Geometrical representation of complex numbers.

Question

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

My attempt:
Multiplying and dividing $f(x)$ by its conjugate. We get,

$\begin{align} f(x) =\frac{x+i}{x²+1}\\ \therefore f(a)=\frac{a+i}{a²+1} \end{align}$

If we assume center of square as origin then angle between $f(a)$ and $f(b)$ in the complex plane will be $90°$. So we can write, $\begin{align} f(a)e^{\frac{iπ}{2}}=f(b)\quad (\because|f(a)|=|f(b)|)\\ \therefore f(b) =if(a) \end{align}$

Now,$|f(a)-f(b)|=|f(a)-if(a)|$ represents distance between two vertices of square i.e side of square.
$\begin{align} \therefore Area=|f(a)-if(a)|^2\\ =2|f(a)|^2\\ =\frac{2}{a²+1} \end{align}$

But the answer given is$\frac{1}{2}$.Please guide me.

Answer 1

First observe that for all real numbers $x$ the distance from $f(x)$ to $i/2$ is $1/2$: $$\begin{align}|f(x)-i/2|^2&=(f(x)-i/2)\overline{(f(x)-i/2)}\\ &=\left(\frac{1}{x-i}-i/2\right)\left(\frac{1}{x+i}+i/2\right)\\ &=\frac14, \end{align}$$

that is the image of $\mathbb R$ under $f$ is (contained in) the circle with center $i/2$ and radius $1/2$. From here it's easy to find the area of the desired square.

Answer 2

@Michael Hoppe already gave you great answer! Here is my try...

I know that all numbers $f(x)$ will lie above $x$-axis, so all vertices of your square (we can give names to them: $A,B,C,D$) will satisfy this case. Without loosing generality, I will assume that the square will have one pair of sides parallel to $x$-axis, for example let sides $AB$ and $CD$ are both parallel to $x$-axis. Now we know that $y$-coordinates of vertices $A$ and $B$ are the same, as well as the $y$-coordinates of vertices $C$ and $D$. So we have: $$\frac{1}{a^2+1} =\frac{1}{b^2+1} \iff a^2+1=b^2+1 \iff a^2=b^2 \iff a=-b.$$

The case $a=b$ is not the solution which we can accept in our problem! Similary, we will have $c=-d$.

Next, it is easy to conclude that this hold: $$\vec{AB}=\vec{OB}-\vec{OA} = \{x_B-x_A, 0\},\\ \vec{BC}=\vec{OC}-\vec{OB} = \{0, y_C-y_B\}.$$

Since we have square $ABCD$, we must have $\left| \vec{AB} \right| = \left| \vec{BC} \right|$, i.e.

$$x_B-x_A = y_C - y_B. \tag{1}$$

$(1)$ we can rewrite:

$$\frac{b}{b^2+1} - \frac{a}{a^2+1} = \frac{1}{c^2+1} - \frac{1}{b^2+1}.$$

Now, we will use that $a=-b$, so we will have:

$$\frac{b}{b^2+1}+\frac{b}{b^2+1}=\frac{1}{c^2+1}-\frac{1}{b^2+1} \iff \frac{2b+1}{b^2+1}=\frac{1}{c^2+1}. \tag{2}$$

It is obvious that the side $BC$ will be parallel to $y$-axis, so we know that $x$-coordinates of vertices $C$ and $B$ are the same, i.e.

$$\frac{c}{c^2+1}=\frac{b}{b^2+1}. \tag{3}$$

From $(3)$ we know this is true: $$b^2+1=\frac{b(c^2+1)}{c}. \tag{4}$$

Put $(4)$ into $(2)$, it will yield to: $$c=\frac{b}{2b+1}.$$

Finally, we can conclude all vertices of our square $ABCD$, i.e. we will have: $$a=-b, \quad c=\frac{b}{2b+1}, \quad d=-c, \quad \text{for some } b \in \mathbb{R}.$$

Figure.

Answer 3

If you know about linear fractional transformations, aka Mobius transformations, then you know that the function $f(z)=1/(z-i)$ maps the extended real line to a circle in the complex plane. Since $f(\infty)=0\in\mathbb{R}$ and $f(z)\not\in\mathbb{R}$ for any other $z$ in the extended real line, the circle is tangent to the real line. Since $f(0)=1/(-i)=i$, we see that the circle is centered at $i/2$ with radius $1/2$. Any four points on this circle forming a square will have area $1/2$.

Note, this is essentially the same answer as Michael Hoppe's, which shows directly that the image of the real line is the circle centered at $i/2$ of radius $1/2$. The virtue of the more advanced standpoint (aside from its own intrinsic beauty) is that knowing about it lets you get to the answer more quickly, with less algebraic fuss.

Remark: Trying to read the Wikipedia entry on Mobius transformations may be like drinking from a firehose, but the key thing here that's easy to remember (and actually not so difficult to prove) is that linear fractional transformations map circles to circles, where lines can be thought of as circles of infinite radius.

Answer 4

After getting a clue from other answers, I found another method to solve the problem.

$$f(t)=\left(\frac{t}{t^2+1}\right)+i\left(\frac{1}{t^2+1}\right)$$

Let $h=\frac{t}{1+t^2}$ and $k=\frac{1}{t^2+1}$. After eliminating $t$, we get,

$$h^2+k^2-k=0$$

It is a circle with center $(0,\frac{1}{2})$ and radius $\frac{1}{2}$ units. Now, in a circle infinite squares can be inscribed but all of them have equal side and hence equal area. In this case $Area=\frac{1}{2}\,sq.unit$