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 September 17th, 2016, 12:06 PM #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 09:30 PM.
 September 17th, 2016, 01:21 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,697 Thanks: 977 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}(\sqrt2-1) \\ &=\frac{2-\sqrt2}{2} \\ &=1-\frac{\sqrt2}{2}\end{align*} Thanks from wortel
 October 31st, 2016, 09:51 PM #3 Global Moderator   Joined: Dec 2006 Posts: 18,427 Thanks: 1462 \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*} Thanks from Joppy

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