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  • Solve it, - Complex numbers and quadratic equations - JEE Main-8

Which of the following can't be argument of z= 3-\sqrt{3}i

  • Option 1)

    \frac{11\pi }{6}

  • Option 2)

    \frac{-13\pi }{6}

  • Option 3)

    \frac{-7\pi }{6}

  • Option 4)

    \frac{23\pi }{6}

 

Answers (1)

\because z lies in 4th quadrant so its argument =-\tan ^{-1}\left | \frac{-\sqrt{3}}{3} \right |=-\tan ^{-1}\left ( \frac{1}{\sqrt{3}} \right )=\frac{-\pi }{6}

\frac{-\pi }{6} is principal argument so \frac{-\pi }{6}+2n\pi ( Where n\, \epsilon \, I)  will be general argument

for  n=1, (A) will be argument

for n=-1 (B) will be argument

for n=2, (D) will be argument

But there is no integer n for which argument is \frac{-7\pi }{6}

\therefore Option (C)

 

General Argument of a Complex Number -

2n\pi+\theta

- wherein

\theta is the principal argument of complex number.

 

 


Option 1)

\frac{11\pi }{6}

This is incorrect

Option 2)

\frac{-13\pi }{6}

This is incorrect

Option 3)

\frac{-7\pi }{6}

This is correct

Option 4)

\frac{23\pi }{6}

This is incorrect

Posted by

subam

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