Question

If $ z $ is a complex number, then the locus of the point $ z $ satisfying $ \arg \left( \dfrac{z-i}{z+i} \right)=\dfrac{\pi }{4} $ is
A. circle with centre $ \left( -1,0 \right) $ and radius $ \sqrt{2} $
B. circle with centre $ \left( 0,0 \right) $ and radius $ \sqrt{2} $
C. circle with centre $ \left( 0,1 \right) $ and radius $ \sqrt{2} $
D. circle with centre $ \left( 1,1 \right) $ and radius $ \sqrt{2} $

Answer 1

Hint: We first assume the complex number $ z=a+ib $ . We find the simplified form of the expression $ \dfrac{z-i}{z+i} $ . We rationalise it and find the argument of the complex form. We take the equation and equate it with the general form of the circle to find the solution.

Complete step-by-step answer:
Let us assume the complex number $ z=a+ib $ where $ a,b\in \mathbb{R} $ . The argument for $ z=a+ib $ can be denoted as $ {{\tan }^{-1}}\left( \dfrac{b}{a} \right) $ .
The function $ \dfrac{z-i}{z+i} $ becomes $ \dfrac{z-i}{z+i}=\dfrac{a+ib-i}{a+ib+i}=\dfrac{a+i\left( b-1 \right)}{a+i\left( b+1 \right)} $ .
We first apply the rationalisation of complex numbers for $ \dfrac{a+i\left( b-1 \right)}{a+i\left( b+1 \right)} $ .
We multiply $ a-i\left( b+1 \right) $ to both the numerator and denominator of the fraction.
So, \[\dfrac{a+i\left( b-1 \right)}{a+i\left( b+1 \right)}\times \dfrac{a-i\left( b+1 \right)}{a-i\left( b+1 \right)}=\dfrac{{{a}^{2}}-{{i}^{2}}\left( b-1 \right)\left( b+1 \right)-ia\left( b+1 \right)+ia\left( b-1 \right)}{{{a}^{2}}-{{i}^{2}}{{\left( b+1 \right)}^{2}}}\].
We used the theorem of \[\left( a+b \right)\times \left( a-b \right)={{a}^{2}}-{{b}^{2}}\].
\[\begin{align}
  & \dfrac{a+i\left( b-1 \right)}{a+i\left( b+1 \right)} \\
 & =\dfrac{\left( {{a}^{2}}+{{b}^{2}}-1 \right)-2ai}{{{a}^{2}}+{{\left( b+1 \right)}^{2}}} \\
 & =\dfrac{\left( {{a}^{2}}+{{b}^{2}}-1 \right)}{{{a}^{2}}+{{\left( b+1 \right)}^{2}}}+i\dfrac{-2a}{{{a}^{2}}+{{\left( b+1 \right)}^{2}}} \\
\end{align}\]
Therefore,
\[\begin{align}
  & \arg \left( \dfrac{z-i}{z+i} \right)=\dfrac{\pi }{4} \\
 & \Rightarrow \arg \left( \dfrac{\left( {{a}^{2}}+{{b}^{2}}-1 \right)}{{{a}^{2}}+{{\left( b+1 \right)}^{2}}}+i\dfrac{-2a}{{{a}^{2}}+{{\left( b+1 \right)}^{2}}} \right)=\dfrac{\pi }{4} \\
 & \Rightarrow {{\tan }^{-1}}\left( \dfrac{-2a}{{{a}^{2}}+{{b}^{2}}-1} \right)=\dfrac{\pi }{4} \\
\end{align}\]
This gives \[\dfrac{-2a}{{{a}^{2}}+{{b}^{2}}-1}=\tan \left( \dfrac{\pi }{4} \right)=1\]. Simplifying we get \[{{a}^{2}}+{{b}^{2}}-1+2a=0\].
This is an equation of the circle of \[{{\left( a+1 \right)}^{2}}+{{b}^{2}}=2\]. The circle is with centre $ \left( -1,0 \right) $ and radius $ \sqrt{2} $ as the general equation of circle is \[{{\left( x-m \right)}^{2}}+{{\left( y-n \right)}^{2}}={{r}^{2}}\] having centre $ \left( m,n \right) $ and radius $ r $ .
The correct option is A.
So, the correct answer is “Option A”.

Note: The integer power value of every even number on $ i $ will give the real number. Odd numbers of power value give back the imaginary part. The relation $ {{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1 $ can also be represented as the unit circle on the complex plane.