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Complex Numbers Part 1

Section: Mathematics

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Question : 22 of 100

Marks: +1, -0

If

z

\omega

are two complex numbers such that

|z \omega|=1

\arg(z)-\arg(\omega)=\frac{3\pi}{2}

\arg\left(\frac{1-2\overline{z}\omega}{1+3\overline{z}\omega}\right)

\arg(z)

denotes the principal argument of complex number z)

[20 Jul 2021 Shift 1]

$\frac{\pi}{4}$
$-\frac{3\pi}{4}$
$-\frac{\pi}{4}$
$\frac{3\pi}{4}$

Solution:

As

|z \omega|=1

\Rightarrow

If

|z|=r

, then

|\omega|=\frac{1}{r}

Let

\arg(z)=\theta

\therefore \arg(\omega)=\left(\theta-\frac{3\pi}{2}\right)

So,

z=r e^{i\theta}

\Rightarrow \overline{Z}=r e^{i(-\theta)}

\omega=\frac{1}{r} e^{i\left(\theta-\frac{3\pi}{2}\right)}

Now, consider

\frac{1-2\overline{z}\omega}{1+3\overline{z}\omega}=\frac{1-2 e^{i\left(-\frac{3\pi}{2}\right)}}{1+3 e^{-\frac{3\pi}{2}}}=\left(\frac{1-2i}{1+3i}\right)

=\frac{(1-2i)(1-3i)}{(1+3i)(1-3i)}=-\frac{1}{2}(1+i)

\therefore

prin

\arg\left(\frac{1-2\overline{z}\omega}{1+3\overline{z}\omega}\right)

=

prin

\arg\left(\frac{1-2\overline{z}\omega}{1+3\overline{z}\omega}\right)

=\left(-\frac{1}{2}(1+i)\right)

=-\left(\pi-\frac{\pi}{4}\right)=-\frac{3\pi}{4}

So, option (2) is correct.

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