Solution 3.2:5c

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Geometrically, the multiplication of two complex numbers means that there magnitudes are multiplied and their arguments are added. The product <math>(\sqrt{3}+i)(1-i)</math> therefore has an argument which is the sum of the argument for the <math>\sqrt{3}+i</math> and <math>1-i</math>, i.e.
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{{Displayed math||<math>\arg \bigl((\sqrt{3}+i)(1-i)\bigr) = \arg (\sqrt{3}+i) + \arg (1-i)\,\textrm{.}</math>}}
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By drawing the factors in the complex plane, we can determine relatively easily the argument using simple trigonometry.
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[[Image:3_2_5_c.gif|center]]
[[Image:3_2_5_c.gif|center]]
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(Because <math>1-i</math> lies in the fourth quadrant, the argument equals
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<math>-\beta</math> and not <math>\beta</math>.)
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Hence,
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{{Displayed math||<math>\arg \bigl((\sqrt{3}+i)(1-i)\bigr) = \arg (\sqrt{3}+i) + \arg (1-i) = \frac{\pi}{6} - \frac{\pi}{4} = -\frac{\pi}{12}\,\textrm{.}</math>}}
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Note: If you prefer to give the argument between <math>0</math> and <math>2\pi </math>, then the answer is
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{{Displayed math||<math>-\frac{\pi}{12}+2\pi = \frac{-\pi+24\pi}{12} = \frac{23\pi}{12}\,\textrm{.}</math>}}

Current revision

Geometrically, the multiplication of two complex numbers means that there magnitudes are multiplied and their arguments are added. The product \displaystyle (\sqrt{3}+i)(1-i) therefore has an argument which is the sum of the argument for the \displaystyle \sqrt{3}+i and \displaystyle 1-i, i.e.

\displaystyle \arg \bigl((\sqrt{3}+i)(1-i)\bigr) = \arg (\sqrt{3}+i) + \arg (1-i)\,\textrm{.}

By drawing the factors in the complex plane, we can determine relatively easily the argument using simple trigonometry.

(Because \displaystyle 1-i lies in the fourth quadrant, the argument equals \displaystyle -\beta and not \displaystyle \beta.)

Hence,

\displaystyle \arg \bigl((\sqrt{3}+i)(1-i)\bigr) = \arg (\sqrt{3}+i) + \arg (1-i) = \frac{\pi}{6} - \frac{\pi}{4} = -\frac{\pi}{12}\,\textrm{.}


Note: If you prefer to give the argument between \displaystyle 0 and \displaystyle 2\pi , then the answer is

\displaystyle -\frac{\pi}{12}+2\pi = \frac{-\pi+24\pi}{12} = \frac{23\pi}{12}\,\textrm{.}
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