If there exist real numbers $a,b,c,d$ for which $f(a),f(b),f(c),f(d)$ form a square on the complex plane.Find the area of the square.

Question

For all real numbers $x$,let the mapping $f(x)=\frac{1}{x-i},\text{where} i=\sqrt{-1}$.If there exist real numbers $a,b,c,d$ for which $f(a),f(b),f(c),f(d)$ form a square on the complex plane.Find the area of the square.

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$f(a)=\frac{1}{a-i}=\frac{a+i}{a^2+1}=(\frac{a}{a^2+1},\frac{1}{a^2+1})$
Similarly,$f(b)=\frac{1}{b-i}=\frac{b+i}{b^2+1}=(\frac{b}{b^2+1},\frac{1}{b^2+1})$
Similarly,$f(c)=\frac{1}{c-i}=\frac{c+i}{c^2+1}=(\frac{c}{c^2+1},\frac{1}{c^2+1})$
Similarly,$f(d)=\frac{1}{d-i}=\frac{d+i}{d^2+1}=(\frac{d}{d^2+1},\frac{1}{d^2+1})$

Now the area of the square$=(\frac{a}{a^2+1}-\frac{b}{b^2+1})^2+(\frac{1}{a^2+1}-\frac{1}{b^2+1})^2$
But this expression simplifies to $\frac{(a-b)^2}{(1+a^2)(1+b^2)}$.
How should i prove the area of the square to be $\frac{1}{2}$.Is my approach not correct?Please help me.Thanks.

Answer 1

$$\displaystyle f(x)=\frac { 1 }{ x-i }=\frac { x+i }{ x^{ 2 }+1 } $$ Now in Argand plane $f(a),f(b),f(c),f(d)$ all lie on a curve whose parametric coordinates

is given by $$\displaystyle x=\frac { t }{ t^{ 2 }+1 } ,y=\frac { 1 }{ t^{ 2 }+1 }$$

Now eliminating variable $t\;,$ We get $$x^{ 2 }+y^{ 2 }=y\Rightarrow x^2+y^2-y=0$$

Or we can write it as $$(x-0)^2+\left(y-\frac{1}{2}\right)^2=\frac{1}{4}$$

So, they lie on a circle of radius $\displaystyle \frac { 1 }{ 2 }.$ So the area of the square is $\displaystyle \frac { 1 }{ 2 } $

Answer 2

Hint: The function $f$ bijectively maps the extended real line into the circle through the origin having centre at $\frac{i}{2}$. The area of a square inscribed in a circle with radius $r$ is $2r^2$.

Answer 3

Let $\Phi(z)=\dfrac{1}{\bar{z}}$, this is the inversion with respect to the unit circle. Now, for a real $x$, we have $f(x)=\Phi(x+i)$.

The image of the line $d=\{z:\Im z=1\}$ under the inversion $\Phi$ is the circle of diameter $[0,i]$, So if $f(a)$, $f(b)$, $f(c)$ and $f(d)$ form a square, it must be inscribed in that circle of radius $\frac12$, and consequently it must have an area equal to $\frac12$.