 -7i , 2 , 3i, 6 + √5 i , 6 - √5 i ,√5
Here the numbers 2, 3, √5 are called purely Real and the numbers 3i, 9i are called Imaginary numbers .

Algebra of complex numbers

Addition of complex numbers

If   Z1 = a1+ ib1 and  Z2 = a2+ib2  are two complex numbers then their addition can be calculated by adding their real parts and imaginary parts separately. 

 Z1 + Z2= (a1 + ib1 ) + (a2  + ib2 )
Z1 + Z2= (a1 + a2  ) + (ib1 + ib2 )

Z1 + Z2= (a1 + a2  ) + i(b1 + b2 )

Here   (a1 + a2  ) is called Real Part and   (b1 + b2 ) is called Imaginary Part of resultant .

Let Z1 = 3+5i  and Z2  = 9 - 6i then

Z1 + Z2  = (3 + 5i) + (9 - 6i)

Z1 + Z2 = (3 + 9) + (5i - 6i)  

Z1 + Z2  = 12 - i

Here  12  is its Real Part   -1 is its Imaginary Part

More Examples 

 $\displaystyle{\left({6}+{7}{j}\right)}+{\left({3}-{5}{j}\right)}=$(6+7i)
$Z1=6+7iand$ Z2 =3−5i
Z1  +  Z2  = (3

Z1  +  Z2   = (6+3)+(7−5)i

Z1  +  Z2 = 9+2i

If   Z1  +  Z2  = (12+6i) + (-4 - 5i) ,
Z1  +  Z2 = (12−4) + (6i−5i),
Z1  +  Z2= (12−4) + (6−5)i  ,
Z1  +  Z2 = 8+i

If   Z1 =  6+ √5 i and   Z2  = 7 - √4 i

then  Z1  +  Z2  =  ( 6 + √5 i ) + ( 7 - √4 i) 

          Z1  +  Z2 = (6 + 7) + (√5 i - √4 i) 

           Z1  +  Z2  = 13 - (√5 - √4)i

Subtracting Complex Numbers

If   Z1 = a1 + ib1 and  Z2 = a2 + ib2  are two complex numbers then their subtraction can be calculated by subtracting their real parts and imaginary parts from each  separately. 

 Z1 - Z2= (a1 +ib1 ) - (a2  + ib2 )

Z1 - Z2= (a1 - a2  ) + (ib1 - ib2 )

Z1 - Z2  = (a1 - a2  ) + i(b1 - b2 )

Here   (a1 - a2  ) is its Real Part  and   (b1 - b2 )  is its Imaginary Part.

Let Z1 = 8+5i  and Z2  = 4-6i then

Z1  - Z2  = (8+5i) - (4-6i) = (8-4) - (5i-6i)  = 4+i
Z1 - Z2  = (3+5i) - (9-6i) = (3+9) - (5i-6i)  = 12+i

Examples

 $\displaystyle{\left({6}+{7}{j}\right)}+{\left({3}-{5}{j}\right)}=$(6+7i)

$Z1=6+7iand$ Z2 = 3−5i

Z1  -  Z2  = (6+7i) - (3-5i)

Z1  -  Z2 = (6-3)+(7+5)i

Z1  -  Z2  =3+ 12i

If  Z1  = 12+6i ,  Z2  = -4-5i

   Z1  -  Z2 = (12+6i)-(-4−5i)
  Z1  -  Z2 = (12+4)+(6i+5i) =12+11i 

  Z1 -  Z2  = (12+4)+(6+5)i
 Z1  -  Z2 = 18+9i

Multiplying Complex Numbers

If   Z1 = a1 + ib1 and  Z2 = a2 + ib2  are two complex numbers then their Multiplication can be done as  follows

 Z1 Z2 = (a1 +ib1 ) (a2  + ib2 )

Z1  Z2 = a1  (a2  + ib2 )+ib1(a2  + ib2 )

Z1  Z2  = a1  a2  + i a1 b2 +ib1a2  + ib2 ib1  (because  i²  = −1)

Z1  Z2 = a1  a2  + i a1 b2 +ib1a2  +  i² b2 b1     

Z1  Z2 = (a1  a2  - b1b2)  +i(a1 b2 + a2b1 )

