I'm stuck on the first step again: $w=\frac{x+jy}{x+jy-1}$
I need to separate out (x and y) and j
The only thing I can come up with at this stage is: $w=\frac{x}{x+jy-1}+\frac{jy}{x+jy-1}$ but that doesn't help me at all.
Aaaaargh!!!
I'm stuck on the first step again: $w=\frac{x+jy}{x+jy-1}$
I need to separate out (x and y) and j
The only thing I can come up with at this stage is: $w=\frac{x}{x+jy-1}+\frac{jy}{x+jy-1}$ but that doesn't help me at all.
Aaaaargh!!!
Write $ w = {x + jy\over (x-1) + jy}$ Multiply both numerator and denominator by $(x-1)-iy$ to obtain $ w = {(x + jy)(x-1-jy)\over (x-1)^2 + y^2}$ Multiply out the numerator and separate into complex and real terms.
This is a game. You are allowed to:
Using these rules, your mission, should you choose to accept it, is to arrange the equation so that on one side, you have $x$ and $y$ terms, but no $j$ terms, and on the other side, you have $j$ terms, but no $x$ or $y$ terms.
Now we need a strategy. Here are some hints:
If we can get the $y$ and $j$ pried apart, we should be in a situation to put all the $x$ and $y$ terms on one side and keep all the $j$ terms on the other side.
Write $\frac1w = \frac{x+jy-1}{x+jy} = 1 - \frac{1}{x+jy},$ so $\frac{1}{x+jy} = 1 - \frac1w = \frac{w-1}{w},$ then $w=(w-1)(x+jy) = wx + wjy - x - jy,$ and $w-wx+x = wjy - jy = (w-1)jy,$ and $ \frac{w-wx+x}{w-1} = jy,$ and finally $ \frac{w-wx+x}{(w-1)y} = j.$