first step..... takin bars we get dat.....
second step thinkin....
first step..... takin bars we get dat.....
second step thinkin....
SECOND STEP IT S CROSSS MUTIPLICATION N THIRD STEP ME ASβ≠0 SO ω≠ω(BAR)
1st purely real means z=z...so..here the same thg is used here z is the whole expression
@richs....can u pls do me d cross multiplyin step?????????
\frac{w- w^cz}{1-z}=\frac{w^c- wz^c}{1-z^c} \\ \Rightarrow (w- w^cz)(1-z^c)=(1-z)(w^c- wz^c) \\ \Rightarrow w-wz^c-w^cz+w^czz^c=w^c-wz^c-w^cz+wzz^c \\ \Rightarrow w+w^czz^c=w^c+wzz^c \\ \Rightarrow w-w^c+w^czz^c-wzz^c=0
hence the result above..
is there a way to put complex conjugate in latex instead of raising to 'c'