Question
Convert the complex number in the polar form: $\sqrt 3 + i$
✓
Answer
Here $z = \sqrt 3 + i = r(\cos \theta + i\sin \theta )$
$ \Rightarrow r\cos \theta = \sqrt 3 $ and $r\;\sin \theta = 1$
Squaring both sides of (i) and adding
${r^2}({\cos ^2}\theta + {\sin ^2}\theta ) = 3 + 1$$ \Rightarrow {r^2} = 4 \Rightarrow r = 2$
$\therefore 2\cos \theta = \sqrt 3 $ and $2\sin \theta = 1$
$\therefore \cos \theta = \frac{{\sqrt 3 }}{2}$ and $\sin \theta = \frac{1}{2}$
Since $\sin \theta $ and $\cos \theta $ are both positive
$\therefore \theta $ lies in first quadrant
$\therefore \theta = \frac{\pi }{6}$
Hence polar form of z is $2\left( {\cos \frac{\pi }{6} + i\;\sin \frac{\pi }{6}} \right)$
$ \Rightarrow r\cos \theta = \sqrt 3 $ and $r\;\sin \theta = 1$
Squaring both sides of (i) and adding
${r^2}({\cos ^2}\theta + {\sin ^2}\theta ) = 3 + 1$$ \Rightarrow {r^2} = 4 \Rightarrow r = 2$
$\therefore 2\cos \theta = \sqrt 3 $ and $2\sin \theta = 1$
$\therefore \cos \theta = \frac{{\sqrt 3 }}{2}$ and $\sin \theta = \frac{1}{2}$
Since $\sin \theta $ and $\cos \theta $ are both positive
$\therefore \theta $ lies in first quadrant
$\therefore \theta = \frac{\pi }{6}$
Hence polar form of z is $2\left( {\cos \frac{\pi }{6} + i\;\sin \frac{\pi }{6}} \right)$
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