Perimeter of the locus represented by $\arg \left( {\dfrac{{z + i}}{{z - i}}} \right) = \dfrac{\pi }{4}$ is equal to
A.$\dfrac{{3\pi }}{2}$
B. $\dfrac{{3\pi }}{{\sqrt 2 }}$
C. $\dfrac{\pi }{{\sqrt 2 }}$
D. None of these

Hint: In this question, we are given the equation $\arg \left( {\dfrac{{z + i}}{{z - i}}} \right) = \dfrac{\pi }{4}$and we have to find the perimeter of the locus. First step is to let $z = x + iy$and solve the given equation using complex number formulas. In the end you will get the equation of the circle, compare it with the standard form of the circle to calculate the value of radius and center. Now put the value of radius in the formula of the perimeter of the circle.

Formula Used:
Complex numbers formula –
$\arg \left( {\dfrac{x}{y}} \right) = \arg x - \arg y$
$\arg \left( {x + iy} \right) = {\tan^{ - 1}}\left( {\dfrac{y}{x}} \right)$
Equation of Circle (Standard form) –
${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$, here $\left( {h,k} \right)$ is the center and $r$ is the radius of the circle
Perimeter of the circle – $2\pi r$where $r$ is the radius.

Complete step by step solution:
Let, $z = x + iy$ be the complex number
Given that,
$\arg \left( {\dfrac{{z + i}}{{z - i}}} \right) = \dfrac{\pi }{4}$
Using the formula, $\arg \left( {\dfrac{x}{y}} \right) = \arg x - \arg y$
$\arg \left( {z + i} \right) - \arg \left( {z - i} \right) = \dfrac{\pi }{4}$
Put the value of $z = x + iy$,
$\arg \left( {x + iy + i} \right) - \arg \left( {x + iy - i} \right) = \dfrac{\pi }{4}$
$\arg \left( {x + i\left( {y + 1} \right)} \right) - \arg \left( {x + i\left( {y - 1} \right)} \right) = \dfrac{\pi }{4}$
Now, using the formula $\arg \left( {x + iy} \right) = {\tan^{ - 1}}\left( {\dfrac{y}{x}} \right)$
${\tan^{ - 1}}\left( {\dfrac{{y + 1}}{x}} \right) - {\tan^{ - 1}}\left( {\dfrac{{y - 1}}{x}} \right) = \dfrac{\pi }{4}$
${\tan^{ - 1}}\left( {\dfrac{{\dfrac{{y + 1}}{x} - \dfrac{{y - 1}}{x}}}{{1 + \dfrac{{y + 1}}{x} \times \dfrac{{y - 1}}{x}}}} \right) = \dfrac{\pi }{4}$
$\left( {\dfrac{{2x}}{{{x^2} + {y^2} - 1}}} \right) = \tan \dfrac{\pi }{4}$
$\dfrac{{2x}}{{{x^2} + {y^2} - 1}} = 1$
${x^2} + {y^2} - 2x - 1 = 0$
${\left( {x - 1} \right)^2} + {y^2} = {\left( {\sqrt 2 } \right)^2}$
Compare the above equation with the standard equation of the circle i.e., ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$
It implies that, the center of the circle is at$\left( {1,0} \right)$ and the radius is$\sqrt 2 $
Therefore, perimeter of the circle$ = 2\pi r = 2\sqrt 2 r$

Option ‘D’ is correct

Note: To solve such problems, one should have a good knowledge of complex numbers and arguments. Also, in geometry, a locus is a set of points that satisfy a given condition or situation for a shape or figure. The locus is pluralized as loci. The region is the location of the loci. The term locus comes from the word location.