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Showing posts with label complex number. Show all posts
Showing posts with label complex number. Show all posts

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

Let us Discuss complex numbers ,complex imaginary numbers,  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

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

Z1 + Z2= (a+ a ) + i(b1 b2 )
Here   (a+ a ) is called Real Part and   (b1 b2 ) is called Imaginary Part of resultant .

Let Z1 = 3+5i  and Z2  = 9 - 6i then

Z1 + Z2  = (5i) + (- 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)
Z2 =35i
Z1  + Z2  = (
Z1  + Z(6+3)+(75)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

## 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)
Z2 = 35i
Z1  - Z2  = (6+7i) - (3-5i)
Z1  - Z2(6-3)+(7+5)i
Z1  - Z =3+ 12i

If  Z1  = 12+6i ,  Z= -4-5i
Z1  - Z2 = (12+6i)-(-45i)
Z1  - Z2 = (12+4)+(6i+5i) =12+11i
Z1 - Z2  (12+4)+(6+5)i
Z1  - Z2 = 18+9i

## Multiplying Complex Numbers

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

Z1 Z= (a+ib1 ) (a2  ib2 )
Z1  Z= a (a2  ib2 )+ib1(a2  ib2 )
Z1  Za a2  i ab2 +ib1a2  ib2 ib (because  i2  = −1)

Z1  Za a2  i ab2 +ib1a2   i2 b2 b

Z1  Z= (a ab1b2 +i(ab2 + a2b1 )

### Examples

If    Z1  = 2 + 6i ,  Z2  = 4 - 5i
Z1Z= (2 + 6i)(45i)
Z1Z=2(45i)+ 6i(45i)

Z1Z= 8-10i+24i−30i2

Z1Z2 = 8-10i + 24i + 30
Z1Z2 = (8 + 30)+ (-10i + 24i)

Z1Z2 = 38 +14i

### Z1Z2 = (-2-i)(4+5i) Z1Z2 = -2(4 + 5i) -i(4 + )  Z1Z2  =  -8-10i-4i−5i2

Z1Z2 =   -8-10i-4i+5

## 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 =   i4×112 . i2  .= 1× (-1) = -1

i451 =   i4×112 . i3  .= 1× (-i) = -i

i452 =   i4×113 . i0  .= 1× 1 = 1

i453 =   i4×113 . 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

## Examples

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

Let  2 - 3i  be any complex number  then

Conjugate of this complex number is
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 | = ( )  =  (13)
If Z = -2-i
Then
|Z|     ( ( )
|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
2 + 3i4 + 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 -11/(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   ....................(1)
$\displaystyle\left\{y\right\}=\left\{r\right\}\ \sin\left\{\theta\right\}$
Then Squaring and adding  (1) and (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

### Convert the complex number

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

$\displaystyle\left\{y\right\}=\left\{r\right\}\ \sin\left\{\theta\right\}$
$\displaystyle{r}=\sqrt{{{x}^{2}+{y}^{2}}$
$\displaystyle=\sqrt{{{7}^{2}+{\left(-{5}\right)}^{2}}}$
$\displaystyle=\sqrt{{{49}+{25}}}$
$\displaystyle=\sqrt{{{74}}}\approx{8.6}$
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 $\displaystyle\left\{7\right\}-\left\{5\right\}\left\{j\right\}$
We need to find r and θ.

$\displaystyle{r}=\sqrt{{{}}}$
$\displaystyle=\sqrt{{{7}^{2}+{\left(-{5}\right)}^{2}}}$
=    1+2=3

r2
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 will be  in the fourth quadrant,

So θ=360 - 45°

θ= 315°
So, expressing 12 in polar form

Z =  2(cos 315° + i sin 315°)

## 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
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

## 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
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
we get         x2 – y2 + x2 + y2 = -15 +17
2 x2 = 2
x2 = 1
x = ±
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
we get  x2 – y2 + x2 + y2 = -3 +5
2 x2 = 2
x2 = 1
x = ±
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

## 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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