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  • Answer please! - Complex numbers and quadratic equations - JEE Main-7

arg \left \{ \left ( \sqrt{3}+i \right )\left ( 1-i \right ) \right \}  equals

  • Option 1)

    \frac{-\pi }{12}

  • Option 2)

    \frac{\pi }{12}

  • Option 3)

    \frac{-\pi }{6}

  • Option 4)

    \frac{-\pi }{3}

 

Answers (1)

best_answer

arg((\sqrt{3}+i) \left ( 1-i \right )) arg \left ( \sqrt3 + i \right )+ arg\left ( 1-i \right )+ 2n\pi

arg((\sqrt{3}+i) \left \right ))= \tan ^{-1}(\frac{1}{\sqrt3})= \frac{\pi }{6}

arg((1-i) \left \right ))= \tan ^{-1}arg((1-i) \left \right ))= \tan ^{-1} \left | -1/1\right | = -\pi /4

required argument = \frac{\pi }{6} - \frac{\pi }{4} +2n\pi

= \frac{-\pi }{12}+2n\pi 

for n=0 , it gives one argument \frac{-\pi }{12}

 

Property of Argument of a Complex Number -

Arg(z.w)=Arg(z)+Arg(w)+2n\pi

- wherein

n\epsilon Integer and it is chosen such that Arg(z.w) lies in the principal value range of Argument.  

 

 


Option 1)

\frac{-\pi }{12}

This is correct

Option 2)

\frac{\pi }{12}

This is incorrect

Option 3)

\frac{-\pi }{6}

This is incorrect

Option 4)

\frac{-\pi }{3}

This is incorrect

Posted by

prateek

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