For complex numbers z and w, prove that $${\left| z \right|^2}w - {\left| w \right|^2}z = z - w$$ if and only if $$ z = w\,or\,z\overline {\,w} = 1$$.

If the roots of the equation $${x^2} - 2ax + {a^2} + a - 3 = 0$$ are real and less than 3, then

If in the expansion of $${\left( {1 + x} \right)^m}{\left( {1 - x} \right)^n},$$ the coefficients of $$x$$ and $${x^2}$$ are $$3$$ and $$-6$$ respectively, then $$m$$ is

Let $$n$$ be any positive integer. Prove that $$$\sum\limits_{k = 0}^m {{{\left( {\matrix{ {2n - k} \cr k \cr } } \right)} \over {\left( {\matrix{ {2n - k} \cr n \cr } } \right)}}.{{\left( {2n - 4k + 1} \right)} \over {\left( {2n - 2k + 1} \right)}}{2^{n - 2k}} = {{\left( {\matrix{ n \cr m \cr } } \right)} \over {\left( {\matrix{ {2n - 2m} \cr {n - m} \cr } } \right)}}{2^{n - 2m}}} $$$

for each non-be gatuve integer $$m \le n.$$ $$\,\left( {Here\left( {\matrix{ p \cr q \cr } } \right) = {}^p{C_q}} \right).$$