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

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= (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

Subtracting Complex Numbers

If   Z1 = a+ ib1 and  Z2 = a+ 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= (a- a ) + (ib1 ib2 )
Z1 - Z = (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 = 38 +14i


 If    Z1  = -2-i ,  Z2  = 4+5i

 Z1Z= (-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 =   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


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  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{y}={r}\ \sin{\theta}
 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

Example

Convert the complex number 

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

\displaystyle{y}={r}\ \sin{\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}
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{7}-{5}{j}
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°) 

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





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 = ±
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 = ±
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




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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