Get Answers to all your Questions

header-bg qa

The real part of  [1+\cos \left( \frac{\pi}{5} \right )+i \sin \left( \frac{\pi}{5} \right )]^{-1}   is:

  • Option 1)

    1

  • Option 2)

    \frac{1}{2}

  • Option 3)

    \frac{1}{2} \cos \left ( \frac{\pi}{10} \right )

  • Option 4)

    \frac{1}{2} \cos \left ( \frac{\pi}{5} \right )

 

Answers (2)

best_answer

As 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

 

 

\left[ 1+\cos \frac {\pi}{5}+ i \sin \frac{\pi}{5} \right ]^{-1} 

\left[ 2\cos^{2} \frac {\pi}{10}+ 2i \sin \frac {\pi}{10} \cos \frac {\pi}{10} \right ]^{-1}

\frac{1}{2 \cos \frac{\pi}{10}}\times \left[ \cos \frac{\pi}{10} - i \sin \frac {\pi}{10} \right ]

 


Option 1)

1

This option is incorrect.

Option 2)

\frac{1}{2}

This option is correct.

Option 3)

\frac{1}{2} \cos \left ( \frac{\pi}{10} \right )

This option is incorrect.

Option 4)

\frac{1}{2} \cos \left ( \frac{\pi}{5} \right )

This option is incorrect.

Posted by

Aadil

View full answer

As 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

 

 

\left[ 1+\cos \frac {\pi}{5}+ i \sin \frac{\pi}{5} \right ]^{-1} 

\left[ 2\cos^{2} \frac {\pi}{10}+ 2i \sin \frac {\pi}{10} \cos \frac {\pi}{10} \right ]^{-1}

\frac{1}{2 \cos \frac{\pi}{10}}\times \left[ \cos \frac{\pi}{10} - i \sin \frac {\pi}{10} \right ]

 


Option 1)

1

This option is incorrect.

Option 2)

\frac{1}{2}

This option is correct.

Option 3)

\frac{1}{2} \cos \left ( \frac{\pi}{10} \right )

This option is incorrect.

Option 4)

\frac{1}{2} \cos \left ( \frac{\pi}{5} \right )

This option is incorrect.

Posted by

divya.saini

View full answer