Examples

  If    Z1  = 2 + 6i ,  Z2  = 4 - 5i

 Z1Z2 = (2 + 6i)(4−5i)

Z1Z2 =2(4−5i)+ 6i(4−5i)

Z1Z2 = 8-10i+24i−30i²

Z1Z2 = 8-10i + 24i + 30  

Z1Z2 = (8 + 30)+ (-10i + 24i)

 Z1Z2 = (38)+ (14i)

Z1Z2 = 38 +14i

 If    Z1  = -2-i ,  Z2  = 4+5i

 Z1Z2 = (-2-i)(4+5i)
Z1Z2 = -2(4 + 5i) -i(4 + )

 Z1Z2  =  -8-10i-4i−5i²

 Z1Z2 =   -8-10i-4i+5

 Z1Z2 = (-8+5)+ (-10i-4i)

 Z1Z2 = (-3)+ (-14i)

Z1Z2 = -3 -14i

Now=(3+3i)(1-7i)
= 3×1 + 3×(-7i) + 3i×1+ 3i×(-7i)

 = 3 - 21i + 3i -21i²

 = 3 - 21i + 3i +21 (because i² = −1)

 = 24 - 18i

 How  to Simplify Huge Power  of Iota  

In order to reduce huge/big power of iota to smallest power . First divide the power if iota with 4 and then write first term as power of (4 × quotient) of iota and second term as power of remainder of iota as follows . we know that i⁴ = 1 , implies the contribution of 1st term reduces to "1" .

Note : In order to get quotient after division by 4  , it is sufficient to divide the last two digits of the given power by 4.

Therefore it is the second term which will contribute to answer.

Before we proceed further Remember these result   i² = -1 ,          i³ = -i ,       i⁴ = 1.

So these are some Simplification 

i⁴⁵⁰ =   i^{4×112 .} i²  .= 1× (-1) = -1

i⁴⁵¹ =   i^{4×112 .} i³  .= 1× (-i) = -i

 i⁴⁵² =   i^{4×113 .} i⁰  .= 1× 1 = 1

i⁴⁵³ =   i^{4×113 .} i¹  .= 1× i = i

 ^{i3523 =   i4×855} i³ = 1×(-i)= -i

 ^{i9998 =   i4×2449} i² = 1×(-1) = -1

 ^{i9997 =   i4×2449} i¹ = 1×(i) = i

^{i2221 =   i4×555} i¹ = 1×(i) =  i

^{i2222 =   i4×555} i² = 1×(-1) = -1

^{i2223 =   i4×555} i³ = 1×(-i) = -i

Complex Conjugate of complex Number 

If  z = + iy be any complex number then  x - iy is called its Conjugate.  Simply change the sign of   i   to -i.

Examples

Let Z= 5-7i then Conjugate of this complex number is 5+7i

Let  5 2 - 7 3i  be any complex number  then

 Conjugate of this complex number is 
5 2 + 7 3i  and
  Z= 2+ i be the complex number then Conjugate of this complex number is 2 - i

Modulus of a Complex Number

If  z = + iy be any complex number then  Positive square root of sum of the squares of its Real and Imaginary Parts is called its Conjugate.It is denoted by |Z|

Then 

e. g 

Let Z= -3+2i

then | Z | = √(${-3)}^2+({2)}^2$ )  = √ (13)

If Z = -2-i

Then 

     |Z|   $=$ √ ( (${-2)}^2+({-1)}^2$ )

     |Z|   = √5

If Z= 6 - 8i
Then |Z| = √( 36+64   )  
          |Z|= √ (100) 

          |Z|= 10

Division of two Complex Numbers

To divide Z1 With Z2 Multiply and divide the numerator and Denominator by the Conjugate of  Denominator and then write in a + ib form   after simplification.

