# 7.2 Imaginary concepts -- complex numbers

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This module introduces the concept of complex numbers in Algebra.

A “complex number” is the sum of two parts: a real number by itself, and a real number multiplied by $i$ . It can therefore be written as $a+\mathrm{bi}$ , where $a$ and $b$ are real numbers.

The first part, $a$ , is referred to as the real part . The second part, $\mathrm{bi}$ , is referred to as the imaginary part .

Examples of complex numbers $a+\mathrm{bi}$ ( $a$ is the “real part”; $\mathrm{bi}$ is the “imaginary part”)
$3+\mathrm{2i}$ $a=3$ , $b=2$
$\pi$ $a=\pi$ , $b=0$ (no imaginary part: a “pure real number”)
$-i$ $a=0$ , $b=-1$ (no real part: a “pure imaginary number”)

Some numbers are not obviously in the form $a+\mathrm{bi}$ . However, any number can be put in this form.

Example 1: Putting a fraction into $a+\mathrm{bi}$ form ( $i$ in the numerator)
$\frac{3-4i}{5}$ is a valid complex number. But it is not in the form $a+\mathrm{bi}$ , and we cannot immediately see what the real and imaginary parts are.
To see the parts, we rewrite it like this:
$\frac{3-4i}{5}$ $=$ $\frac{3}{5}$ $\frac{4}{5}i$
Why does that work? It’s just the ordinary rules of fractions, applied backward. (Try multiplying and then subtracting on the right to confirm this.) But now we have a form we can use:
$\frac{3-4i}{5}$ $a=$ $\frac{3}{5}$ , $b=–\frac{4}{5}$
So we see that fractions are very easy to break up, if the $i$ is in the numerator. An $i$ in the denominator is a bit trickier to deal with.
Example 2: Putting a fraction into $a+\mathrm{bi}$ form ( $i$ in the denominator)
$\frac{1}{i}$ $=$ $\frac{1\cdot i}{i\cdot i}$ Multiplying the top and bottom of a fraction by the same number never changes the value of the fraction: it just rewrites it in a different form.
$=$ $\frac{i}{-1}$ Because $i•i$ is ${i}^{2}$ , or –1.
$=\mathrm{-i}$ This is not a property of $i$ , but of –1. Similarly, $\frac{5}{-1}$ $=–5$ .
$\frac{1}{i}$ : $a=0$ , $b=-1$ since we rewrote it as $\mathrm{-i}$ , or $0-\mathrm{1i}$

Finally, what if the denominator is a more complicated complex number? The trick in this case is similar to the trick we used for rationalizing the denominator: we multiply by a quantity known as the complex conjugate of the denominator .

## Definition of complex conjugate

The complex conjugate of the number $a+\mathrm{bi}$ is $a-\mathrm{bi}$ . In words, you leave the real part alone, and change the sign of the imaginary part.

Here is how we can use the “complex conjugate” to simplify a fraction.

Example: Using the Complex Conjugate to put a fraction into $a+\mathrm{bi}$ form
$\frac{5}{3-4i}$ The fraction: a complex number not currently in the form $a+bi$
$=$ $\frac{5\left(3+4i\right)}{\left(3-4i\right)\left(3+4i\right)}$ Multiply the top and bottom by the complex conjugate of the denominator
$=$ $\frac{\text{15}+\text{20}i}{{3}^{2}-\left(4i{\right)}^{2}}$ Remember, $\left(x+a\right)\left(x–a\right)={x}^{2}{\mathrm{–a}}^{2}$
$=$ $\frac{\text{15}+\text{20}i}{9+\text{16}}$ ${\left(\mathrm{4i}\right)}^{2}={4}^{2}{i}^{2}=16\left(–1\right)=–16$ , which we are subtracting from 9
$=$ $\frac{\text{15}+\text{20}i}{\text{25}}$ Success! The top has $i$ , but the bottom doesn’t. This is easy to deal with.
$=$ $\frac{\text{15}}{\text{25}}$ $+$ $\frac{\text{20}i}{\text{25}}$ Break the fraction up, just as we did in a previous example.
$=$ $\frac{3}{5}$ $+$ $\frac{4}{5}i$ So we’re there! $a=$ $\frac{3}{5}$ and $b=\frac{4}{5}$

Any number of any kind can be written as $a+\mathrm{bi}$ . The above examples show how to rewrite fractions in this form. In the text, you go through a worksheet designed to rewrite $\sqrt[3]{-1}$ as three different complex numbers. Once you understand this exercise, you can rewrite other radicals, such as $\sqrt{i}$ , in $a+\mathrm{bi}$ form.

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20/(×-6^2)
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Salomon
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Salomon
I got X =-6
Salomon
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