I have the next question:
Find $z\in \mathbb{C}$ such that $\bar{z}=i(z-1)$
This is my solution:
$\overline{(x+iy)}=i(x+iy-1)$
$x-iy =ix +i^{2}y-i$
$x-iy=-y+i(x-1)$
Now I have compared the real parts of the equation and the imaginary parts of the equation to find x and y.
Re(): $x=-y$
Img(): $-y=x-1$
$\Rightarrow -y=-y-1 \rightarrow 0y=-1$
Where is my mistake? Thank you all for the help in advance.
If you didn't want to convert into parts, conjugating the identity gives:
$$\bar {\bar z} = \bar{i(z-1)} = -i (\bar z -1)$$
Here we use the identity again.
$$= - i( i(z-1) -1) = z-1+i$$
So $z= z -1+i$ which implies $0= -1+i,$ which is absurd.