Question

If Z and W are two non zero complex numbers such that | ZW | = 1 and Arg(Z) - Arg(W) = π/2, then ( Z bar × W ) IS equal to... ​

Answer 1

Answer:

Answer...................

Answer 2

Given:

$Z$  and ω are two non zero complex numbers

| zω | = 1   and Arg(z) - Arg(ω) =π/2

To find:

We have to find the value of $\bar{z}\omega$

Solution:

Answer:  (a) -i

Let $z=r_{1}e^{i\theta}$  and  $\omega=r_{2}e^{i\phi}$

It is given that $|z\omega|=1$

Now  $|r_{1}e^{i\theta}.r_{2}e^{i\phi} | = 1$

      $|r_{1}r_{2}|.|e^{i\theta}| .|e^{i\phi} } | = 1$

Here $e^{i\theta} =1$ because

$e^{i\theta}=cos{\theta}+i sin{\theta}\\| e^{i\theta}|=\sqrt{cos^{2}{\theta} +sin^{2}{\theta} } \\| e^{i\theta}| = \sqrt{1}=1$

similarly $|e^{i\phi} | =1$

So $r_{1}r_{2}=1$ _ _ _ (1)

And  $arg(z)-arg(\omega)=\frac{\pi}{2}$

                  $\theta-\phi=\frac{\pi}{2}$ _ _ _(2)

Now $\bar{z}=r_{1}e^{-i\theta}$

So, $\bar{z}\omega=r_{1}e^{-i\theta} .r_{2}e^{i\phi}$

     $\bar{z}\omega=r_{1}.r_{2}.e^{-i\theta+i\phi}$

     $\bar{z}\omega=1.e^{-i(\theta-\phi)}$      [ ∵ from (1)]

     $\bar{z}\omega=e^{-i\frac{\pi}{2} }$                [∵ from (2) ]

     $\bar{z}\omega = cos\frac{\pi}{2} -i sin\frac{\pi}{2}$

     $\bar{z}\omega=0-i(1)$

     $\bar{z}\omega=-i$

∴ $\bar{z}\omega=-i$