Example

Divide  2+3i with 4+7i  

4 + 7i

Multiply Numerator And Denominator  by the conjugate of 4+7i 

2 +3i4 +7i×4 - 7i4 - 7i  =  8 -14i + 12i - 21i²16 + 49

=  29 - 2i65

Now change  into a +ib form

=  29 65 - i265

Example

Divide  3+4i with 2+3i

  3+4i2+3i

Multiply Numerator And Denominator by the conjugate of2+3i 

3+4i2+3i × 2-3i2-3i  =  6 -9i +8i -12i²4+9

=  18- i13

Now change  into a +i b form:

=  18 13 - i113

Multiplicative Inverse of complex Number

If Z = x + iy be any complex number then 1/Z is called it Multiplicative Inverse.

Example

If Z = 1 - ithen Z -1= 1/(1-i)
If  Z = 4+3i 
then Z -1= 1/(4+3i)




If  Z = 1-√ 2i
then Z -1= 1/(1-√ 2 i)

Polar Form of a Complex Number

To Convert  Z = x + iy complex number into Polar Form 

Put $x=r\cos \theta$  ....................(1)

$\backslash displaystyle\{y\}=\{r\}\backslash \; \backslash sin\{\backslash theta\}$

 Then Squaring and adding  (1) and (2)
$we\; get$


 
 
 $r^2=x^2+y^2$ and 

upon dividing  (2)equation  by (1) we get

$\displaystyle \tan{\ }\theta=\frac{y}{{x}}$$\displaystyle{x}={r}\ \cos{\theta}$

Then Z = r( cos θ + i sinθ ) is in Polar Form

Example

Convert the complex number 

7 − 5i    into Polar Form , We need to find r and θ.

Put $x=r\cos \theta$  

$\backslash displaystyle\{y\}=\{r\}\backslash \; \backslash sin\{\backslash theta\}$

\displaystyle{r}=\sqrt{{{x}^{2}+{y}^{2}}
​

$\displaystyle=\sqrt{{{7}^{2}+{\left(-{5}\right)}^{2}}}$

$\displaystyle=\sqrt{{{49}+{25}}}$

$\displaystyle=\sqrt{{{74}}}\approx{8.6}$
8

To find θ, we first find the acute angle α 

which is equal to

$\displaystyle={{\tan}^{{-{1}}}{\left(\frac{5}{{7}}\right)}}$

As the Points (7,-5) lies in fourth Quadrant in Argand Plane ,So   $\displaystyle{7}-{5}{j}$ will be  in the fourth quadrant, 

  θ= 360° - α  = 324.5°

So, expressing $\displaystyle{7}-{5}{j}$ in polar form

7-5i = 8.59 ( cos 324.5° + i sin 324.5° )

Example

Convert the complex number $\backslash displaystyle\{7\}-\{5\}\{j\}$

We need to find r and θ.

$\displaystyle{r}=\sqrt{{{}}}$

$\displaystyle=\sqrt{{{7}^{2}+{\left(-{5}\right)}^{2}}}$

       =    1+2=3

   r = √2

 

To find θ, we first find the acute angle α 

which is equal to

$\displaystyle={{\tan}^{{-{1}}}{\left(\frac{5}{{7}}\right)}}$

α = 45°

As the Points (1,-√2) lies in fourth Quadrant in Argand Plane ,So  1−√2i will be  in the fourth quadrant, 

So θ=360 - 45°

θ= 315°

So, expressing 1$-$√2i  in polar form


Z =  √2(cos 315° + i sin 315°) 

Also Watch   HOW TO CONVERT ANY COMPLEX NUMBER TO IT POLAR FORM 

Square root of complex number

Example 1

Find the square root of 8 + 6i

 Let        z² = (x + yi)²  = 8+6i   Squaring  Both Sides

          \    (x² – y²) + (2xy)i     = 8+6i

          Compare real parts and imaginary parts,we have 

       x² – y² = 8   .............................    (1)

        2xy = 6          .............. ...............  (2)

         

 Using   Identity

  ( x² + y² )²  = (   x² - y² )²  - 4 x²  y² 

    ( x² + y² )²  = 8² + 6²

   ( x² + y² )²  =100

     x² + y² = √(8² + 6²) = 10                  

   x² + y² = 10    .............................. (3)

To find the values of  x and y 

 Adding   (1) and (3) 

 we get  x² – y² + x² + y² = 10 +8   

           2 x² = 18 

          x² = 9 

           x = ±3  

Put   x = 3  and x = -3 and    in equation (2)    

we get     y = 1 when x=3  

and          y = -1 when x= -3

From (2) we concluded that  x and y are of same  sign,

 (x = 3 and y = 1)    or   (x = -3 or y =-1)

