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September 17th, 2016, 11:06 AM  #1 
Newbie Joined: Jun 2012 Posts: 11 Thanks: 0  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. 
September 17th, 2016, 12:21 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,429 Thanks: 868 Math Focus: Elementary mathematics and beyond 
$\displaystyle \begin{align*}\frac{z}{(z+1)^2}&=\frac{e^{i\pi/4}}{(e^{i\pi/4}+1)^2} \\ &=\frac{e^{i\pi/4}}{e^{i\pi/2}+2e^{i\pi/4}+1} \\ &=\frac12\frac{\sqrt2+\sqrt2i}{i+\sqrt2+\sqrt2i +1} \\ &=\frac{\sqrt2}{2}\frac{1+i}{(\sqrt2+1)(1+i)} \\ &=\frac{\sqrt2}{2}\frac{1}{\sqrt2+1} \\ &=\frac{\sqrt2}{2}(\sqrt21) \\ &=\frac{2\sqrt2}{2} \\ &=1\frac{\sqrt2}{2}\end{align*}$ 
October 31st, 2016, 08:51 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 16,766 Thanks: 1231 
$\displaystyle \begin{align*}\frac{z}{(z+1)^2}&=\frac{1}{z + 2 + 1/z} \\ &= \frac{1}{2 + \sqrt2} \\ &= \frac{2  \sqrt2}{2} \\ &= 1  \frac{\sqrt2}{2}\end{align*}$ 

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complex, form, number, real, write 
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