# Algebra: Imaginary Numbers

1.
$$\sqrt{-13}$$
Solution
2.
$$(2+3i)+(1-6i)-(8+7i)$$
Solution
3.
$$(8+2i)+(4+7i)$$
Solution
4.
$$2i+3i$$
Solution
5.
$$3i\times 4i$$
Solution
6.
$$-i\times 2i\times 3i$$
Solution
7.
$${i}^{99}$$
Solution
8.
$${i}^{17}$$
Solution
9.
$${i}^{120}$$
Solution
10.
$${i}^{64002}$$
Solution
11.
$$(3i+4)i$$
Solution

# Imaginary Numbers - Introduction

An imaginary number is a complex number that can be written in the form of a real number multiplied by an imaginary part, named i. This imaginary part i is defined by the property
$${i}^{2}=-1$$
. Although it might be difficult to intuitively map imaginary numbers to the physical world, they do easily result from common math operations. The classic way of obtaining an imaginary number is when we try to take the square root of a negative number, like
$$\sqrt{-5}$$
.

# What to Do when $${x}^{2}=-1$$?

There is no way of solving this equation by using real numbers since
$$x$$
would be equal to
$$\sqrt{-1}$$
, and we know that it is illegal to take the square root of a negative number. Therefore, mathematicians proposed a solution: substitute that value by a number, call it
$$i$$
.
$$\sqrt{-1}=i$$
And, given this equality, it follows that:
$${i}^{2}=ii=-1$$
Or, to put this another way:
$${i}^{2}=\sqrt{-1}\sqrt{-1}=-1$$
$${i}^{2}={\sqrt{-1}}^{2}=-1$$

# Problems with Imaginary Numbers

Now, let's see how to solve a more elaborate problem using the imaginary numbers.
First, we have:
$$\sqrt{-9}\sqrt{-4}+7$$
Let's rewrite the expression in a way that we can isolate
$$\sqrt{-1}$$
from the other terms:
$$\sqrt{9}\sqrt{-1}\sqrt{4}\sqrt{-1}+7$$
Next, substitute
$$\sqrt{-1}$$
with
$$i$$
and simplify the roots that are now positive:
$$3i\times 2i+7$$
$$6{i}^{2}+7$$
Substitute
$${i}^{2}$$
with
$$-1$$
:
$$6\times -1+7$$
$$-6+7$$
$$1$$
$$1$$