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November 16th, 2017, 02:31 AM   #1
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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.
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November 16th, 2017, 03:09 AM   #2
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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.
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November 16th, 2017, 08:07 AM   #3
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Why use the conjugate? Just fill in w and z in the given expression.
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November 16th, 2017, 10:10 AM   #4
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I suggest, aga150, that you correct the mistake in the statement of the problem.
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January 9th, 2018, 04:20 AM   #5
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I suspect that should be $w= 1+ i\sqrt{3}$ NOT "$1+ \sqrt{3}$".
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January 9th, 2018, 11:29 AM   #6
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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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