Is it possible to rewrite $f(z) = z^{n}(z^*)^{m}$ with the use of the binomial function.
The binomial function is defined as $(x+y)^{n} = \sum\limits_{k=0}^n \binom{n}{k} x^{n-k}y^{k} = \sum\limits_{k=0}^n \binom{n}{k} x^{k}y^{n-k}$
For the equation $f(z) = z^{n}(z^*)^{m}= (x+iy)^{n}(x-iy)^{m}$
[lets call $(x+iy)^{n}$ term 1 and $(x-iy)^{m}$ term 2]
we get:
$ 1: \sum\limits_{k=0}^n \binom{n}{k} x^{k}iy^{n-k}$
$ 2: \sum\limits_{k=0}^m \binom{m}{k} x^{k}(-i)y^{m-k}$
the reason why i want to rewrite f(z) is to show that f(z) is not anylical in any point, but am not sure how to draw this conclusion from here.
any help on how i should approach this? And is the binomial function applied correctly?
thanx
added Cauchy-Riemann approach as advised by David:
$f(z) = z^{n}(z^*)^{m}$
write in polar form:
$f(z) = (e^{i\theta})^{n}(e^{-i\theta})^{m} = e^{i\theta n} e^{-i \theta m}$
$f(z) = (cos(\theta n) +i sin(\theta n))(cos(\theta m) -i sin(\theta m)) = cos(\theta n)cos(\theta m) + icos(\theta n)sin(\theta m)+ i cos(\theta m)sin(\theta n) - sin(\theta n)sin(\theta m) $
f(z) = u(n,m) + iv(n,m)
$u = cos(\theta n)cos(\theta m) - sin(\theta n)sin(\theta m) $
$v = icos(\theta n)sin(\theta m)+ i cos(\theta m)sin(\theta n) $
$u_{n} = -\theta sin(\theta n)cos(\theta m) - \theta con(\theta n)sin(\theta m)$
$u_{m} = -\theta cos(\theta n)sin(\theta m)- \theta sin(\theta n)sin(\theta m)$
$v_{n} = - i\theta sin(\theta n)sin(\theta m)+ i \theta cos(\theta m)cos(\theta n)$
$v_{m} = i\theta cos(\theta n)cos(\theta m) - i \theta sin(\theta m)sin(\theta n)$
does not satisfy Cauchy-Riemann eq; not analytical.