Complex Algebra

**Haven** · October 18th 2010, 09:40 PM

I'm a little rusty in complex algebra I was wondering how i would go about solving this equation for $w$

$\frac{w+i}{w-i} = \frac{-1+i}{1-i}\left(\frac{z+1}{z-1}\right)$

The correct answer being $w = \frac{i-z}{i+z}$

**Gusbob** · October 18th 2010, 09:56 PM

Multiply both sides by (w-i). Then move both terms to one side. Expand the term with (w-i) and factorise w.

**Haven** · October 18th 2010, 10:23 PM

I'm not sure exactly what you're getting at, could you write some of the steps out?

**Unknown008** · October 18th 2010, 10:27 PM

The long way round?

$\dfrac{w+i}{w-i} = \dfrac{-1+i}{1-i}\left(\dfrac{z+1}{z-1}\right)$

$\dfrac{w+i}{w-i} = \dfrac{-z-1+iz+i}{z-1-iz+i}$

$(w+i)(z-1-iz+i) = (-z-1+iz+i)(w-i)$

$w(z-1-iz+i) + i(z-1-iz+i) = w(-z-1+iz+i) - i(-z-1+iz+i)$

$w(z-1-iz+i) - w(-z-1+iz+i) = - i(-z-1+iz+i) - i(z-1-iz+i)$

$w(z-1-iz+i +z+1-iz-i) = - i(-z-1+iz+i + z-1-iz+i)$

$w(2z-2iz) = - i(-2+2i)$

$2w(z-iz) = - 2i(-1+i)$

$w(z-iz) = i +1$

$w = \dfrac{i + 1}{z-iz} = \dfrac{i}{z}$

Are you sure this is the correct question?

**Gusbob** · October 18th 2010, 10:39 PM

Thanks for taking the time to work it out, Unknown008!

Mathematica agrees with your answer.

Original poster, please check either your question or answer.

**Haven** · October 19th 2010, 07:44 AM

Sorry I was off by -1 on the RHS. it should be $-\frac{1+i}{1-i}\frac{z+1}{z-1}$ , but i have the general method now. thanks

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