Divide Imaginary Numbers

Introduction about imaginary numbers:

The term "numbers" are used to measure the quantity. The term "numbers" are a fixed value, if it is integers or constants. It is also refer to a complex numbers, real number, imaginary numbers, etc. The complex numbers are numbers involving, there is no number that when squared equals -1. The square root of minus one is denoted by mathematical symbol ' i '. A number contains, i  am the co-efficient of that numbers are called imaginary numbers. Having problem with What is Composite Number keep reading my upcoming posts, i will try to help you.


Evaluation of Imaginary Numbers:

In the definition (`sqrt(-1)` )2should equal -1, but it does not, therefore the symbol i =`sqrt(-1)` is introduced,

i.e.,         `sqrt(-x)`      = i `sqrt(x)`  Where x is a positive number and (i)2 = -1.

The square roots of other than -1 of the negative numbers format is, `sqrt(-n)` = i`sqrt(n)` .

The important rules for symbol 'i' :

i-1 = `1/ i` = `i / (i^2)` = `i /(-1)` =  -i      ( reciprocal of symbol ' i ' is ' -i ' )

The powers of symbol ' i ' :

i1 = i

i2 = -1

i3 = i2 i = (-1)(i) = -i

i4 = i3 i= (-i)(i) = -(i)2 = -(-1) = 1

i5 = i4 i = (1)( i) = i

..........

Complex numbers are numbers involving i and are generally in the form:

iy `hArr` (real number)(i)  `hArr` imaginary number

x + iy   (real number) + (i)(real number) `hArr` complex number

Dividing Imaginary numbers

In the division of complex number operations performed through learn complex numbers online, first find out the complex conjugate of the denominator. Second we multiply the conjugate of complex number to both numerator and denominator of complex number. The conjugate of complex number of (x + iy) is (x - iy) and vice versa.

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Example 1:dividing Imaginary Numbers.

(a). `(2 + i3) /(4-i2)`

Solution:

Given:

` (2 + i3) /(4-i2)`

Dividing imaginary numbers:

`(2 + i3) /(4-i2)` = `(2 + i3) /(4-i2)` . `(4 + i2) /(4 +i2)`

= `((2 + i3)(4 + i2))/((4-i2)(4 + i2))`

= `((2 + i3)(4 + i2)) / (4^2 + 2^2)`

= `(8 + i4 + i12 -6)/(16 + 4)`

= `(2 + i16) / (20)`

= `(2)/(20)` + `i(16)/(20)`

= `1/(10)` + `i (4/5)`

Example 2:dividing Imaginary Numbers.
(a). `(6 + i2)/(-2+i3)`

Solution:

Given:

`(6 + i2)/(-2+i3)`

Dividing imaginary numbers:

`(6 + i2)/(-2+i3)`   = `(6 + i2)/(-2 + i3)` . `(-2-i3)/(-2-i3)`

= `((6 + i2)(-2-i3))/((-2 + i3)(-2-i3))`

= `(-12 +i18-4i -6)/(-2^2 + 3^2)`

= `(-20 + i14)/(4+9)`

= `(-20 + i14)/(13)`

=`(-20)/(13)` +`i(14)/(13)`    [Answer.]