Show that C is a circle and find its radius and centre. 
A transformation $T$ from the $z$ plane to the $w$ plane is given by$$w=\frac1{z+1},z\ne-1$$Show that $T$ maps the imaginary axis in the $z$ plane onto a circle $C$ in the $w$ plane and find its centre and radius.

This is from a June 2019 FP2 IAL edexcel paper. Not sure about my answer. Please help!


 A: Let $z=it$ where $t$ is a real number  we get $$w=\frac{1}{1+it} = \frac{1}{1+t^2}-\frac{it}{1+t^2}$$
With $$x= \frac{1}{1+t^2}$$ and $$y=-\frac{t}{1+t^2}$$ we have $$x^2 + y^2 = x$$ which is a circle with center at $\big(\frac{1}{2}, 0\big)$ and radius of $\frac{1}{2}.$
A: This is not correct. $z=x+iy$ is a variable point, so $x,y$ are variable. $\left(\frac1{1-y^2},0\right)$ does not describe a fixed centre neither does $\frac1{1-y^2}$ describe a fixed radius.

To solve your question, equate the real and imaginary parts of the equality you obtained:$$iuy-vy=(1-u)-iv\implies uy=-v\wedge vy=u-1$$Isolate $y$ from the first equation and substitute for it in the second equation:$$uy=-v\implies uvy=-v^2\implies u(u-1)+v^2=0$$Thus you have shown that the image of the imaginary axis in the $z$ plane is a subset of the circle $C:|w-1/2|=1/2$ in the $w$ plane.

For proving that the image is onto, you will need to prove $\forall w\in C$, there exists a purely imaginary $z$ such that $T(z)=w$. You have already figured out the inverse transformation. For $w\ne0$:$$wz=1-w\iff z=\frac1w-1=\frac{-[u(u-1)+v^2]-iv}{u^2+v^2}$$Can you verify that $\mathfrak{Re}(z)=0$ whenever $|w-1/2|=1/2$?

As a final step, notice that $0\in C$. What purely imaginary $z$ satisfies $T(z)=0$?
