Can someone show me how $\frac{1-\bar{z}}{(1-z)(1-\bar{z})}=\frac{1-\bar{z}}{2(1-Re(z))}.$
Equation involving complex conjugate
1
$\begingroup$
complex-numbers
-
1Do you wish to include that $|z|=1$? If you let $z=3$, then $(1-3)(1-3)=4$, but $2(1-3)=-4$. – 2012-11-06
3 Answers
0
Let $z=x+iy$
Cancelling the numerators we get, $\{1-(x+iy)\}\{1-(x-iy)\}=2(1-x)$
or, $(1-x)^2-(iy)^2=2-2x$
or, $x^2+y^2=1\implies \mid z\mid =1$
1
Hint: Use the fact that $Re(z) = \frac{z + \bar{z}}{2}$
0
For every $z \in\mathbb{C}\setminus \{1\}$: $\frac{1}{1-z}=\frac{1-\bar{z}}{(1-z)(1-\bar{z})}=\frac{1-\bar{z}}{1-z-\bar{z}+z\bar{z}}=\frac{1-\bar{z}}{1-2\Re{z}+|z|^2},$ therefore $\frac{1-\bar{z}}{(1-z)(1-\bar{z})}=\frac{1-\bar{z}}{2(1-\Re(z))}$ if $|z|=1.$
-
0@Mercy Thanks for remark! – 2012-11-06