Imaginary Numbers
An imaginary number is the square root of a negative number. How is that imaginary? It's something that can happen when you're doing math. For instance, what if you have the equation:
$x^2=-16$
You'll square root both sides.
$\eqalign{x^2&=-16\\\sqrt{x^2}&=\sqrt{-16}\\x&=\sqrt{-16}}$
$x$ equals the square root of -16 is an answer that you could get. But, it's not a mathematically possible number. Why not? What number, times itself, is equal to a negative number?
$\eqalign{4\times4&=16\\-4 \times -4&=16}$
Think about it. The only way to get a negative number is to multiple a negative times a positive -- and that's not a perfect square. So, the square root of a negative number is not possible in real math. It's not a real number.
But it can happen. Math deals with this with imaginary numbers. Imaginary numbers are written as $i$.
$i=\sqrt{-1}$
Officially, the definition of $i$ is that $i^2=-1$.
Because squares and square roots are inverse operations $i^2=-1$ and $i=\sqrt{-1}$ are equivalent statements.
The most interesting thing about $i$ is that you can work with it in ways that can make it turn back into a real number.
Critically, know that working with exponents of $i$ results in a pattern:
$\eqalign{i&=\sqrt{-1}\\i^2&=\sqrt{-1}\times \sqrt{-1}=-1\\i^3&=\sqrt{-1}\times \sqrt{-1}\times \sqrt{-1}=-1\sqrt{-1}\\i^4&= \sqrt{-1}\times \sqrt{-1}\times \sqrt{-1}\times\sqrt{-1}=-1 \times -1=1}$
This pattern continues:
$\eqalign{i&=\sqrt{-1}\\i^2&=-1\\i^3&=-1\sqrt{-1}\\i^4&=1\\i^5&=\sqrt{-1}\\i^6&=-1\\i^7&=-1\sqrt{-1}\\i^8&= 1\\i^9&=\sqrt{-1}\\i^10&=-1\\i^11&=-1\sqrt{-1}\\i^12&=1}$
You can figure out the value of $i$ to any power by dividing the exponent by 4.
If the remainder is 0, $i^x=1$
If the remainder is 1, $i^x=\sqrt{-1}$
If the remainder is 2, $i^x=-1$
If the remainder is 3, $i^x=-1\sqrt{-1}$
Let's try it.
Example:
$i^{88}=$
$88\div4=11$
There is no remainder, so $i^{88}=1$
How can you use $i$? First thing of it as a radical. So anything you need to do with a radical (for instance, rationalize it out of the denominator of a fraction) you need to do with $i$. But, you can also use it in problems like the one we started with. Rather than stopping at an "impossible" answer, you can use $i$ to finish the problem:
$\eqalign{x^2&=-16\\\sqrt{x^2}&=\sqrt{-16}\\x&=\sqrt{-16}\\x&=4i}$
Practice Problems:
Imaginary Numbers
Write the following numbers without using $i$.
- $i$
- $i^2$
- $i^3$
- $i^4$
- $i^5$
- $4i^3$
- $6i^9$
- $8i^{12}$
- $2i^2$
- $5i^3$
- $7i^6$
- $11i^9$
- $21i^{14}$
- $56i^{16}$
- $7i^{20}$
- $9i^{33}$
- $3i^{32}$
- $4i^{30}$
- $(5i^2)(6i^2)$
- $(7i^3)(8i^7)$
Answer Key: