Question

The amplitude of $1 - \cos \theta - i\sin \theta$ is
A.$\pi + \dfrac{\theta }{2}$
B.$\dfrac{{\left( {\pi - \theta } \right)}}{2}$
C.$\dfrac{{\left( {\theta - \pi } \right)}}{2}$
D.$\dfrac{\theta }{2}$

Answer 1

Hint : Complex number is a number generally represented as$z = a + ib$, where $a$ and $b$is real number represented on real axis whereas $i$is an imaginary unit represented on imaginary axis whose value is$i = \sqrt { - 1}$. Modulus of a complex number is length of line segment on real and imaginary axis generally denoted by $\left| z \right|$ whereas angle subtended by line segment on real axis is argument of matrix denoted by argument (z) calculated by trigonometric value. Argument of complex numbers is denoted by$\arg (z) = \theta = {\tan ^{ - 1}}\dfrac{b}{a}$.

Complete step-by-step answer :
In this question, we need to determine the amplitude of $1 - \cos \theta - i\sin \theta$ for which we will use the properties of the complex numbers as discussed above.
Let $z = 1 - \cos \theta - i\sin \theta - - (i)$
We know the double angle identities theorem of $\cos 2x = 1 - 2{\sin ^2}x$, for angle $x$this identity can be written as $1 - \cos x = 2{\sin ^2}\dfrac{x}{2} - - (ii)$,
Also we know the double angle identities theorem of$\sin 2x = 2\sin x\cos x$, for angle $x$this identity can be written as $\sin x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2} - - (iii)$
Now substitute the obtained identities (ii) and (iii) in equation (i), so we get
$z = 2{\sin ^2}\dfrac{\theta }{2} - i2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$
Now we take $2\sin \dfrac{\theta }{2}$as common hence we get
$z = 2\sin \dfrac{\theta }{2}\left( {\sin \dfrac{\theta }{2} - i\cos \dfrac{\theta }{2}} \right)$
Now again we take $i$ as common, hence we get
$z = 2i\sin \dfrac{\theta }{2}\left( {\dfrac{1}{i}\sin \dfrac{\theta }{2} - \cos \dfrac{\theta }{2}} \right) - - (iv)$
As we know the value of$i = \sqrt { - 1}$, so this can be written as

$${i^2} = {\left( {\sqrt { - 1} } \right)^2} \\\ {i^2} = - 1 \\\ 1 = - {i^2} \\\$$

So by using this in equation (iv), we can write

$$z = 2i\sin \dfrac{\theta }{2}\left( {\dfrac{{ - {i^2}}}{i}\sin \dfrac{\theta }{2} - \cos \dfrac{\theta }{2}} \right) \\\ = 2i\sin \dfrac{\theta }{2}\left( { - i\sin \dfrac{\theta }{2} - \cos \dfrac{\theta }{2}} \right) \\\$$

Now we take $-$ as common term, so we get
$z = - 2i\sin \dfrac{\theta }{2}\left( {\cos \dfrac{\theta }{2} + i\sin \dfrac{\theta }{2}} \right) - - (v)$
Now we use the Euler’s equation$\cos x + i\sin x = {e^{ix}}$, hence by using this we can write equation (v) as
$z = - 2i\sin \dfrac{\theta }{2}{e^{i\dfrac{\theta }{2}}}$
Now we find the argument of the obtained equation
$\arg \left( z \right) = \arg \left( { - 2i\sin \dfrac{\theta }{2}{e^{i\dfrac{\theta }{2}}}} \right) - - (vi)$
As we know$\arg \left( {{z_1} \times {z_2}} \right) = \arg \left( {{z_1}} \right) + \arg \left( {{z_2}} \right)$, so we can write the equation (vi) as
$\arg \left( z \right) = \arg \left( { - 2i} \right) + \arg \left( {\sin \dfrac{\theta }{2}} \right) + \arg \left( {{e^{i\dfrac{\theta }{2}}}} \right)$
This is equal to

$$\arg \left( z \right) = - \dfrac{\pi }{2} + 0 + \dfrac{\theta }{2} \\\ = \dfrac{{\theta - \pi }}{2} \\\$$

So, the correct answer is “Option C”.

Note : Complex numbers are always written in the form of $z = a + ib$where $a$ and $b$are real numbers whereas $i$being imaginary part.
We can convert a degree into radian by multiplying it by $\dfrac{\pi }{{180}}$.