Find the equation of the image of the line $x+y=1$ by Möbius transformation $$w=\dfrac{z+1}{z-1}$$
My approach, if $x=\Re(z)$ , and $y=\Im(z)$, then $x+y=(\frac{1}{2}-\frac{i}{2})z+(\frac{1}{2}+\frac{i}{2})\overline{z}-1=0$, then if $$w=\dfrac{z+1}{z-1}\implies z=\dfrac{w+1}{w-1}$$ And then we have a equation for a line, this is correct?
On
The line $L: x+y=1$ corresponds to $z = x+(1-x)i$.
A Mobius transform $m$ maps generalizd circles onto generalized circles. Note, a generalized circle in the complex plane is either a circle or a straight line (a circle passing through $\infty$ on the Riemann sphere).
We need three points on $L$ and their images under $m$. $$z_1 = 1,z_2= i, z_3 = \infty$$
$$m(z_1) = \infty,\, m(z_2) = \dfrac{i+1}{i-1}=-i,\,m(z_3) = 1$$
S0, $$m(L): w = x + (x-1)i \mbox{ or } y-x=-1 $$
Note that $x+y=1$ if and only if $z=x+i(1-x)$, so in fact we have that
\begin{align*}\frac{x+i(1-x)+1}{x+i(1-x)-1}&=\frac{[x+1+i(1-x)]\cdot[(x-1)+i(x-1)]}{(x-1)^2+(1-x)^2}\\ &=\frac{x^2-1+i(x^2-1)-i(1-x)^2+(1-x)^2}{2(x-1)^2}\\ &=\frac{x^2-1+(1-x)^2}{2(x-1)^2}+\frac{i(x^2-1)-i(1-x)^2}{2(x-1)^2}\end{align*}
and note that this is on the line $y=x$.
Now suppose that
we have some $a+ai \in \mathbb C$. Then you need to find some $x+i(1-x)$ so that the mobius function applied to it lands you at $a+ai$ to see that the line surjects onto the line $y=x$. I plugged it in symbollically here, but it seems quite complicated. On the other hand, it is continuous so it sends connected components to connected components, and it is not bounded since it is a nontrivial map, so surjectivity follows from the theory.