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The Argument of 1 − I √ 3 1 + I √ 3 is

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प्रश्न

The argument of \[\frac{1 - i\sqrt{3}}{1 + i\sqrt{3}}\] is

विकल्प

  •  60°

  • 120°

  • 210°

  • 240°

MCQ
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उत्तर

240°

\[\frac{1 - i\sqrt{3}}{1 + i\sqrt{3}}\]

\[\text { Rationalising the denominator,} \]

\[\frac{1 - i\sqrt{3}}{1 + i\sqrt{3}} \times \frac{1 - i\sqrt{3}}{1 - i\sqrt{3}}\]

\[ = \frac{1 + 3 i^2 - 2\sqrt{3} i}{1 - 3 i^2}\]

\[ = \frac{- 2 - 2\sqrt{3} i}{4} \left( \because i^2 = - 1 \right)\]

\[ = \frac{- 1}{2} - i\frac{\sqrt{3}}{2}\]

\[\tan \alpha = \left| \frac{Im (z)}{Re (z)} \right|\]

\[\text { Then,} \tan \alpha = \left| \frac{\frac{- \sqrt{3}}{2}}{\frac{- 1}{2}} \right|\]

\[ = \sqrt{3} \]

\[ \Rightarrow \alpha = 60°\]

\[\text { Since the points } \left( \frac{- 1}{2}, \frac{- \sqrt{3}}{2} \right) \text { lie in the third quadrant, the argument is given by}: \]

\[\theta = 180° + 60°\]

\[ = 240°\]

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अध्याय 13: Complex Numbers - Exercise 13.6 [पृष्ठ ६५]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 13 Complex Numbers
Exercise 13.6 | Q 18 | पृष्ठ ६५

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