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If z= \frac{-2}{1+i \sqrt 3}  ,     then the value of     \arg (z) is:

  • Option 1)

    \pi

  • Option 2)

    \frac{\pi}{3}

  • Option 3)

    \frac{2\pi}{3}

  • Option 4)

    \frac{\pi}{4}

 

Answers (1)

best_answer

As we learnt in 

Definition of Argument/Amplitude of z in Complex Numbers -

\theta =tan^{-1}|\frac{y}{x}|, z\neq 0

\boldsymbol{\theta,\pi-\theta,-\pi+\theta,-\theta} are Principal Argument if z lies in first, second, third or fourth quadrant respectively.

- wherein

 

 Z=\frac{-2}{1+i\sqrt{3}}=\frac{-2}{1+i\sqrt{3}}\times \frac{1-i\sqrt{3}}{1-i\sqrt{3}}\\*\\*= \frac{-2\left ( 1-i\sqrt{3} \right )}{1^{2}+\sqrt{3}^{2}}=\frac{\left ( 1-i\sqrt{3} \right )}{2}\\*\\*= \frac{-1}{2}+\frac{i\sqrt{3}}{2}

Thus \theta=\frac{2\pi}{3}


Option 1)

\pi

Incorrect

Option 2)

\frac{\pi}{3}

Incorrect

Option 3)

\frac{2\pi}{3}

Correct

Option 4)

\frac{\pi}{4}

Incorrect

Posted by

Aadil

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