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

Option 1)
$-1$

Option 2)
$i$

Option 3)
$-i$

Option 4)
$1$

Answers (2)

As we learnt in

Euler's Form of a Complex number -

$z=re^{i\theta}$

- 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$

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

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

$=\frac{1}{r_{1}.r_{2}.e^{i(\alpha-\beta)}}=\frac{1}{r_{1}.r_{2}e^{\pi/2}}=\frac{1}{ir_{1}.r_{2}}$

$=\frac{1}{i}=\frac{i}{5^{2}}=\frac{i}{-1}=-i$

Since $e^{\alpha i}\frac{\pi}{2}=1$

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

$|z\omega|=r_{1}r_{2}=1$

Correct option is 3.

Option 1)

$-1$

This is an incorrect option.

Option 2)

$i$

This is an incorrect option.

Option 3)

$-i$

This is the correct option.

Option 4)

$1$

This is an incorrect option.

Posted by

Sabhrant Ambastha

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