Golden ratio in complex number squares 
In the Argand diagram shown below the complex numbers
$– 1 + i, 1 + i, 1 – i, – 1 – i$ represent the vertices of a square ABCD.
The equation of its diagonal BD is $y = x$. The complex number $k + ki$
where $– 1 < k < 0$ represents the point E which is in the fourth quadrant and
lies on the line $y = x.$ 
EFGC is a square such that F lies on AB. The line GE meets the line CD
produced at H such that H is represented by the complex number $– 2 – i$.
(i) Show that $F=\frac{-k}{k+2}+i$
(ii) Hence show that $\frac{AF}{FB}=\varphi$
For (i), I haven't been able to arrive at the result. I've tried the following:
$M_{HG}=\frac{k+1}{k+2}$, therefore $M_{CF}=-\frac{k+2}{k+1}$ 
From the diagram, we know that $\Im{F}=1$. Let $f$ denote the real part. Therefore line $CF$ is given by $(y-1)=-\frac{k+2}{k+1}(x-f)$.
Similarly, line $EF$ is given by $(y-1)=-\frac{k-f}{1-k}(x-f)$
$\therefore -\frac{k-f}{1-k}=-\frac{k+2}{k+1}$, which gives $f=\frac{2k^2+2k-2}{k+1}$, which is clearly wrong. Could someone show me the correct way or at least give a hint as to the right lines to consider and if the path I'm going down is somewhat correct?
For (ii) I did get it through a rather convoluted way so if anyone has a quicker way that would be great to know! My way was:
$EF=F-E$ 
$=\frac{-k}{k+2}+i-k-ki$
$=\frac{-k}{k+2}-k+i(1-k)$
$EC=C-E$
$=1-i-k-ki$
$=1-k-i(1+k)$
Since $EFGC$ is a square, $EF=iEC$.
$\therefore \frac{-k}{k+2}-k+i(1-k)=i(1-k-i(1+k))$
$\frac{-k}{k+2}-k+i(1-k)=(1+k)+i(1-k)$
Equating the Real part gives $\frac{-k}{k+2}=(1+k)$, that is $k^2+3k+1=0$
$\therefore k=\frac{-3±\sqrt{5}}{2}$
Since $k>-1$, reject the negative solution
$AF=F-A$
$=\frac{-k}{k+2}+i-(-1+i)$
$=\frac{-k}{k+2}+1$
$FB=B-F$
$=1+i-\left(\frac{-k}{k+2}+i\right)$
$=1+\frac{k}{k+2}$
$\therefore \frac{AF}{FB}=\frac{\frac{-k}{k+2}+1}{1+\frac{k}{k+2}}$
$=\frac{1}{1+k}$
$=\frac{1}{1+\frac{-3+\sqrt{5}}{2}}$
$=\frac{1+\sqrt{5}}{2}$
$=\varphi$

(i)
The slope of the line $EH$ is $\dfrac{k+1}{k+2}$. So, the equation of the line $CF$ is given by $y+1=-\dfrac{k+2}{k+1}(x-1)$. Since $F(f+i)$ is on this line, one has
$$1+1=-\dfrac{k+2}{k+1}(f-1)\implies f=\frac{-k}{k+2}$$
(ii)
Using the fact that $|EF|=|EC|$, one has
$$\begin{align}|EF|^2=|EC|^2&\implies (f-k)^2+(1-k)^2=(1-k)^2+(-1-k)^2 
\\\\&\implies (f-k)^2=(k+1)^2
\\\\&\implies f-k=\pm (k+1)
\\\\&\implies f=2k+1,-1\end{align}$$
Since $f=\dfrac{-k}{k+2}=-1+\dfrac{2}{k+2}\gt -1$, one has
$$(f=)\ \ 2k+1=\frac{-k}{k+2}\implies k^2+3k+1=0\implies k=\frac{-3+\sqrt 5}{2}$$
from which you can show that $\dfrac{AF}{FB}=\varphi$.
