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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?

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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

enter image description here

Quanto
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  • +1: nice. I credited you, in my just edited answer, re mapping to a portion of the unit circle, and mapping to $x + y = -1$. – user2661923 Apr 01 '21 at 04:45
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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 $z$.

user2661923
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  • I'm now trying to graph the image of the triangle and its interior. So I know that on the real axis, as $x$ goes from 0 to 1, the value of $g$ becomes unbounded. And that's quite easy to plot. I also have the points $g(0) = i$ and $g(i) = -1$. But the apart from that, I don't see an easy way to graph the rest. Is there some graphing software that would make this easier to see? –  Apr 01 '21 at 01:59
  • @user1321233212 Yes, there certainly is, but I'm not familiar with such software, apart from Latex + Asymptote. However, I question what the point of the assignment is. Isn't your teacher going to be looking for an analytical solution, rather than the output from canned software? One approach for the interior is to consider the ray going from $(0,0)$ to somewhere on the hypotenuse, along the line $y = kx ~:~ k > 0.$ You can assume that $k$ is some fixed positive value, and use equation (1) above, plugging in the value $y = kx$, to express $f(z)$ in terms of $x \in [0,1)$ and fixed $k$. – user2661923 Apr 01 '21 at 02:05
  • @user1321233212 By the way, my specification that $x \in [0.1)$ in the previous comment, was something of a misnomer. Not only would you have the constraint $y = kx$, but since the equation for the hypotenuse is $(x + y) = 1$, that translates to the constraint $(x + kx) = 1 \implies 0 \leq x \leq \frac{1}{k+1}.$ – user2661923 Apr 01 '21 at 02:09