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$a=\frac{1+i}{\sqrt2}$

$b=\frac{\sqrt3+i}{2}$

$a,b,z\in\mathbb{C}$

What is the real part of $z=\frac{a-b}{1+ab}$ ?

The answer is $0$ but i do not know why.

I tried simply substituting $a, b$ but i didn't get anything in a simple form. Then i tried using trigonometric form because i had $\frac{1}{\sqrt2},\frac{\sqrt3}{2},\frac{1}{2}$ which are common cosine and sine values, but again i got nothing. I have also noted that $a^2=i$ but that didn't seem to help much.

Do you have any tips ? Thanks in advance !

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as $|a|=|b|=1\implies a=\dfrac{1}{ \bar a },b=\dfrac{1}{\bar b}$ thus $$Z=\frac{a-b}{1+ab}=\frac{\dfrac{1}{ \bar a }-\dfrac{1}{ \bar b }}{1+\dfrac{1}{ \bar a }\cdot \dfrac{1}{ \bar b }}=-\left(\frac{\bar a-\bar b}{1+\bar a\bar b}\right)=-\bar Z $$ so $\text{Re}( Z)=0$

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If $|a|=|b|=1$ (and this is your case) $$\frac{a-b}{1+ab}= \frac{(a-b)\overline{1+ab}}{(1+ab)\overline{1+ab}} = \frac{(a-b)(1+\bar a\bar b)}{|1+ab|^2} = \frac{a-b+a\bar a\bar b-\bar a b\bar b}{|1+ab|^2}= \frac{a-b+|a|^2\bar b-\bar a |b|^2}{|1+ab|^2}=$$ $$=\frac{(a-\bar a)-(b-\bar b)}{|1+ab|^2}=\frac{2i\Im(a)-2i\Im(b)}{|1+ab|^2}=i\cdot 2\frac{\Im(a)-\Im(b)}{|1+ab|^2}$$

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With the complex exponential:

  • $a=\mathrm e^{\tfrac{i\pi}4},\quad b=\mathrm e^{\tfrac{i\pi}6}$.
  • For any $z\in\mathbf C$, $\;\operatorname{Re}z=\frac12(z+\bar z)$.

Therefore \begin{align} 2\operatorname{Re}\Bigl(\frac{a-b}{1+ab}\Bigr)&=\frac{\mathrm e^{\tfrac{i\pi}4}-\mathrm e^{\tfrac{i\pi}6}}{1+\mathrm e^{\tfrac{5i\pi}{12}}}+\frac{\mathrm e^{-\tfrac{i\pi}4}-\mathrm e^{-\tfrac{i\pi}6}}{1+\mathrm e^{-\tfrac{5i\pi}{12}}}\\ &=\frac{\Bigl(\mathrm e^{\tfrac{i\pi}4}-\mathrm e^{\tfrac{i\pi}6}\Bigr)\Bigl(1+\mathrm e^{-\tfrac{5i\pi}{12}}\Bigr)+\Bigl(\mathrm e^{-\tfrac{i\pi}4}-\mathrm e^{-\tfrac{i\pi}6}\Bigr)\Bigl(1+\mathrm e^{\tfrac{5i\pi}{12}}\Bigr)}{\Bigl(1+\mathrm e^{\tfrac{5i\pi}{12}}\Bigr)\Bigl(1+\mathrm e^{-\tfrac{5i\pi}{12}}\Bigr)}, \end{align} and expanding the numerator, you can check one obtains $0$.

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