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  • Solve this problem - Complex numbers and quadratic equations - JEE Main

Let  z= \frac{\left ( \sqrt{3}+i \right )\left ( 1+\sqrt{3}i \right )}{\left ( -1+i \right )\left ( 2-2i \right )} then 

  • Option 1)

    arg\left ( z \right )=\frac{\pi }{3}

  • Option 2)

    arg\left ( z \right )=\frac{\pi }{6}

  • Option 3)

    z is purely imaginary 

  • Option 4)

    z is purely real

 

Answers (1)

best_answer

arg\left ( \sqrt3+i \right )+arg\left ( 1+\sqrt3i \right )- arg \left ( -1+i \right )- arg (2-2i)

arg(z)= \tan ^{-1}(\frac{1}{\sqrt3})+ \tan ^{-1}(\frac{\sqrt3}{1})- \left \{ \pi -\tan ^{-1\left | \frac{1}{-1} \right |} \right \}

-\left \{ -\tan \left | \frac{-2}{2} \right | \right \}

arg(z)= \frac{\pi }{6}+\frac{\pi }{3}-\left (\pi -\frac{\pi }{4} \right )+\frac{\pi }{4}

arg=0

z is purely real 

 

Properties of Argument of a Complex Number -

Arg(z)=0\Rightarrow z\ is\ real\Rightarrow z=\bar{z}

- wherein

Arg(z) denotes Argument of z and \bar{z} denotes conjugate of z

 

 

 


Option 1)

arg\left ( z \right )=\frac{\pi }{3}

This is incorrect

Option 2)

arg\left ( z \right )=\frac{\pi }{6}

This is incorrect

Option 3)

z is purely imaginary 

This is incorrect

Option 4)

z is purely real

This is correct

Posted by

Aadil

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