0
$\begingroup$

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)

  • 7
    $y=\frac{z-\bar z}{2i}$ – 2017-02-01
  • 0
    Is that possible only use Z and real number? – 2017-02-01
  • 2
    Consider $Z-x$? – 2017-02-01
  • 0
    @Parting What is $\bar z$ when $z=x+iy$? – 2017-02-01
  • 0
    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 $b=y$. – 2017-02-01
  • 0
    Since we're putting arbitrary restrictions on which numbers we use in the formulas, just for fun let's find the real part of $z$ using only $z,$ $\bar z,$ and the number $\pi.$ – 2017-02-01

2 Answers 2

4

Just juggle the variables. Do it $$z=x+iy\Rightarrow z-x=iy$$ We know that $x=\frac{z+\bar{z}}{2}$. Then $$z-x=iy\Rightarrow z-\frac{z+\bar{z}}{2}=iy \Rightarrow \frac{2z}{2}-\frac{z+\bar{z}}{2}=iy$$ $$\Rightarrow iy=\frac{z-\bar{z}}{2}\Rightarrow y=\frac{z-\bar{z}}{2i}.$$

3

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