Solution 3.2:6d

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Current revision (13:56, 29 October 2008) (edit) (undo)
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We can determine the number's magnitude directly using the distance formula,
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<center> [[Bild:3_2_6d.gif]] </center>
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{{Displayed math||<math>\begin{align}
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\bigl|\sqrt{10}+\sqrt{30}i\bigr|
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&= \sqrt{\bigl(\sqrt{10}\,\bigr)^2+\bigl(\sqrt{30}\,\bigr)^2}\\[5pt]
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&= \sqrt{10+30}\\[5pt]
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&= \sqrt{40}\\[5pt]
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&= \sqrt{4\cdot 10}\\[5pt]
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&= 2\sqrt{10}\,\textrm{.}
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\end{align}</math>}}
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In addition, the number lies in the first quadrant and we can therefore determine the argument using simple trigonometry.
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[[Image:3_2_6_d_bild.gif]] [[Image:3_2_6_d_bildtext.gif]]
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The polar form is
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{{Displayed math||<math>2\sqrt{10}\Bigl( \cos\frac{\pi}{3} + i\sin\frac{\pi}{3} \Bigr)\,\textrm{.}</math>}}

Current revision

We can determine the number's magnitude directly using the distance formula,

\displaystyle \begin{align}

\bigl|\sqrt{10}+\sqrt{30}i\bigr| &= \sqrt{\bigl(\sqrt{10}\,\bigr)^2+\bigl(\sqrt{30}\,\bigr)^2}\\[5pt] &= \sqrt{10+30}\\[5pt] &= \sqrt{40}\\[5pt] &= \sqrt{4\cdot 10}\\[5pt] &= 2\sqrt{10}\,\textrm{.} \end{align}

In addition, the number lies in the first quadrant and we can therefore determine the argument using simple trigonometry.

Image:3_2_6_d_bild.gif Image:3_2_6_d_bildtext.gif

The polar form is

\displaystyle 2\sqrt{10}\Bigl( \cos\frac{\pi}{3} + i\sin\frac{\pi}{3} \Bigr)\,\textrm{.}
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