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Question
Find the principal argument of \[\left( 1 + i\sqrt{3} \right)^2\] .
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Solution
\[z = \left( 1 + i\sqrt{3} \right)^2 \]
\[ = 1 + 3 i^2 + 2\sqrt{3}i\]
\[ = 1 - 3 + 2\sqrt{3}i\]
\[ = - 2 + 2\sqrt{3}i\]
\[\text { Let } \beta \text { be an acute angle given by } \tan\beta = \frac{\left| Im\left( z \right) \right|}{\left| Re\left( z \right) \right|} . \text { Then }, \]
\[\tan\beta = \frac{\left| 2\sqrt{3} \right|}{\left| 2 \right|} = \left| \sqrt{3} \right|\]
\[ \Rightarrow \tan\beta = \left| \tan\frac{\pi}{3} \right|\]
\[ \Rightarrow \beta = \frac{\pi}{3}\]
\[\text { Clearly, z lies in the second quadrant . Therefore, }\arg\left( z \right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3} . \]
\[\text { Hence, the principal argument of z is } \frac{2\pi}{3} .\]
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