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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!!!

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Does it matter how tangled $w$ gets? If it doesn't then clear the denominator by multiplying through and you have a linear equation in $x$ and $y$ and can express $y$ in terms of $x$ or $x$ in terms of $y$. If $j$ is a square root of minus 1 and you need to find the real and imaginary parts of $w$ in terms of $x$ and $y$ then ncmathsadist has shown the method. –  Mark Bennet Jun 28 '12 at 21:35

3 Answers 3

up vote 2 down vote accepted

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.

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I don't think $j$ here is the complex unit ... –  Neal Jun 28 '12 at 21:09
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Eek. It is often used as a complex unit by engineers. Gineer, is that correct? Say so and I will delete this post as irrelevant. –  ncmathsadist Jun 28 '12 at 21:12
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The question is ambiguous. This is a helpful attempt to discern what is meant and answer it. +1. (Note the complex analysis tag, which makes this a likely interpretation) –  Mark Bennet Jun 28 '12 at 22:38
    
correct: Here $j=\sqrt{-1}=i$. –  Gineer Jun 29 '12 at 4:35

This is a game. You are allowed to:

  • multiply the left- and right-hand sides of the equation by the same expression,
  • add the same expression to both sides of the equation,
  • add an expression equivalent to zero to either side of the equation, and
  • multiply either side of the equation by an expression equivalent to one.

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 you're like me, you really don't like dealing with fractions. Maybe we should try to somehow get rid of the fractions.
  • Since the $j$ terms are also $y$ terms, perhaps we might be able to get all the $yj$ terms together and factor out a $j$.

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.

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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 $$ \frac{w-wx+x}{(w-1)y} = j.$$

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