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  • Give answer! - Complex numbers and quadratic equations - JEE Main-2

Polar representation of complex number z= \left ( 1-\sqrt{3}i \right )\left ( -1+i \right ) will be

  • Option 1)

    z= 2\sqrt{2}\left ( \cos \frac{5\pi }{12}+i\sin\frac{5\pi }{12} \right )

  • Option 2)

    z= 2\sqrt{2}\left ( \cos \frac{\pi }{3}+i\sin\frac{\pi }{3} \right )

  • Option 3)

    z= 2\sqrt{2}\left ( \cos \left (\frac{-5\pi }{12} \right )+i\sin\left (\frac{-5\pi }{12} \right ) \right )

  • Option 4)

    z= 2\sqrt{2}\left ( \cos \frac{\pi }{6}+i\sin\frac{\pi }{6} \right )

 

Answers (1)

r= \left | z \right |= \left | \left ( 1-\sqrt3i \right ) \cdot \left ( -1+i \right )\right |

r= \left | \left ( 1-\sqrt3i \right ) \cdot \left ( -1+i \right )\right |= 2\times \sqrt{2}

r=2\sqrt2

argz = -\tan ^{-1}\left | \frac{-\sqrt3}{1} \right | + \pi - \tan ^{-1}\left | \frac{1}{-1} \right | + 2n\pi

argz = \frac{-\pi }{3}+\pi - \frac{\pi }{4}+ 2n\pi

argz = \frac{5\pi }{12}+ 2n\pi

which gives arg(z) 

5\pi /12 for n= 0

 

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)

z= 2\sqrt{2}\left ( \cos \frac{5\pi }{12}+i\sin\frac{5\pi }{12} \right )

This is correct

Option 2)

z= 2\sqrt{2}\left ( \cos \frac{\pi }{3}+i\sin\frac{\pi }{3} \right )

This is incorrect

Option 3)

z= 2\sqrt{2}\left ( \cos \left (\frac{-5\pi }{12} \right )+i\sin\left (\frac{-5\pi }{12} \right ) \right )

This is incorrect

Option 4)

z= 2\sqrt{2}\left ( \cos \frac{\pi }{6}+i\sin\frac{\pi }{6} \right )

This is incorrect

Posted by

subam

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