Image of a region under a Mobius transformation I have a function $$f(z) = \frac{z+1}{i(z-1)}.$$ And I have a triangle enclosed by the vertices $0$, $1$ and $i$. I am trying to determine the image of the triangle under $f$ on the standard complex plane.
I also know that $f$ is a mobius transformation. Is there a trick to determining the image?
 A: With $z=x+iy$, the triangle boundaries are
$$x=0,\>\>\>\>\>y=0,\>\>\>\>\>x+y =1$$
Given $f(z)= \frac{z+1}{i(z-1)}$, it can be verified that the boundary $x=0$ maps to the unit circle centered at origin; $y=0 $ maps to the imaginary axis; and $x+y =1$ maps to the line $x+y=-1$. Thus, the triangle area maps to an open region in the second quadrant, as shown in the graph

A: 
I have a function $$f(z) = \frac{z+1}{i(z-1)}.$$
And I have a triangle enclosed by the vertices $0$, $1$ and $i$.
I am trying to draw the image of the triangle under $f$ on the
standard complex plane.

I am assuming that you intend the vertices $(0,0), (1,0), (0,1)$, so that
you have a right triangle.  As you have indicated $(1,0)$, which represents
$[1 + i(0)]$ is not in the domain of $f$.
Let $z$ be denoted $(x + iy)$.
Then,
$$f(z) = \frac{(x + 1) + i(y)}{(-y) + i(x-1)} \times
\frac{(-y) - i(x-1)}{(-y) - i(x-1)}
~=~
\frac{(-2y)+ i(1 - x^2 - y^2)}{y^2 + (x-1)^2}. \tag1$$
You can now use equation (1) above to discuss the mapping of $f(z)$ on the triangle.
Along the real axis, with $y=0$ and $x$ going from $0$ to $1$
$$f(x) = \frac{i(1 - x^2)}{(x - 1)^2}.$$
This will map along the positive $y$ axis, going from $(0,1)$ and growing unbounded.

Edit
I couldn't resist the sanity check.
$$\frac{d}{dx}\left[\frac{1 - x^2}{(x - 1)^2}\right] = 
\frac{(x-1)^2(-2x) - (1 - x^2)2(x-1)}{(x-1)^4} = 
\frac{(x-1)^2(2)}{(x-1)^4}.$$
Therefore, the value of
$\displaystyle \frac{(1 - x^2)}{(x - 1)^2}$ is strictly increasing.

Along the imaginary axis, with $x=0$ and $y$ going from $0$ to $1$
$$f(x) = \frac{(-2y) + i(1 - y^2)}{y^2 + 1}.$$

Edit
Imagine my surprise to read Quanto's answer that $x=0$ maps to a portion of the unit circle.  Verifying:
$\displaystyle (-2y)^2 + (1 - y^2)^2 = (1 + y^2)^2.$

At $(0,0)$, where $(y=0)$, as mentioned before, this evaluates to $(i)$.
At $(0,1)$, where $(y=1)$, this evaluates to $\displaystyle \frac{(-2) + i(0)}{1 + 1} = -1.$
As you travel along the line segment going from $(0,1)$ to $(1,0)$ you can
use equation (1) above to map the function, via the constraint that $x = (1-y)$.
Plugging that value of $x$ into equation (1) above, you have that
$$f([1-y],y) = \frac{(-2y) + i(2y - 2y^2)}{2y^2}.$$

Edit
After reading Quanto's answer, I realized that I had a typo in the above fraction, that I have corrected, and that the fraction simplifies to 
$\displaystyle x + y = -1.$

It is unclear what discussion you are looking for, with respect to values of $z$ inside the
triangle.  Equation (1) may be used to map any particular value of $z$.
