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I'm having trouble finding the real and imaginary part of $z/(z+1)$ given that z=x+iy. I tried substituting that in but its seems to get really complicated and I'm not so sure how to reduce it down. Can anyone give me some advice?

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    $\begingroup$ I would probably rewrite as $1-\frac{1}{z+1}$. So we want to invert $x+1+iy$. Multiply top and bottom by $x+1-iy$. $\endgroup$ – André Nicolas Jan 19 '15 at 18:32
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Just multiply the fraction by the complex conjugate of $z+1$, that is, \begin{equation*} \frac{z}{z+1} = \frac{x+iy}{1+x+iy} = \frac{1+x-iy}{1+x-iy} \frac{x+iy}{1+x+iy} = \frac{(1+x)x+y^2+iy }{(1+x)^2+y^2} = \end{equation*} \begin{equation*} = \frac{(1+x)x+y^2}{(1+x)^2+y^2} + i \frac{y}{(1+x)^2+y^2} \end{equation*}

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$$\begin{align} &\frac{z}{z+1}\\ =& \frac{x+iy}{x+iy+1}\\ =&\frac{x+iy}{x+iy+1}\times \frac{x-iy+1}{x-iy+1} \end{align}$$ I'm sure you can take the rest from here.

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Usually if $z'$ is a non-zero complex number, $\frac{1}{z'} = \frac{\overline{z'}}{|z'|^2}$. Here $z' = 1+z$. Use this.

You can also use same type of idea to put inverse of numbers of the form $a+b\sqrt{d}$ in the same form...

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