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 !


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 $\text{Re}( Z)=0$


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


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 $0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.