As w moves between $-i$ and $i$ on $w = \frac {1+zi}{z+i}$, prove z moves across positive imaginary axis

Question

I'm having trouble with this: As w moves between $-i$ and $i$ on $w = \frac {1+zi}{z+i}$, prove z moves across positive imaginary axis.

I've got the two cartesian equations:

$u = \frac {2x}{x^2 + (y+1)^2}$ and $v = \frac {x^2 + y^2 -1}{x^2 + (y+1)^2}$

and I cannot seem to get to a y>0 if I work from w to z, nor can I seem to get $-1<v<1$ if I move from z towards w. I'm not after the solution but any tips on how to proceed would be greatly appreciated.

Closest I've come is by starting with $v = \frac {y^2 -1}{(y+1)^2}$ (as $x = 0$). Rearranging and solving for y, I get $v < \pm 1$, but I can't seem to get $-1<v<+1$

Thanks.

Answer 1

I believe the method I'm using below is easier than what you're trying to do. First, you can determine $z$ in terms of $w$ as follows:

$$\begin{equation}\begin{aligned} w & = \frac{1 + zi}{z + i} \\ w(z + i) & = 1 + zi \\ wz + wi & = 1 + zi \\ wz - zi & = 1 - wi \\ z(w - i) & = 1 - wi \\ z & = \frac{1 - wi}{w - i} \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

Since you asked for hints, I trust this will do. Nonetheless, just in case you would like to see the rest of the solution, I've placed it below in spoilers.

It asks about $w$ moving between $-i$ and $i$. Thus, you have $w = ki$ for real $k$ where $-1 \lt k \lt 1$.

Substituting this into \eqref{eq1A} gives

$$z = \frac{1 - (ki)i}{ki - i} = \frac{1 + k}{(k - 1)i} = \left(\frac{1 + k}{1 - k}\right)i \tag{2}\label{eq2A}$$

The remainder of the solution is:

Let the coefficient of $i$ be $j = \frac{1 + k}{1 - k}$. As $k$ goes from $-1$ to $1$, note $j$ is always positive, but with it going from just above $0$ when $k$ is close to $-1$ to approach positive infinity as $k$ gets close to $1$. Thus, as asked, this shows $z$ is starting from the origin and moving across the positive imaginary axis.

Update: Using Cartesian coordinates instead, your correctly get to

$$v = \frac{y^2 - 1}{(y+1)^2} \tag{3}\label{eq3A}$$

As you stated, you next need to solve for $y$ in terms of $v$. Doing this gives

$$\begin{equation}\begin{aligned} & v = \frac{y^2 - 1}{y^2 + 2y + 1} \\ & v(y^2 + 2y + 1) = y^2 - 1 \\ & vy^2 + 2vy + v = y^2 - 1 \\ & (v - 1)y^2 + 2vy + (v + 1) = 0 \end{aligned}\end{equation}\tag{4}\label{eq4A}$$

This is a quadratic equation, so using the quadratic formula gives

$$\begin{equation}\begin{aligned} y & = \frac{-2v \pm \sqrt{(2v)^2 - 4(v - 1)(v + 1)}}{2(v - 1)} \\ & = \frac{-2v \pm \sqrt{4v^2 - (4v^2 - 4)}}{2(v - 1)} \\ & = \frac{-v \pm 1}{v - 1} \end{aligned}\end{equation}\tag{5}\label{eq5A}$$

One solution of $y = \frac{-v + 1}{v - 1} = -1$ gives in \eqref{eq3A} that $v = \frac{0}{0}$, so it's not valid. Instead, you have

$$y = \frac{-v - 1}{v - 1} = \frac{1 + v}{1 - v} \tag{6}\label{eq6A}$$

This is the same as \eqref{eq2A}, in the spoiler, with $k$ replaced by $v$.

Answer 2

The locus of $z$ is an upward hyperbola existing for $y\ge 1$

We can write $$|w+i|+|w-i|=2~~~(1)$$ Introducing $$w=\frac{1+iz}{z+i}$$ in (1), we get $$\left |\frac{2+2iz}{z+i}\right|+ \left| \frac{2}{z+i}\right|=2 \implies |z+i|-|z-i|=1$$ Which is the upward parabola existing for $y \ge 1$.