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 November 16th, 2017, 02:31 AM #1 Newbie   Joined: Nov 2017 From: Poland Posts: 1 Thanks: 0 Proof that real part of complex number is equal 0 Given that $\displaystyle w = 1 + \sqrt{3}$ and $\displaystyle z = 1 + i$ show that $\displaystyle \Re(\frac{\sqrt{2}z + w}{\sqrt{2}z - w}) = 0$ I only found that $\displaystyle \Re(w) = \frac{1}{2} (w + \bar{w})$ but still I have no idea how to solve it. November 16th, 2017, 03:09 AM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 642 Thanks: 406 Math Focus: Dynamical systems, analytic function theory, numerics I suspect you should have $w - \bar{w}$ in your simplification. The solution then follows by thinking about the real part of any complex number minus its own conjugate. November 16th, 2017, 08:07 AM #3 Senior Member   Joined: Dec 2015 From: holland Posts: 162 Thanks: 37 Math Focus: tetration Why use the conjugate? Just fill in w and z in the given expression. November 16th, 2017, 10:10 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,965 Thanks: 2214 I suggest, aga150, that you correct the mistake in the statement of the problem. January 9th, 2018, 04:20 AM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 I suspect that should be $w= 1+ i\sqrt{3}$ NOT "$1+ \sqrt{3}$". January 9th, 2018, 11:29 AM #6 Member   Joined: Aug 2011 From: Nouakchott, Mauritania Posts: 85 Thanks: 14 Math Focus: Algebra, Cryptography Good afternoon ! As Country Boy has said, we suspect that it should be $\displaystyle w=1+i\sqrt3$. In this case , let $\displaystyle A=\frac{\sqrt{2}z + w}{\sqrt{2}z - w}$. Then : $\displaystyle \overline A=\frac{\sqrt{2}\bar z + \bar w}{\sqrt{2}\bar z - \bar w}$. We have : \displaystyle \begin{align*} A\times\frac1{\overline A}&=\frac{\sqrt{2}z + w}{\sqrt{2}z - w}\times\frac{\sqrt{2}\bar z - \bar w}{\sqrt{2}\bar z + \bar w}=\frac{2|z|^2-|w|^2+\sqrt2(w\bar z-z\bar w)}{2|z|^2-|w|^2-\sqrt2(w\bar z-z\bar w)}\\& =\frac{2\times2-4+\sqrt2(w\bar z-\overline{w\bar z})}{2\times2-4-\sqrt2(w\bar z-\overline{w\bar z})}\\& =\frac{\sqrt2\times 2i\Im(w\bar z)}{-\sqrt2\times 2i\Im(w\bar z)}\\&=-1 \end{align*} Thus : $\displaystyle A=-\overline A$ and this means that : $\displaystyle \Re(A)=0$ Tags complex, equal, number, part, proof, real Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Bogdano Complex Analysis 2 September 3rd, 2014 02:19 AM archer18 Complex Analysis 3 March 2nd, 2014 12:46 PM rayman Complex Analysis 4 February 22nd, 2013 09:21 PM Punch Complex Analysis 4 April 1st, 2012 09:23 AM TsAmE Complex Analysis 1 October 18th, 2010 04:38 PM

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