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September 17th, 2016, 11:06 AM   #1
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write complex number in a+bi form with a and b real

$\frac{z}{(z+1)^2}$, with $z=\frac{1}{2}\sqrt{2}+\frac{1}{2}\sqrt{2}i$

I know the answer is $1-\frac{\sqrt{2}}{2}$, but I don't know how to get there. Anybody know which steps I should take?

Last edited by skipjack; October 31st, 2016 at 08:30 PM.
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September 17th, 2016, 12:21 PM   #2
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$\displaystyle \begin{align*}\frac{z}{(z+1)^2}&=\frac{e^{i\pi/4}}{(e^{i\pi/4}+1)^2} \\
&=\frac{e^{i\pi/4}}{e^{i\pi/2}+2e^{i\pi/4}+1} \\
&=\frac12\frac{\sqrt2+\sqrt2i}{i+\sqrt2+\sqrt2i +1} \\
&=\frac{\sqrt2}{2}\frac{1+i}{(\sqrt2+1)(1+i)} \\
&=\frac{\sqrt2}{2}\frac{1}{\sqrt2+1} \\
&=\frac{\sqrt2}{2}(\sqrt2-1) \\
&=\frac{2-\sqrt2}{2} \\
&=1-\frac{\sqrt2}{2}\end{align*}$
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October 31st, 2016, 08:51 PM   #3
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$\displaystyle \begin{align*}\frac{z}{(z+1)^2}&=\frac{1}{z + 2 + 1/z} \\
&= \frac{1}{2 + \sqrt2} \\
&= \frac{2 - \sqrt2}{2} \\
&= 1 - \frac{\sqrt2}{2}\end{align*}$
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