Imaginary numbers are numbers which are not real. They are only because of our imagination. The imaginary numbers are the result of them. We will see here about Imaginary numbers and how it works in Algebra.

Take any number positive or negative. When we square, result will also be a positive number. 6 x 6 = +36. Also - 6 x -6 = +36. Even for negative numbers square is always positive. So square can never be negative for any real number.

But consider the quadratic equation, ax^{2 }+ bx + c = 0. We all know that the discriminant is b square - 4ac. And also the root consists of $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Whenever this discriminant is < 0, finding square root becomes necessary for us. For this we assume root of -1 = i. This i is the base for all imaginary numbers. Solution is written by using this imaginary number in the form a+bi. The number written like this is called a complex number. In other words a complex number is one which includes both real and imaginary numbers.

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In Algebra, notation used for imaginary number is i. "i" is defined by square root of -1. With this i, square root of any negative number can be expressed in some real number multiplied by i.

Whenever a number has imaginary number in that, it is called a complex number. For example, a+bi is a complex number. For a +bi, conjugate pair is a-bi. Whenever a complex root exists for real function, then its conjugate pair also is a root. Then only the function is real. In other words, complex roots exist in pairs so that when multiplied become equations with real coefficients.

Consider the pure quadratic equation:* **x*^{ 2} = * a ,* where * a - *a known value. Its solution may be presented as: x = $\sqrt{a}$

Here the three cases are possible:

1) | If a = 0 , then x = 0. |

2) | If |

3) | If a is a negative number, then its square root has two values: one positive imaginary number and one negative imaginary number. This is often written as the root with double sign before + or - 5i. i is added to denote it is imaginary. |

Now i = root of -1. So i^{2}=-1, i^{4} =+1. i^{3 }= -i: In general, we can write as i^{2 }= -1, i^{4n} =1, i^{4n-1}= -i.

** ****Addition of numbers having imaginary numbers**

Here the answer is (a+c) + i(b+d).

** ****Subtraction of numbers having imaginary numbers**

Here (a+bi)-(c+di) = (a-c) +i(b-d).

** ****Multiplication of numbers having imaginary numbers :**

= ac+bci+adi+bdi^{2}

= (ac-bd)+i(bc+ad)

** Division of numbers having imaginary numbers: ** Consider division of one imaginary number by another.

$\frac{a+bi}{c+di}$ . Here what we apply for irrational numbers, apply here also. Multiply the denominator by its conjugate pair,

and make it real. Example , $\frac{a+bi}{c+di}$ = $\frac{(a+bi)(c-di)}{(c+di)(c-di)}$ = (ac+bd)+ i(bc-ad).

Thus for so many simplifications, numbers involving imaginary numbers follow the rules as applied to Algebra.

In contrast to imaginary numbers all the rest numbers (positive and negative, integers and fractional, rational and irrational ones) are called ** real numbers .** A sum of a real and an imaginary number is called a

a + b i ,

where *a, b* – real numbers, ** i **– an imaginary unit.

**Examples of complex numbers: ** 3 + 4 *i *, 7 – 13.6 *i* , 0 + 25 *i* = 25 *i* , 2 + *i*.

**Examples of Real numbers :** 3, 7, 0,-2

**Examples of Imaginary Numbers : ** 3i, 7i, -2i, root A

Now we see addition, subtraction, multiplication and division of only imaginary numbers:

**1. ** Addition: ** ** 2i + 7i = 9i

**2. ** Subtraction: 2i - 5i = -3i.

**3. **Multiplication: 5i x 9i = 45i^{2} = -45

**4. **Division: $\frac{6i}{2i}$ = 3.

**5.** Multiplication of a real no by imaginary no say 3 x 2i = 6i

**6.** Division of a real no by imaginary no $\frac{5}{i}$ = $\frac{5i}{i^2}$ = -5i.

So imaginary numbers also follow the same rule as for others in Algebra.

Now one question arises in the mind of every one. The name Imaginary itself suggests that the number is our imagination and not real.

