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Amplitude of the complex number $i \sin \left(\frac{\pi}{19}\right)$ is
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Verified Answer
The correct answer is:
$\frac{\pi}{2}$
Let $z=i \sin \frac{\pi}{19}$
or $\quad z=0+i \sin \frac{\pi}{19}$
Let amplitude of $z$ be $\theta$, then
$$
\sin \theta=\frac{\sin \frac{\pi}{19}}{\sqrt{0^{2}+\left(\sin \frac{\pi}{19}\right)^{2}}}\left[\because \sin \theta=\frac{b}{\sqrt{a^{2}+b^{2}}}\right]
$$
or $\sin \theta=\frac{\sin \frac{\pi}{19}}{\sin \frac{\pi}{19}}=1 \Rightarrow \theta=\frac{\pi}{2}$
or $\quad z=0+i \sin \frac{\pi}{19}$
Let amplitude of $z$ be $\theta$, then
$$
\sin \theta=\frac{\sin \frac{\pi}{19}}{\sqrt{0^{2}+\left(\sin \frac{\pi}{19}\right)^{2}}}\left[\because \sin \theta=\frac{b}{\sqrt{a^{2}+b^{2}}}\right]
$$
or $\sin \theta=\frac{\sin \frac{\pi}{19}}{\sin \frac{\pi}{19}}=1 \Rightarrow \theta=\frac{\pi}{2}$
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