Find the image of a set under a complex function I have the function $$f(z) = \frac{z-1}{z+1}$$ and I have to find the image of the set
$$D= \{ z\in \mathbb C | \operatorname{Re} z<0 \text{ and } \operatorname{Im} z<0\}$$
And it's given that $f(-1)=\infty$ and $f(\infty) = 1$.
I started by writing $z = x + yi$ and I plugged it into the function, the result is:$$ f(z) = \frac{x^2+y^2-1}{(x+1)^2+y^2} + \frac{2yi}{(x+1)^2+y^2} $$
and i realized that all the points will have the image in the 4th quadrant except those that are in the unit circle and inside D and those would have the image between -1 and 1 with y < 0. So the image will be all the points with $x\in(-1, \infty)$ and $y < 0$.
Am I right?
 A: According to Wiki any Mobius transformation $f(z) = \frac{az+b}{cz+d}$ could be represented as a composition of four simple transformations (here we have $a=1, b=-1, c=1, d=1$)


*

*$f_1(z) = z + \frac{d}{c}$, i.e. translation by $\frac{d}{c}$ of the plane.

*$f_2(z) = \frac{1}{z}$, i.e. inversion of the plane

*$f_3(z) = \frac{bc-ad}{c^2}z$, i.e. rotation, then scaling

*$f_4(z) = z + \frac{a}{c}$, another translation


in our case


*

*$f_1(z) = z + 1$

*$f_2(z) = \frac{1}{z}$

*$f_3(z) = -2z$

*$f_4(z) = z + 1$


The interesting part is to understand, how $\frac{1}{z}$ deforms the "translated quadrant". Basically, the inversion is first invert with respect to the unit circle, than reflect with respect to Real axis. Inside of the unit circle $D$ becomes the plane except $D$.


See how the inside of the unit circle got inverted with respect to the unit circle!
The line $1 + i\cdot t$ (the line containing the right boundary) is transformed into the circle that passes through $f_2(1) = 1$, $f_2(\infty) = 0$ and $f_2(1-i) = \frac{1}{2} + \frac{1}{2}i$ -i.e. the circle centered at $(0, \frac{1}{2})$ with radius $\frac{1}{2}$. Check out the anything with $Re(z) > 1$ is outside the unit circle, hence is sent to the inside, i.e. the circle above is not the image of $D$ under $f_2 \circ f_1$.


And the real line is preserved.
Overall, after the final reflection with respect to the Real axis, we have the next image

$f_3$ - rotates the plane by $\pi$, this is the multiplication by $-1$, 

then scales everything by $2$, i.e. the image is now $Im(z) < 0$, except the circle centered at $(-1,0)$ with radius $1$.

And $f_4$ then translates everything by $1$.

All in all, $f(D) = \{z\; |\; Im(z) <0 \} \setminus D$
A: The mapping $f$ takes the negative imaginary axis to the unit semi-circle below the real axis and takes the negative real axis to the two segments on the real line: $x>1$ and $x<1$.  We now have the boundary for the mapping.  Take a point in the interior of $\cal{D}$, for example $p=-1-i$.  $f(p)= 1-2i$ which is below the boundary.  Thus, $f$ maps $\cal{D}$ below the boundary.
Since the boundary of $\cal{D}$ is not included in the domain, the image of the boundary of $\cal{D}$ is not included in the range of $f(\cal{D})$. 
A: Given that $Re(z)<0$ and $Im(z)<0$, we can deduce follows successively,
$$Re(z+1)<1,\>\>\>\>\>Im(z+1)<0$$
$$Re(\frac1{z+1})<1 ,\>\>\>\>\> Im(\frac1{z+1})>0$$
$$Re(\frac2{z+1})<2 ,\>\>\>\>\> Im(\frac2{z+1})>0$$
$$Re(-\frac2{z+1})>-2 ,\>\>\>\>\> Im(-\frac2{z+1})<0$$
$$Re(1-\frac2{z+1})>-1 ,\>\>\>\>\> Im(1-\frac2{z+1})<0$$
Recognizing $f(z) = \frac{z-1}{z+1}=1- \frac2{z+1}$, we have the image $Re f(z) >-1$ and $Im f(z) <0$.
