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If z and are two non­zero complex numbers such that then is equal to

  • Option 1)


  • Option 2)


  • Option 3)


  • Option 4)



Answers (2)

As we learnt in

Euler's Form of a Complex number -


- wherein

r denotes modulus of z and \theta denotes argument of z.




\left |z\omega \right |=1  and \alpha-\beta=\frac{\pi}{2}

\therefore\ \; \beta=\alpha-\frac{\pi}{2}

Let A=r_{1}(cos\alpha + i sin \alpha)

    \omega=r_{2}(cos\beta + i sin \beta)

So that |z|^{2}=z \overline{z}

Now \left |z\omega \right |=1

\Rightarrow\ \; \left |z\omega \right |^{2}=1

\Rightarrow\ \; \(z\omega){(\bar{z}\bar{\omega}})=1

\therefore\ \;{\bar{z}\bar{\omega}}=\frac{1}{\bar{\omega}z}

\Rightarrow\ \; \frac{1}{r_{2}(cos\beta- i sin\beta).r_{1}(cos\alpha +i sin\alpha)}



Since e^{\alpha i}\frac{\pi}{2}=1

Since z=r_{1}e^{i\alpha},   \omega=r_{2}e^{i\beta}


Correct option is 3.

Option 1)


This is an incorrect option.

Option 2)


This is an incorrect option.

Option 3)


This is the correct option.

Option 4)


This is an incorrect option.

Posted by

Sabhrant Ambastha

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