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Question
Write the argument of \[\left( 1 + i\sqrt{3} \right)\left( 1 + i \right)\left( \cos\theta + i\sin\theta \right)\].
Disclaimer: There is a misprinting in the question. It should be \[\left( 1 + i\sqrt{3} \right)\] instead of \[\left( 1 + \sqrt{3} \right)\].
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Solution
Let the argument of \[\left( 1 + i\sqrt{3} \right)\] be α. Then,
\[\tan\alpha = \frac{\sqrt{3}}{1} = \tan\frac{\pi}{3}\]
\[ \Rightarrow \alpha = \frac{\pi}{3}\]
Let the argument of \[\left( 1 + i \right)\] be β. Then,
\[\text { tan }\beta = \frac{1}{1} = \tan\frac{\pi}{4}\]
\[ \Rightarrow \beta = \frac{\pi}{4}\]
Let the argument of \[\left( cos\theta + isin\theta \right)\] be γ. Then,
\[\text { tan }\gamma = \frac{sin\theta}{cos\theta} = \text { tan }\theta\]
\[ \Rightarrow \gamma = \theta\]
∴ The argument of
\[\left( 1 + i\sqrt{3} \right)\left( 1 + i \right)\left( cos\theta + isin\theta \right) = \alpha + \beta + \gamma = \frac{\pi}{3} + \frac{\pi}{4} + \theta = \frac{7\pi}{12} + \theta\]
Hence, the argument of
\[\left( 1 + i\sqrt{3} \right)\left( 1 + i \right)\left( cos\theta + isin\theta \right) is \frac{7\pi}{12} + \theta\]
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