Problem on bilinear transformation Under the transformation $\displaystyle w=\frac{z-1}{z+1}$, show that the map of the straight line x=y is a circle and find its centre and radius.
My attempt: Putting z=x+ix, $\displaystyle w=\frac{(x-1)+ix}{(x+1)+ix}$
$\displaystyle w=\frac{[(x-1)+ix][(x+1)-ix]}{[(x+1)+ix][(x+1)-ix]}$
$\displaystyle w=\frac{(2x^2-1)+ix}{(x+1)^2+x^2}$
This is another complex number. I know my approach is wrong here. Please help.
 A: You should use the exponential notation as it is very simple here: $M: \{x=y\} \Leftrightarrow M: \{\rho \exp(i\frac{\pi}{4}), \rho \in \mathbb{R}\}$.
Then you know that the equation of a circle in polar coordinates is $M'=\{r\exp(it) + c,t \in [0,2\pi]\}$
So you need to rewrite $w$ in that form.
EDIT: (Elements of solutions)
You're looking for the image of the set $M=\{\rho (1+i),\rho \in \mathbb{R} \} $
$f(z) = \frac{z-1}{z+1} = 1 - \frac{2}{z+1}$
With the above parametrization: $f(\rho(1+i))= 1 - \frac{2}{\rho(1+i)+1} = 1 - 2 \frac{\rho(1-i)+1}{(\rho +1) ^2+\rho^2} = \frac{2\rho ^2 -1}{2\rho^2+2\rho+1} + i \frac{2 \rho}{2\rho^2+2\rho+1} = \frac{2\rho ^2 -1}{2\rho^2+2\rho+1} + i \frac{2\rho^2+4\rho+1}{2\rho^2+2\rho+1} -i$
Now you recognize the fact that as $(\frac{2\rho^2 -1}{(\rho +1) ^2+\rho^2})^2 + (\frac{2\rho^2+4\rho+1}{(\rho +1) ^2+\rho^2})^2 = 2$, the above equation can be rewritten $\sqrt 2 e^{i\theta}-i$, which is the equation of a circle of center $-i$ and radius $\sqrt2$.
Drawing a figure should tell you that you need to separate $-i$. You also need to justify that both the real and imaginary part map to $[-1,1]$ so that the full circle is actually the image.
A: There are several ways to answer your question. There is MattB.'s approach with complex numbers in polar form, and below another one with complex numbers in cartesian form. However, the least cumbersome seems to be a geometric approach, but you have to know some facts about plane transformations, and especially inversion.
Geometric method
First, it's clear that the mapping of the line $y=x$ is a circle by the transformation
$$z\rightarrow\frac{z-1}{z+1}=2+\left(-1-\frac{2}{\overline{\bar z+1}}\right)$$
Indeed, it's the composition of a reflection (inner conjugate) an inversion (inside parentheses, center $-1$ and coefficient $-2$) and a translation (the first term, $2$). And the inverse of the line is here a circle, since the line $y=x$ does not pass through the center of inversion $z=-1$.
The only nontrivial component is the inversion. Let's show it looks like with complex numbers, in the case the center of inversion is $O$ (corresponding to complex number $0$). The definition states that $\overline{OM}\cdot \overline{OM'}=k$, with $M:z=re^{i\theta}$, $M':z'=r'e^{i\theta'}$.
Thus $rr'=k$, or $r'=\frac{k}{r}$ and since $O$, $M$ and $M'$ are colinear, you must have $\theta'=\theta$ or $\theta'=\theta+\pi$, according to the sign of $k$.
Then, you can write $\displaystyle{z'=\frac kr e^{i\theta}=\frac k{\bar z}}$. (notice the complex conjugate in denominator)
Inversion at center $A$ (corresponding to complex $a$) is similar, you just have to do a translation before and after to place the center at the origin, and use the formula above for the inversion:
$$z'-a=\frac k{\overline{z-a}}$$
Here, $a=-1$ is real, so
$$z'=-1+\frac k{\overline{z}+1}$$
And $k=-2$. To remove the complex conjugate, you have to do a reflection $z'=\bar z$, before the inversion.
All in all, with


*

*a translation $f(z)=z+2$

*an inversion $g(z)=-1-\frac{2}{\bar z+1}$

*a reflection $h(z)=\bar z$


You have
$$f\circ g\circ h(z)=2+-1-\frac{2}{\overline{\bar z}+1}=1-\frac{2}{z+1}=\frac{z-1}{z+1}$$
Now, the image of line $z=t+it$ by the reflection $h$ is the line $(\Delta)$ of parametric equation $z=t-it$ (or equation $y=-x$), and you have to find the image of this line by the inversion $g$. It's a circle (center $c$, radius $r$), and the translation will change only the center.
To get the circle that the image of $(\Delta)$ the inversion, it's enough to have two opposite points (i.e. a diameter). We have a trivial point, the center of inversion, $A$, because it's the image of the point of the line "at infinity", and one may prove that $AB$ is a diameter, where $B$ is the image of the projection of $A$ on the line. Here, $A$ corresponds to complex number $-1$, and the projection is $H$, which corresponds to complex number $-\frac12+\frac12i$, and $AH=\frac{\sqrt2}{2}$. The image of $H$ is $B$, such that $\overline{AH}\cdot\overline{AB}=-2$, so $AB=2\sqrt{2}$ and $B:-3-2i$ (opposite to $H$ w.r.t $A$, since $k=-2<0$). The center of the circle is then the midpoint of $-1$ and $-3-2i$, that is $-2-i$, and the radius is $r=\frac12(AB)=\sqrt{2}$.
Then the image of the circle by translation $f$ is simply the circle with center at $-i$ and radius $\sqrt{2}$. This is the answer to your question.

Calculatory method
You approach is right (modulo a little mistake that I did too at first), and it gives you a parametric equation of what should be a circle: real and imaginary parts are the $x$- and $y$-coordinates expressions of your parametric curve. Up to this point, it's ok.
Now, if $w$ describes a circle, then if you find the largest and the lowest $y$-coordinates, you have two opposite points of your circle.
The $y$-coordinate is $\displaystyle{\frac{2x}{2x^2+2x+1}}$, and its derivative is $\displaystyle{2\frac{1-2x^2}{(2x^2+2x+1)^2}}$.
Hence, min/ax are obtained for $\displaystyle{x=\pm\frac{\sqrt{2}}2}$.
And the values of $y$ are respectively $\displaystyle{\frac{\sqrt{2}}{\sqrt{2}\pm2}}$. You can check that the $x$-coordinate is zero in both cases. Hence your center should be at
$$x=0$$
$$y=\frac12\left(\frac{\sqrt{2}}{\sqrt{2}+2}+\frac{\sqrt{2}}{\sqrt{2}-2}\right)=-1$$
Then, your circle must have equation $x^2+(y+1)^2=r^2$
Now, replacing $x$ and $y$ with $\displaystyle{x=\frac{2t^2-1}{2t^2+2t+1}}$ and $\displaystyle{y=\frac{2t}{2t^2+2t+1}}$, you can check that $r=\sqrt{2}$.
