Convert the complex number $z = \dfrac{{2 - \sqrt 3 i}}{{5 - \sqrt 3 i}}$ in polar form.

Hint: First, multiply and divide by the conjugate of the denominator to remove the imaginary part from the denominator and simplify it. After that find the modulus of z. Then, find the argument of z. then substitute the value in $z = r\left( {\cos \theta + i\sin \theta } \right)$ to get the polar form.

Complete step by step answer:
Given: - $z = \dfrac{{2 - \sqrt 3 i}}{{5 - \sqrt 3 i}}$
Multiply and divide z with the conjugate of the denominator,
$\Rightarrow z = \dfrac{{2 - \sqrt 3 i}}{{5 - \sqrt 3 i}} \times \dfrac{{5 + \sqrt 3 i}}{{5 + \sqrt 3 i}}$
Multiply the terms on the right to get real numbers in the denominator,
$\Rightarrow z = \dfrac{{10 - 3{i^2} - 5\sqrt 3 i + 2\sqrt 3 i}}{{25 - 3{i^2}}}$
As we know ${i^2} = - 1$, substitute the value in the equation,
$\Rightarrow z = \dfrac{{10 - 3\left( { - 1} \right) - 5\sqrt 3 i + 2\sqrt 3 i}}{{25 - 3\left( { - 1} \right)}}$
Open the brackets and add the like terms,
$\Rightarrow z = \dfrac{{13 - 3\sqrt 3 i}}{{28}}$
Separate the real part and imaginary parts,
$\Rightarrow z = \dfrac{{13}}{{28}} - i\dfrac{{3\sqrt 3 }}{{28}}$
The formula of modulus is,
$\Rightarrow \left| z \right| = \sqrt {{a^2} + {b^2}} $
Here $a = \dfrac{{13}}{{28}}$ and $b = - \dfrac{{3\sqrt 3 }}{{28}}$. Then,
$\Rightarrow \left| z \right| = \sqrt {{{\left( {\dfrac{{13}}{{28}}} \right)}^2} + {{\left( { - \dfrac{{3\sqrt 3 }}{{28}}} \right)}^2}} $
Square the terms in the bracket,
$\Rightarrow \left| z \right| = \sqrt {\dfrac{{169}}{{784}} + \dfrac{{27}}{{784}}} $
Since the denominator is the same. So, add the numerator,
$\Rightarrow \left| z \right| = \sqrt {\dfrac{{196}}{{784}}} $
Cancel out the common factors from the numerator and denominator,
$\Rightarrow \left| z \right| = \sqrt {\dfrac{1}{4}} $
Take square root on the right side,
$\Rightarrow \left| z \right| = \dfrac{1}{2}$
Now, $\tan \alpha = \left| {\dfrac{b}{a}} \right|$. Then,
$\Rightarrow \tan \alpha = \left| {\dfrac{{ - \dfrac{{3\sqrt 3 }}{{28}}}}{{\dfrac{{13}}{{28}}}}} \right|$
Cancel out the common factor,
$\Rightarrow \tan \alpha = \dfrac{{3\sqrt 3 }}{{13}}$
Take ${\tan ^{ - 1}}$ on both sides,
$\Rightarrow \alpha = {\tan ^{ - 1}}\dfrac{{3\sqrt 3 }}{{13}}$
As the real part of the complex number is positive and the imaginary part is negative. The number will lie in 4th quadrant. Then,
$\Rightarrow \arg \left( z \right) = - \alpha $
Substitute the value of $\alpha $,
$\Rightarrow \arg \left( z \right) = - {\tan ^{ - 1}}\left( {\dfrac{{3\sqrt 3 }}{{13}}} \right)$
So, the complex number in the polar form will be,
$\Rightarrow z = r\left( {\cos \theta + i\sin \theta } \right)$
where, $r = \left| z \right|$ and $\theta = \arg \left( z \right)$
Then,
$\Rightarrow z = \dfrac{1}{2}\left[ {\cos \left( { - \tan \left( {\dfrac{{3\sqrt 3 }}{{13}}} \right)} \right) + i\sin \left( { - \tan \left( {\dfrac{{3\sqrt 3 }}{{13}}} \right)} \right)} \right]$
Hence, the polar form of the complex number is $z = \dfrac{1}{2}\left[ {\cos \left( { - \tan \left( {\dfrac{{3\sqrt 3 }}{{13}}} \right)} \right) + i\sin \left( { - \tan \left( {\dfrac{{3\sqrt 3 }}{{13}}} \right)} \right)} \right]$.

Note:
The complex numbers are the field C of numbers of the form $x + iy$, where x and y are real numbers and i is the imaginary unit equal to the square root of -1. When a single letter z is used to denote a complex number. It is denoted as $z = x + iy$.