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 \[\dfrac{\pi }{{180}}\].