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4-1.Complex numbers
easy
જો $|z|\, = 4$ અને $arg\,\,z = \frac{{5\pi }}{6},$તો $z =$
A
$2\sqrt 3 - 2i$
B
$2\sqrt 3 + 2i$
C
$ - 2\sqrt 3 + 2i$
D
$ - \sqrt 3 + i$
Solution
(c)$|z| = 4$and $arg\,z = \frac{{5\pi }}{6} = {150^o}$
Let $z = x + iy$, then $|z| = r = \sqrt {{x^2} + {y^2}} = 4$
and $\theta = \frac{{5\pi }}{6} = {150^o}$
$\therefore $ $x = r\cos \theta = 4\cos \,\,{150^o} = – 2\sqrt 3 $.
and $y = r\sin \theta = 4$$\sin {150^o} = 4\frac{1}{2} = 2$
$\therefore $ $z = x + iy = – 2\sqrt 3 + 2i$
Trick : Since $arg\,z = \frac{{5\pi }}{6} = {150^o}$,
here the complex number must lie in second quadrant, so $(a) $ and $(b) $ rejected. Also $|z| = 4$ which satisfies $(c)$ only.
Standard 11
Mathematics
