Is $f(z)=z^n$ holomorphic? Is $f(z)=z^n$ holomorphic?
I have tested a number of other functions using the Cauchy Riemann equations $u_x=v_y$, $v_x=-u_y$. However in the case of $f(z)=z^n$ I cannot think of a way to find the functions $u(x,y)$ and $v(x,y)$ without using a binomial expansion of $(x+iy)^n$. 
Any help or pointers is appreciated.
edit - the problem requires the use of the Cauchy - Riemann equations and not the formal definition of complex differentiation.

HINT: Show that if $f=u+iv$ and $g=\hat{u}+i\hat{v}$ satisfy the Cauchy-Riemann equations, then so does $fg$. Use this together with the fact that $z^n$ is a finite product of holomorphic functions (since $z$ is holomorphic).
