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Step-by-Step Solution
Step 1: Understand the given information
We have two complex numbers, $z$ and $\omega$, with the conditions:
$|z\,\omega| = 1$
$\arg(z) - \arg(\omega) = \frac{3\pi}{2}$
We want to find the principal argument of the expression:
\[
\arg\Biggl(\frac{1 - 2\,\overline{z}\,\omega}{1 + 3\,\overline{z}\,\omega}\Biggr).
\]
Step 2: Express $z$ and $\omega$ in polar form
Let $|z| = r$. Since $|z\,\omega| = 1$, we have
\[
|z|\;|\omega| = 1 \quad \Rightarrow \quad r \,\cdot |\omega| = 1
\]
so
\[
|\omega| = \frac{1}{r}.
\]
Let $\arg(z) = \theta$. Then according to the given condition,
\[
\arg(\omega) = \arg(z) - \frac{3\pi}{2} = \theta - \frac{3\pi}{2}.
\]
Hence, in polar form,
\[
z = r\,e^{i\theta}, \quad \omega = \frac{1}{r}\,e^{\,i\left(\theta - \frac{3\pi}{2}\right)}.
\]
Step 3: Notice how $\overline{z}$ multiplies with $\omega$
Recall that if $z = r\,e^{i\theta}$, then its conjugate is
\[
\overline{z} = r\,e^{-\,i\theta}.
\]
Hence,
\[
\overline{z}\,\omega
= \Bigl(r\,e^{-\,i\theta}\Bigr) \Bigl(\tfrac{1}{r} e^{\,i(\theta - \tfrac{3\pi}{2})}\Bigr)
= e^{\,i\left(\theta - \frac{3\pi}{2} - \theta\right)}
= e^{\,i\left(-\,\frac{3\pi}{2}\right)}.
\]
But
\[
e^{\,i\left(-\,\tfrac{3\pi}{2}\right)}
= \cos\Bigl(-\,\tfrac{3\pi}{2}\Bigr) + i\,\sin\Bigl(-\,\tfrac{3\pi}{2}\Bigr)
= 0 + i \cdot 1
= i.
\]
Thus,
\[
\overline{z}\,\omega = i.
\]
Step 4: Substitute into the target expression
Substitute $\overline{z}\,\omega = i$ into
\[
\frac{1 - 2\,\overline{z}\,\omega}{1 + 3\,\overline{z}\,\omega}
= \frac{1 - 2i}{1 + 3i}.
\]
Step 5: Simplify the complex fraction
Multiply numerator and denominator by the conjugate of the denominator $(1 - 3i)$:
\[
\frac{1 - 2i}{1 + 3i} \times \frac{1 - 3i}{1 - 3i}
= \frac{(1 - 2i)(1 - 3i)}{(1 + 3i)(1 - 3i)}.
\]
Compute each part separately:
Denominator:
\[
(1 + 3i)(1 - 3i)
= 1 - 9i^2
= 1 - 9 \times (-1)
= 1 + 9
= 10.
\]
Numerator:
\[
(1 - 2i)(1 - 3i)
= 1(1 - 3i) - 2i(1 - 3i)
= (1 - 3i) - 2i + 6i^2
= 1 - 5i - 6
= -5 - 5i
= -5(1 + i).
\]
Putting them together:
\[
\frac{-5(1 + i)}{10}
= -\frac{5}{10}(1 + i)
= -\frac{1}{2}(1 + i).
\]
Step 6: Determine the principal argument
We now have
\[
\frac{1 - 2\,\overline{z}\,\omega}{1 + 3\,\overline{z}\,\omega}
= -\tfrac{1}{2}(1 + i).
\]
The complex number $(1 + i)$ can be written in polar form as
\[
\sqrt{2}\,e^{\,i\frac{\pi}{4}}
\]
since $1 + i$ has magnitude $\sqrt{1^2 + 1^2} = \sqrt{2}$ and argument $\frac{\pi}{4}$. Multiplying by $- \frac{1}{2}$ introduces a factor of $\frac{1}{2}$ in magnitude and an additional rotation by $\pi$ (since $-1$ corresponds to an argument of $\pi$). Hence,
\[
-\tfrac{1}{2}(1 + i)
= \tfrac{1}{2} \cdot \sqrt{2}\,e^{\,i\frac{\pi}{4}} \times e^{\,i\pi}
= \frac{\sqrt{2}}{2}\,e^{\,i\left(\frac{\pi}{4} + \pi\right)}
= e^{\,i\left(\frac{\pi}{4} + \pi\right)}
\]
up to a scaling factor in magnitude. The argument here becomes:
\[
\frac{\pi}{4} + \pi
= \frac{5\pi}{4}.
\]
Since $\frac{5\pi}{4}$ lies outside the principal interval $(-\pi,\,\pi]$, we convert it to its principal value by subtracting $2\pi$, obtaining:
\[
\frac{5\pi}{4} - 2\pi = \frac{5\pi}{4} - \frac{8\pi}{4} = -\frac{3\pi}{4}.
\]
Therefore, the principal argument is:
\[
-\frac{3\pi}{4}.
\]
Step 7: Conclude the correct answer
Hence, the required principal argument is
\[
-\frac{3\pi}{4},
\]
which matches option (2).
