The Imaginary Unit i Lesson

Up to this point, we have not been able to find the solution for a problem in which we had to take the square root of a negative number. The problem occurs because squaring a number always yields a non-negative result. Let's suppose we saw the following equation: $$\require{color}x^2=-9$$ We know that there is no real number that can be squared and give a result of -9. To deal with this situation, we will introduce the imaginary unit "i". The imaginary unit "i" is a number whose square is (-1). $$i^2=-1$$ If we take the square root of each side: $$\sqrt{i^2}=\sqrt{-1}$$ $$i=\sqrt{-1}$$ We can define i as the square root of -1. We can use this property to deal with our equation above. $$x^2=-9$$ Take the square root of each side: $$\sqrt{x^2}=\sqrt{-9}$$ $$x=\sqrt{-9}$$ Let's use i to solve our problem. $$x=\sqrt{-1}\cdot \sqrt{9}$$ We know the square root of (-1) is defined as i so we can replace this in our problem: $$x=i\sqrt{9}$$ Since the square root of 9 is 3, we can write our answer as: $$x=3i$$ Does this solution actually work in the original equation? Let's try it out. $$x^2=-9$$ $$(3i)^2=-9$$ $$3^2 \cdot i^2=-9$$ Remember i squared is defined as (-1). $$9(-1)=-9$$ $$-9=-9 \hspace{.2em}\color{green}{✔}$$ It's worth noting that -3i works as a solution as well. When solving this equation, we would take the principal and negative square root of (-9). We will learn this rule in a few lessons when we talk about the square root property. Let's look at a few examples.
Example 1: Write each as the product of i and a real number. $$\sqrt{-5}$$ Let's begin by writing the square root of -5 as the square root of -1 times the square root of 5. $$\sqrt{-1}\cdot \sqrt{5}$$ We can replace the square root of (-1) with i. $$i\sqrt{5}$$ Example 2: Write each as the product of i and a real number. $$\sqrt{-44}$$ Let's begin by writing the square root of -44 as the square root of -1 times the square root of 4 times the square root of 11 $$\sqrt{-1}\cdot \sqrt{4}\cdot \sqrt{11}$$ We can replace the square root of (-1) with i and the square root of 4 with 2. $$2i\sqrt{11}$$

Multiplying Square Roots of Negative Numbers

Up to this point, we have been using the product rule for radicals to simplify. When square roots of negative numbers are involved, we can't use the same rule. We have to change the form first before we can multiply. To see this in action, let's suppose we had the following problem: $$\sqrt{-6}\cdot \sqrt{-3}$$ Based on our prior experience, one may try to solve the problem in this manner: $$\sqrt{-6}\cdot \sqrt{-3}=\sqrt{-6 \cdot -3}=\sqrt{18}=3\sqrt{2}\hspace{.2em}\color{red}{✖}$$ The product rule for radicals is only valid for real numbers. To solve this problem, we can change the form first: $$\sqrt{-6}\cdot \sqrt{-3}=i\sqrt{6}\cdot i\sqrt{3}=i^2\sqrt{18}=-1 \cdot 3\sqrt{2}=-3\sqrt{2}$$ Let's look at an example.
Example 3: Simplify each. $$\sqrt{-2}\cdot \sqrt{-14}$$ First, we will change our form. $$i\sqrt{2}\cdot i\sqrt{14}$$ Now we can use our product rule for radicals. $$i\sqrt{2}\cdot i\sqrt{14}=i^2\sqrt{4 \cdot 7}=-2\sqrt{7}$$ This method will also apply to a quotient. Let's look at an example.
Example 4: Simplify each. $$\frac{\sqrt{-75}}{\sqrt{-3}}$$ First, we will change our form. $$\frac{i\sqrt{75}}{i\sqrt{3}}$$ Let's cancel common factors. $$\require{cancel}\frac{i \cdot \sqrt{3}\cdot \sqrt{25}}{i \cdot \sqrt{3}}=\frac{\cancel{i}\cdot \cancel{\sqrt{3}}\cdot \sqrt{25}}{\cancel{i}\cdot \cancel{\sqrt{3}}}=5$$

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