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An extension of Fundamental Mathematics/Arithmetic/Arithmetic Number/Imaginary Number (Imaginary numbers), complex numbers allow us to combine real numbers with imaginary numbers to produce meaningful results from manipulating and changing things around. This page covers very basic imaginary number ideas designed for high school students.

The Basics - Complex Number
A complex number is part of the set of complex numbers which we represent as the symbol, $$\Complex$$. Thus any complex number is part of this set $$\Complex$$. Suppose we have complex number, $$z$$. Now since its a complex number we can say that '$$z$$ is part of the set of complex numbers, $$\Complex$$', which we write as $$z \in \Complex$$.

Now the general formula for our complex number, or any complex number, can be written as:

$$z = a + bi$$

where $$a$$ and $$b$$ are real numbers i.e $$a$$ and $$b$$ in the set of real numbers, $$\Reals$$, which we can again right as $$a,b \in \Reals$$.

The complex number is made up of two parts. They are:


 * The real part, which is $$a$$ (We can show this as $$Re(z)$$ = $$a$$ )
 * The imaginary part, the number in front of the $$i$$, which is $$b$$ (We can show this as $$Im(z) = b$$)

This is how we connect the real part with the imaginary part. There are different forms of expressing a complex number, but this form is called the Cartesian Form (because its like how we show real numbers on the Cartesian plane (x,y))

Conjugate
An important idea in complex numbers is the conjugate of a complex number. This is simply the opposite of a complex number and is rather easy to figure out. Lets take our previous complex number, $$z$$. We already know what this is ($$z = a + bi$$), so the conjugate of it would be:

$$\bar z = a-bi$$

We simply change the sign in front of the imaginary part to get our conjugate for $$z$$. Pretty simple. We the conjugate of $$z$$, is shown with a special symbol, $$\bar z$$. Its the variable for our complex number with a bar over it. This is the conjugate.

Operations involving 2 complex numbers
Doing operations with complex numbers such as addition, subtraction, multiplication and division are rather simple, borrowing the same rules as regular algebra. We shall use two complex numbers in the upcoming examples:

$$z_1 = a + bi$$

$$z_2 = x + yi$$

Addition
Adding complex numbers is just like adding two different expressions in algebra. No tricks here.

$$z_1 + z_2 = a + bi + x + yi$$

$$= (a+x) + (b+y)i$$

Note: We put the real parts in one bracket and the imaginary parts in another. This is just done to make it simpler to understand. In a real question, add them together

Subtraction
Subtracting is the opposite process. Be wary of the negative sign however.

$$z_1 - z_2 = (a+bi) - (x+yi) = a + bi -x - yi$$

$$= (a-x)+(b-y)i$$

Multiplication
Unlike, addition or subtraction, multiplying imaginary numbers isn't as straightforward although we still use the same idea of FOIL to multiply the two numbers (treat them like expressions). What usually happens is that we times an imaginary part by another imaginary part which results in a negative. Remember: $$i^2 = -1$$

$$z_1 \times z_2 = (a+bi)\times (x+yi) = a(x+yi)+bi(x+yi)$$

$$= ax + ayi + bx + byi^2 = ax + ayi + bxi - by$$

$$= (ax - by) + (ay+bx)i$$

We can show the multiplication of $$z_1 \times z_2$$ as $$z_1z_2$$

Remember to be careful when squaring $$i$$.

Division
Division is not like any of the previous operations in that it requires knowledge of the conjugate of a complex numbers. We can express the division of two complex numbers as:

$$\frac{z_1}{z_2} = \frac{x+yi}{a+bi}$$

Using this we must make the denominator (the number at the bottom) a real number. That is we need to times the fraction by the conjugate of $$z_2 $$. Here is a step by step process:

$$, which is $$a + bi$$. $$\bar z_2$$ is $$a - bi$$. That is our conjugate
 * 1) Find the conjugate of the denominator: In our case, the denominator is $$z_2
 * 1) Times the top and bottom by the conjugate: $$\frac{x+yi}{a+bi} \times \frac{a-bi}{a-bi}$$
 * 2) Multiply like normal: $$\frac{x+yi}{a+bi} \times \frac{a-bi}{a-bi} = \frac{x(a-bi)+yi(a-bi)}{a(a-bi)+bi(a-bi)}$$
 * 3) Answer: $$\frac{ax-bxi+ayi-byi^2}{a^2-abi+abi-b^2i^2} = \frac{(ax-by)+(ay-bx)i}{a^2 + b^2}$$

Note the denominator in our answer. It has some significance later.