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I have

$$L: \omega = z(1 + i) \hspace{1.5cm} M: \omega = \frac{1}{1-z} \hspace{1.5cm} L^{-1}: \omega = \frac{z}{i + 1}$$

I need to do the composition

$$L \circ M \circ L^{-1}.$$

So, I first did $M \circ L^{-1}$ which gave me

$$\frac{1}{{1 - \frac{z}{1 +i}}}.$$

From here, I changed $\frac{z}{1 + i}$ to $\frac{z(1 - i)}{2}$ and so my composition becomes

$$\frac{2}{2 - z(1 - i)}.$$

Now, doing the whole composition $L \circ M \circ L^{-1}$ I get

$$\frac{2}{2 - z(1 - i)} \cdot (1 + i).$$

Which I then multiplied out to get

$$\frac{2 + 2i}{2 - z + i}.$$

But this answer is wrong. It says it should be $\frac{2i}{-z + (1 + i)}$. Where have I gone wrong?

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1 Answer 1

up vote 1 down vote accepted

$$z\mapsto \frac{z}{1+i}\mapsto \frac{1}{1-\frac{z}{1+i}}=\frac{1}{\frac{1+i-z}{1+i}}=\frac{1+i}{1+i-z}\mapsto \frac{(1+i)^2}{1+i-z}=\frac{2i}{1+i-z}$$

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Yeah, that was a typo, sorry. –  Kaish Jan 12 '13 at 15:25
    
Because see on the next line I've got it as $(1 -i)$ again –  Kaish Jan 12 '13 at 15:25
1  
Ok, thanks for that. Why is it different though from changing $1 \rightarrow \frac{1 + i}{1 + i}$ and from rationalising the denominator. Shouldn't both compositions give me the same answer? –  Kaish Jan 12 '13 at 15:35

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