If |z|, = 4 and arg,,z = frac{{5pi }}{6},then | vedclass

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(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 $	heta = \frac{{5\pi }}{6}

Vedclass · Accepted Answer

$ - 2\sqrt 3 + 2i$

Vedclass · Answer

(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 $	heta = \frac{{5\pi }}{6} = {150^o}$
$	herefore $ $x = r\cos 	heta = 4\cos \,\,{150^o} = - 2\sqrt 3 $.
and $y = r\sin 	heta = 4$$\sin {150^o} = 4\frac{1}{2} = 2$
$	herefore $ $z = x + iy = - 2\sqrt 3 + 2i$
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.

If $|z|\, = 4$ and $arg\,\,z = \frac{{5\pi }}{6},$then $z =$

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