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Let $Z$ and $W$ be complex numbers such that $|Z|=|W|$, and $\arg Z$ denotes the principal argument of $Z$.
Statement 1:If $\arg Z+\arg W=\pi$, then $Z=-\bar{W}$.
Statement 2: $|Z|=|W|$, implies arg $Z-\arg \bar{W}=\pi$.
Options:
Statement 1:If $\arg Z+\arg W=\pi$, then $Z=-\bar{W}$.
Statement 2: $|Z|=|W|$, implies arg $Z-\arg \bar{W}=\pi$.
Solution:
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Verified Answer
The correct answer is:
Statement 1 is true, Statement 2 is false.
Statement 1 is true, Statement 2 is false.
Let $|Z|=|\mathrm{W}|=r$
$$
\Rightarrow Z=r e^{i \theta}, \mathrm{W}=r e^{i \phi}
$$
where $\theta+\phi=\pi$
$$
\therefore \quad \bar{W}=r e^{-i \phi}
$$
Now, $Z=r e^{i(\pi-\phi)}=r e^{i \pi} \times e^{-i \phi}=-r e^{-i \phi}$
$$
=-\bar{W}
$$
Thus, statement-1 is true but statement- 2 is false.
$$
\Rightarrow Z=r e^{i \theta}, \mathrm{W}=r e^{i \phi}
$$
where $\theta+\phi=\pi$
$$
\therefore \quad \bar{W}=r e^{-i \phi}
$$
Now, $Z=r e^{i(\pi-\phi)}=r e^{i \pi} \times e^{-i \phi}=-r e^{-i \phi}$
$$
=-\bar{W}
$$
Thus, statement-1 is true but statement- 2 is false.
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