Therefore required Square roots of   8+6i are

Z = 3+i and Z = -3-i

Example 2

Find the square root of -7+24i

 Let        z² = (x + yi)²  = -7+24i   Squaring  Both Sides

  (x² – y²) + (2xy)i     = -7+24i

  Compare real parts and imaginary parts,we have 

 x² – y² = -7  .............    (1)

  2xy = 24      ..............  (2)

         

 Using   Identity

   ( x² + y² )²  = (   x² - y² )²  - 4 x²  y² 

   ( x² + y² )²  = (-7)² + (24)²

   ( x² + y² )²  = 49 + 576=625

    x² + y² = √(25)                  

   x² + y² = 25    ...................   (3)           

 To find the values of  x and y 

 Adding   (1) and (3) 

 we get         x² – y² + x² + y² = -7 +24   

 2 x² = 18 

  x² = 9 

 x = ±3  

Put   x = 3  and x = -3 and    in equation (2)    

we get     y = 4 when x=3  

and          y = -4when x= -3

  From (2) we concluded that  x and y are of same  sign,

   (x = 3 and y = 4)    or   (x = -3 or y = -4)

Therefore required Square roots of   8 + 6i are

Z= 3+4i and Z = -3-4i

Example 3

Find the square root of -15 - 8i

 Let        z² = (x + yi)²  = -15 - 8i

  Squaring  Both Sides

 (x² – y²) + (2xy)i     = -15 - 8i

 Compare real parts and imaginary parts,we have 

 x² – y² = -15   .............................    (1)

  2xy = -8         .............. ...............  (2)

         

 Using   Identity

  ( x² + y² )²  = (   x² - y² )²  - 4 x²  y² 

  ( x² + y² )²  = -15² + -8²

  ( x² + y² )²  =225+64

   x² + y² = √289                   

 x² + y² = 17    ..............................   (3)           

 To find the values of  x and y 

 Adding   (1) and (3) 

 we get         x² – y² + x² + y² = -15 +17   

 2 x² = 2 

   x² = 1 

  x = ±1 

Put   x = 1  and x = -1 and    in equation (2)    

we get     y =  -4  when x = 1  

and          y = 4 when x = -1

  From (2) we concluded that  x and y are of same  sign,

  (x = 1 and y = -4)    or   (x = -1 or y = 4)

Therefore required Square roots of   -15 - 8i are

Z=1 - 4i  and Z  = -1+ 4i

Example 4

Find the square root of -3 + 4i

 Let        z² = (x + yi)²  =  -3 + 4i

  Squaring  Both Sides
  (x² – y²) + (2xy)i     = -3 + 4i

          Compare real parts and imaginary parts,we have 

                        x² – y² = -3   .............................    (1)

                        2xy =   4         .............. ...............  (2)

         

 Using   Identity

    ( x² + y² )²  = (   x² - y² )²  - 4 x²  y² 

      ( x² + y² )²  = -3² + 4²

       ( x² + y² )²  =9+4

        x² + y² = √25                  

    x² + y² = 5    ..............................   (3)           

 To find the values of  x and y 

  Adding   (1) and (3) 

 we get  x² – y² + x² + y² = -3 +5   

          2 x² = 2 

          x² = 1 

         x = ±1 

Put   x = 1  and x = -1 and    in equation (2)    

we get     y =  2  when x = 1  

and          y = -2 when x = -1

          From (2) we concluded that  x and y are of same  sign,

                  \ (x = 1 and y = 2)    or   (x = -1 or y =-2)

Therefore required Square roots of   -15 - 8i are

Z=1 +2i  and Z  = -1- 2i

HOW TO FIND THE SQUARE ROOTS OF ANY COMPLEX NUMBER IN HINDI

Conclusion

In this post I have discussed Complex number, rational numbers,
real numbers ,prime numbers, modulus ,inverse, polar form,square roots of complex numbers. If this post helped you little bit, then share it with your friends and like this post to boost me to do better, and follow this Blog .We shall meet in next post till then Bye .

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