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 \(~\dfrac{1}{\sqrt{3x}}\)
We multiply the numerator and denominator by the denominator, \(\sqrt{3x}\text{.}\)
Example 7.31.
Rationalize the denominator \(~\dfrac{x}{\sqrt{2}+\sqrt{x}}\)
When there are two terms in the denominator, we multiply numerator and denominator by the conjugate of the denominator, in this case \(~\sqrt{2}-\sqrt{x}\text{.}\)
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
Checkpoint 7.32.
Rationalize the denominator \(~\dfrac{2\sqrt{3}}{\sqrt{2k}}\)
Checkpoint 7.33.
Simplify the radical, then rationalize the denominator: \(~\sqrt{\dfrac{27x}{20}}\)
Checkpoint 7.34.
Rationalize the denominator \(~\dfrac{3}{7-\sqrt{2}}\)
Checkpoint 7.35.
Rationalize the denominator \(~\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
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 \(~w^2\text{.}\)
Use the formula \(~(a-b)^2=a^2-2ab+b^2~\) with \(~a=\dfrac{i\sqrt{3}}{2}~\) and \(~b=\dfrac{1}{2}.~\)
Subsubsection Exercises
Checkpoint 7.37.
Use the values of \(w\) and \(w^2\) from the Example to calculate \(w^3\text{.}\)
Checkpoint 7.38.
Show that \(~\sqrt{i} = \dfrac{\sqrt{2}}{2} +i \dfrac{\sqrt{2}}{2}\)
Square \(~\dfrac{\sqrt{2}}{2} +i \dfrac{\sqrt{2}}{2}.\)