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Suppose that $Z = x + iy$, we know that $x= \frac{z +\bar{z}}{2}$

Is that a similar way to get y?(Only use Z and real numbers)

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closed as unclear what you're asking by Namaste, user223391, Daniel W. Farlow, Vladhagen, Shailesh Feb 2 '17 at 0:04

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    $\begingroup$ $y=\frac{z-\bar z}{2i}$ $\endgroup$ – Tim B. Feb 1 '17 at 15:19
  • $\begingroup$ Is that possible only use Z and real number? $\endgroup$ – Parting Feb 1 '17 at 15:26
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    $\begingroup$ Consider $Z-x$? $\endgroup$ – Piotr Benedysiuk Feb 1 '17 at 15:31
  • $\begingroup$ @Parting What is $\bar z$ when $z=x+iy$? $\endgroup$ – Error 404 Feb 1 '17 at 15:32
  • $\begingroup$ Your followup question "Is that possible only use Z and real number" is too vague, because you have not specified what kind of formulas you would allow. For example, you could ask more specifically whether it is possible to find real values $a,b$ such that $aZ+b=y$, in which case the definitive answer is "no", unless you use $a=0$ and $b=y$. $\endgroup$ – Lee Mosher Feb 1 '17 at 15:42
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Just juggle the variables. Do it $$z=x+iy\Rightarrow z-x=iy$$ We know that $x=\frac{z+\bar{z}}{2}$. Then $$z-x=iy\Rightarrow z-\frac{z+\bar{z}}{2}=iy \Rightarrow \frac{2z}{2}-\frac{z+\bar{z}}{2}=iy$$ $$\Rightarrow iy=\frac{z-\bar{z}}{2}\Rightarrow y=\frac{z-\bar{z}}{2i}.$$

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Hint: $\require{cancel} z=x+iy \;\implies\; \bar z = x - iy \;\implies\; z-\bar z = \bcancel{x}+iy-(\bcancel{x}-iy)=2iy\,$.

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