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$$ \frac{\bar{z}+i}{1-i\bar{z}} $$

I wanted to know what is the conjugate to simplify the equation from above, i mean:

Would it be like this: $$ \overline{(1-i\bar{z})}=1+i\bar{z} $$ Or: $$ \overline{(1-i\bar{z})}=1+iz $$ since $\overline{\bar{z}}=z$

J. W. Tanner
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James
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1 Answers1

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$$\overline{(1-i\bar{z})}=1+iz $$ is the correct way

So the fraction will now become

$$ \frac{z-i}{1+iz} $$

thedumbkid
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  • i don t how you got that fraction, what i have been doing is multiplicating the original fraction for the denominator conjugate, i arrive to $\frac{i(1Im(z)+|z|^2+1)}{2Re(zi)+|z|^2+1}$ i don´t know what i have done wrong – James Dec 15 '20 at 21:51
  • The fraction that I have mentioned is just another version of the same expression. Now multiplying both numerator and denominator by the denominator of the original fraction will make the denominator real and hence simplify the expression. – thedumbkid Dec 15 '20 at 21:55
  • Ok i get it, it´s just it seems such a strange answer to be the final part , couldn´t i find a relation betwen Im(z) and Re(zi) to simplify more ? $$\frac{i(2Im(z)+|z|^2+1)}{2Re(zi)+|z|^2+1}$$ – James Dec 15 '20 at 21:58
  • No, it's pretty simplified to work with now actually. I haven't checked your calculations, but if that's what it is, that's what it is! – thedumbkid Dec 15 '20 at 22:09
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    I had an error it´s actually $$ \frac{i(-2Im(z)+|z|^2+1)}{2Re(zi)+|z|^2+1} $$ and since Im(z)=b and Re(zi)=-b for z=a+bi we have that Im(z)=-Re(zi) so the fraction is actually equal to i.The funnny part is that if i consider $\overline{1-i\bar{z}}=1+i\bar{z}$ i get the same result, i am speechless – James Dec 15 '20 at 22:26