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Section 7.3 Complex Numbers

Subsection 1. Rationalize denominators

Irrational numbers are harder to approximate if there is a radical in the denominator. So we remove those radicals by "rationalizing the denominator."

Subsubsection Examples

Example 7.30.

Rationalize the denominator \(~\dfrac{1}{\sqrt{3x}}\)

Solution

We multiply the numerator and denominator by the denominator, \(\sqrt{3x}\text{.}\)

\begin{equation*} \dfrac{1}{\sqrt{3x}} \cdot \alert{\dfrac{\sqrt{3x}}{\sqrt{3x}}} = \dfrac{\sqrt{3x}}{3x} \end{equation*}
Example 7.31.

Rationalize the denominator \(~\dfrac{x}{\sqrt{2}+\sqrt{x}}\)

Solution

When there are two terms in the denominator, we multiply numerator and denominator by the conjugate of the denominator, in this case \(~\sqrt{2}-\sqrt{x}\text{.}\)

\begin{equation*} \dfrac{x\blert{(\sqrt{2}-\sqrt{x})}}{(\sqrt{2}+\sqrt{x})\blert{(\sqrt{2}-\sqrt{x})}}=\dfrac{x({\sqrt{2}-\sqrt{x})}}{2-x} \end{equation*}

We have used the formula for the differnce of two squares, \(~(a+b)(a-b)=a^2-b^2,~\) to simplify \(~(\sqrt{2}+\sqrt{x})(\sqrt{2}-\sqrt{x})~\) to \(~2-x~\text{.}\)

Subsubsection Exercises

Rationalize the denominator \(~\dfrac{2\sqrt{3}}{\sqrt{2k}}\)

Answer
\(\dfrac{\sqrt{6k}}{k}\)

Simplify the radical, then rationalize the denominator: \(~\sqrt{\dfrac{27x}{20}}\)

Answer
\(\dfrac{3\sqrt{15x}}{10}\)

Rationalize the denominator \(~\dfrac{3}{7-\sqrt{2}}\)

Answer
\(\dfrac{7+\sqrt{2}}{15}\)

Rationalize the denominator \(~\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)

Answer
\(\dfrac{x+2\sqrt{xy}+y}{x-y}\)

Subsection 2. Work with radicals

Remember that \(~~\blert{i = \sqrt{-1}}~~\text{,}\) and thus \(~~\blert{i^2 = -1}\text{.}\)

Subsubsection Example

Example 7.36.

Define \(~w=\dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2}.~\) Calculate \(~w^2\text{.}\)

Solution

Use the formula \(~(a-b)^2=a^2-2ab+b^2~\) with \(~a=\dfrac{i\sqrt{3}}{2}~\) and \(~b=\dfrac{1}{2}.~\)

\begin{align*} (a-b)^2 \amp = a^2-2ab+b^2\\ \left(\dfrac{i\sqrt{3}}{2}-\dfrac{1}{2}\right)^2 \amp = \left(\dfrac{i\sqrt{3}}{2}\right)^2-2\left(\dfrac{i\sqrt{3}}{2}\right)\left(\dfrac{1}{2}\right)+\left(\dfrac{1}{2}\right)^2\\ \amp = \dfrac{-3}{4}-\dfrac{i\sqrt{3}}{2}+\dfrac{1}{4}\\ \amp = \dfrac{-1}{2}-\dfrac{i\sqrt{3}}{2} \end{align*}

Subsubsection Exercises

Use the values of \(w\) and \(w^2\) from the Example to calculate \(w^3\text{.}\)

Answer
\(1\)

Show that \(~\sqrt{i} = \dfrac{\sqrt{2}}{2} +i \dfrac{\sqrt{2}}{2}\)

Answer

Square \(~\dfrac{\sqrt{2}}{2} +i \dfrac{\sqrt{2}}{2}.\)