Question:
Amplitude of the complex number $i\, \sin \bigg( \frac{\pi}{19} \bigg)$ is
Amplitude of the complex number $i\, \sin \bigg( \frac{\pi}{19} \bigg)$ is
Updated On: May 12, 2024
- $ \frac{\pi}{19}$
- $- \frac{\pi}{19}$
- $ \frac{\pi}{2}$
- $ \frac{\pi}{2} - \frac{\pi}{19}$
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The Correct Option is C
Solution and Explanation
Let $z = i \, \sin \frac{\pi}{19}$
or $z = 0 + i \, \sin \frac{\pi}{19}$
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[\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 $z = 0 + i \, \sin \frac{\pi}{19}$
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[\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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Concepts Used:
Complex Numbers and Quadratic Equations
Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.
Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.