So Do Imaginary Numbers really exist?

We think it is a complete mathematical fiction and not real because even for negative numbers, square is always positive.

But it is not correct. Imaginary numbers are part of our mathematical system. The theory of complex numbers, fully dealing with imaginary numbers also is nearly half of total theories in Math. So we cannot set aside imaginary numbers as a fiction or imagination.

The imaginary numbers also form a group, follow rules, has multiplicative and additive identity, and follows commutative, associative and distributive property.

More or less complex numbers form the pattern of irrational numbers. For each complex no, a conjugate pair exists.

Hence though name suggests** "Imaginary"**, Imaginary numbers are not real imagination and a very important of branch of mathematics.

Z, the group of complex numbers, actually is very important and a full topic is covered learning about them.

In conclusion, Imaginary numbers are like algebraic expressions, for addition, multiplication, etc., like real numbers, for additive inverse, identity, multiplicative inverse, identity. They resemble irrational numbers, for the part of conjugate pairs, etc.

We can be simplified remembering the main following important things:

We know i = root of -1.

Now i^{2 }= -1

i^{4} = +1

In other words, we can say that for any integer n, i^{4n} = +1, i^{4n}-1 = -i, I^{4n-2} = -1, i^{4n-3} = i.

These four can be straight away applied when doing polynomials of greater degrees with imaginary numbers.

Imaginary numbers can thus be considered as complex numbers where the actual division is zero. The square of an imaginary number is an nonconstructive real number.

i ^{-3} = i

i ^{-2} = -1

i ^{-1} = -i

i ^{0} = 1

i ^{1} = i

i ^{2} = -1

i ^{3} = -i

i ^{4} = 1

i ^{5} = i

i ^{6} = -1

i ^{n} = i ^{n mod 4}

Here we learn about exponents of imaginary numbers. When numbers are purely imaginary, the exponents finding is very easy.

**For example:** (ai)^{n} = a^{n}(i^{n}) and i^{n} can be found out using the above formula namely i^{2} =-1, i^{4} =+1, i^{3} =-i.

When the number is complex i.e. when it involved both real and imaginary, the number will be of the form

(a+bi) so it exponent will be of the form (a+bi)^{n}.

(a+bi)^{2} = a^{2} +2abi +(bi)^{2} = a^{2} b^{2}+ 2abi

Same formula applies as in Algebra except that it involves i and rules for i we can apply here.

(a+bi)^{3} = a^{3} +(bi)^{3} +3ab(a+bi) = a^{3} +b^{3} +3a^{2b} + 3ab^{2i}

Now consider e power l where l is an imaginary number of the form bi.

For normal x, e^{x} = $\frac{1+x+x^2}{2!}$ + $\frac{x^3}{3!}$ + $\frac{x^4}{4!}$ + $\frac{x^5}{5!}$+.........

Similarly. e^{bi} = $\frac{1+bi}{1!}$ + $\frac{b^2i^2}{2!}$ + $\frac{b^3i^3}{3!}$ + $\frac{b^4i^4}{4!}$ + $\frac{b^5i^5}{5!}$ +.....

= 1+bi-$\frac{b^2}{2!}$-$\frac{b^3i}{3!}$+$\frac{b^4}{4!}$+........................

So here first, fourth, eighth, twelfth terms after 1 will be positive, whereas first, fifth, ninth terms after 1 will be positive i,

second sixth, tenth, will have negative real numbers, third, seventh, eleventh items after 1 will have negative i.

The above factors are the main things to know about** imaginary Numbers in Algebra** for our doing problems.

Step 1: The given imaginary number is i^{7}

Step 2: i^{7} = i^{2} x i^{2} x i^{2} x i

= -1 x i

= -i

Step 1: The given imaginary number is $\sqrt{-16}$

Step 2: $\sqrt{-16}$ = $\sqrt{16x-1}$

Step 3: $\sqrt{-16}$ = $\sqrt{16x}$ $\sqrt{-1}$

Step 4: $\sqrt{4^2}$ x i

= 4i

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