4-1.Complex numbers
easy

$ - 1 - i\sqrt 3 $ का कोणांक है

A

$\frac{{2\pi }}{3}$

B

$\frac{\pi }{3}$

C

$ - \frac{\pi }{3}$

D

$ - \frac{{2\pi }}{3}$

Solution

(d) माना $z =  – 1 – i\sqrt 3 $ तब $\alpha  = {\tan ^{ – 1}}\left| {\,\frac{b}{a}\,} \right| = {\tan ^{ – 1}}\left| {\, – \frac{{\sqrt 3 }}{1}\,} \right| = \frac{\pi }{3}$

स्पष्टत: $z$ तृतीय चतुर्थाश में स्थित है

अत: कोणाक $\theta  =  – (\pi  – \alpha ) =  – (\pi  – \pi /3) = \frac{{ – 2\pi }}{3}$.

 

Standard 11
Mathematics

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