MCQ
If $ - 1 + \sqrt { - 3} = r{e^{i\theta }},$then $\theta $ is equal to
  • A
    $\frac{\pi }{3}$
  • B
    $ - \frac{\pi }{3}$
  • $\frac{{2\pi }}{3}$
  • D
    $ - \frac{{2\pi }}{3}$

Answer

Correct option: C.
$\frac{{2\pi }}{3}$
c
(c) Here $ - 1 + \sqrt { - 3} = r{e^{i\theta }}$==> $ - 1 + i\sqrt 3 = r{e^{i\theta }}$
$ = r\cos \theta + ir\sin \theta $
Equating real and imaginary parts, we get
$r\cos \theta = - 1$and $r\sin \theta = \sqrt 3 $
Hence$\tan \theta = - \sqrt 3 \,\, \Rightarrow \tan \theta = \tan \frac{{2\pi }}{3}$.Hence $\theta = \frac{{2\pi }}{3}$.

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