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Question: If \( z \) is a complex number, then the locus of the point \( z \) satisfying \( \arg \left( \dfrac...

If zz is a complex number, then the locus of the point zz satisfying arg⁡(z−iz+i)=π4\arg \left( \dfrac{z-i}{z+i} \right)=\dfrac{\pi }{4} is
A. circle with centre (−1,0)\left( -1,0 \right) and radius 2\sqrt{2}
B. circle with centre (0,0)\left( 0,0 \right) and radius 2\sqrt{2}
C. circle with centre (0,1)\left( 0,1 \right) and radius 2\sqrt{2}
D. circle with centre (1,1)\left( 1,1 \right) and radius 2\sqrt{2}

Explanation

Solution

Hint : We first assume the complex number z=a+ibz=a+ib . We find the simplified form of the expression z−iz+i\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+ibz=a+ib where a,b∈Ra,b\in \mathbb{R} . The argument for z=a+ibz=a+ib can be denoted as tan⁡−1(ba){{\tan }^{-1}}\left( \dfrac{b}{a} \right) .
The function z−iz+i\dfrac{z-i}{z+i} becomes z−iz+i=a+ib−ia+ib+i=a+i(b−1)a+i(b+1)\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 a+i(b−1)a+i(b+1)\dfrac{a+i\left( b-1 \right)}{a+i\left( b+1 \right)} .
We multiply a−i(b+1)a-i\left( b+1 \right) to both the numerator and denominator of the fraction.
So, a+i(b−1)a+i(b+1)×a−i(b+1)a−i(b+1)=a2−i2(b−1)(b+1)−ia(b+1)+ia(b−1)a2−i2(b+1)2\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 (a+b)×(a−b)=a2−b2\left( a+b \right)\times \left( a-b \right)={{a}^{2}}-{{b}^{2}}.

& \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{aligned}$$ Therefore, $$\begin{aligned} & \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{aligned}$$ This gives $$\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.