The amplitude of $1-\cos \theta -i\sin \theta $ is:
A. $\dfrac{\pi -\theta }{2}$
B. $\dfrac{\theta }{2}$
C. $\dfrac{\theta -\pi }{2}$
D. $\dfrac{\pi +\theta }{2}$
Answer
636.3k+ views
Hint: First we are going to state the meaning of the amplitude of a complex number and then we will use it to find the amplitude and then we will use some trigonometric formula to simplify it further to convert it in the form of options that are given.
Complete Step-by-Step solution:
Let’s first write the definition of amplitude of complex number,
The complex number is referred to as the complex amplitude, a polar representation of the amplitude and the initial phase of the complex exponential signal. The complex amplitude is also called a phasor as it can be represented graphically as a vector in the complex plane.
So, if x + iy is a complex number then it’s amplitude $\alpha $ will be:
$\tan \alpha =\dfrac{y}{x}$
Now using the same formula on $1-\cos \theta -i\sin \theta $ we get,
$\tan \alpha =\dfrac{-\sin \theta }{1-\cos \theta }$
Now we are going to use some trigonometric formula,
$\begin{align}
& \sin 2x=\dfrac{2\tan x}{1+{{\tan }^{2}}x} \\
& \cos 2x=\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x} \\
\end{align}$
Now using these formula we get,
$\begin{align}
& \tan \alpha =\dfrac{\dfrac{-2\tan \dfrac{\theta }{2}}{1+{{\tan }^{2}}\dfrac{\theta }{2}}}{1-\dfrac{1-{{\tan }^{2}}\dfrac{\theta }{2}}{1+{{\tan }^{2}}\dfrac{\theta }{2}}} \\
& \tan \alpha =\dfrac{-2\tan \dfrac{\theta }{2}}{2{{\tan }^{2}}\dfrac{\theta }{2}} \\
& \tan \alpha =\dfrac{-1}{\tan \dfrac{\theta }{2}} \\
& \tan \alpha =-\cot \dfrac{\theta }{2} \\
\end{align}$
Now we will use the formula for,
$\tan \left( \dfrac{\pi }{2}+x \right)=-\cot x$
Using this formula we get,
$\tan \alpha =\tan \left( \dfrac{\pi +\theta }{2} \right)$
Hence, from this we can say that the correct answer is $\left( \dfrac{\pi +\theta }{2} \right)$ .
Hence option (d) is correct.
Note: It is important to understand the concept of amplitude in complex numbers. And students might get confused between amplitude and argument of complex numbers, these two are different things and give different answers. The trigonometric formulas that we have used should also be kept in mind as it helps to simplify the given equation in the form that is given in the answer.
Complete Step-by-Step solution:
Let’s first write the definition of amplitude of complex number,
The complex number is referred to as the complex amplitude, a polar representation of the amplitude and the initial phase of the complex exponential signal. The complex amplitude is also called a phasor as it can be represented graphically as a vector in the complex plane.
So, if x + iy is a complex number then it’s amplitude $\alpha $ will be:
$\tan \alpha =\dfrac{y}{x}$
Now using the same formula on $1-\cos \theta -i\sin \theta $ we get,
$\tan \alpha =\dfrac{-\sin \theta }{1-\cos \theta }$
Now we are going to use some trigonometric formula,
$\begin{align}
& \sin 2x=\dfrac{2\tan x}{1+{{\tan }^{2}}x} \\
& \cos 2x=\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x} \\
\end{align}$
Now using these formula we get,
$\begin{align}
& \tan \alpha =\dfrac{\dfrac{-2\tan \dfrac{\theta }{2}}{1+{{\tan }^{2}}\dfrac{\theta }{2}}}{1-\dfrac{1-{{\tan }^{2}}\dfrac{\theta }{2}}{1+{{\tan }^{2}}\dfrac{\theta }{2}}} \\
& \tan \alpha =\dfrac{-2\tan \dfrac{\theta }{2}}{2{{\tan }^{2}}\dfrac{\theta }{2}} \\
& \tan \alpha =\dfrac{-1}{\tan \dfrac{\theta }{2}} \\
& \tan \alpha =-\cot \dfrac{\theta }{2} \\
\end{align}$
Now we will use the formula for,
$\tan \left( \dfrac{\pi }{2}+x \right)=-\cot x$
Using this formula we get,
$\tan \alpha =\tan \left( \dfrac{\pi +\theta }{2} \right)$
Hence, from this we can say that the correct answer is $\left( \dfrac{\pi +\theta }{2} \right)$ .
Hence option (d) is correct.
Note: It is important to understand the concept of amplitude in complex numbers. And students might get confused between amplitude and argument of complex numbers, these two are different things and give different answers. The trigonometric formulas that we have used should also be kept in mind as it helps to simplify the given equation in the form that is given in the answer.
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