Question

Amplitude of is $\dfrac{{1 + \sqrt 3 i}}{{\sqrt 3 + i}}$ is

Answer 1

Hint:We simply have to multiply and divide by the conjugate of the denominator. We have to also know the complex number $i\,where\,{i^2} = - 1$



Complete step by step solution:
To find the amplitude of the function at first we have to convert this into the simplest form. then we can solve the problem. to convert this into the simplest form we have to rationalize this fraction. then we get the proper form. so the solve is written below
$z = \dfrac{{1 + \sqrt 3 i}}{{\sqrt 3 + i}}$
$ \Rightarrow \left( {\dfrac{{1 + \sqrt 3 i}}{{\sqrt 3 + i}}} \right) \times \left( {\dfrac{{\sqrt 3 - i}}{{\sqrt 3 - i}}} \right)$
$ \Rightarrow \dfrac{{\sqrt 3 + 3i - i - \sqrt {3{i^2}} }}{{{{(\sqrt 3 )}^2} - {i^2}}}$
$ \Rightarrow \dfrac{{2\sqrt 3 + 2i}}{4}$
\[ \Rightarrow \dfrac{{\sqrt 3 }}{2} + \dfrac{1}{2}i\]
Comparing with $x + iy$ form we get
$x = \dfrac{{\sqrt 3 }}{2}\,\& \,y = \dfrac{1}{2}$

Now, amplitude of
\[\begin{gathered}
  z\, = \,{\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right) \\
  \,\,\, = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{1}{2}}}{{\dfrac{{\sqrt 3 }}{2}}}} \right) \\
  \,\,\, = {\tan ^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }}} \right) \\
  \,\,\, = \dfrac{\pi }{6} \\
\end{gathered} \]


Note: In the above question, conjugate of denominator is the main step to find the answer.amplitude of a function denotes an angular quantity,basically it denotes the trigonometric function tan..by inverse of tan we get the amplitude of the complex numbers.