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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: 444 Thanks: 254 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: 163 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: 19,542 Thanks: 1751 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: 3,261 Thanks: 894 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 : $\displaystyle \Re(A)=0$

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