For $f(z) = z^n$ and $z=x+iy$. How do I expand this using binomial theorem?

Question

For $f(z) = z^n$ with $n \in \mathbb{N}$, obtain expressions for the real valued functions $u$ and $v$ so that $f(x + iy) = u(x,y) + iv(x,y)$ where $z=x + iy$. Is this function differentiable?

I need to use binomial and split it into real and imaginary parts, but I'm not entirely sure how to do this.

Answer 1

$$z^n=(x+iy)^n=\sum_{r=0}^n\binom{n}{r}x^{n-r}(iy)^r$$$$=\sum_{r=0}^{\lfloor\frac{n}{2}\rfloor}\binom{n}{2r}x^{n-2r}(-1)^ry^{2r}+i\sum_{r=0}^{\lfloor\frac{n}{2}\rfloor\pm 1}\binom{n}{2r+1}x^{n-(2r+1)}(-1)^ry^{2r+1}$$$$=u(x,y)+iv(x,y).$$The $\pm$ sing in second summation occur as, $n$ is even and $-$ when $n$ is odd.

$f$ is differentiable on $\Bbb C$. This is because $$\lim_{h\to 0}\frac{f(z+h)-f(z)}{h}=\lim_{h\to 0}\frac{(z+h)^n-z^n}{h}$$$$=\lim_{h\to 0}\frac{\big(z^n+\binom{n}{1}z^{n-1}h+\binom{n}{2}z^{n-2}h^2+...+\binom{n}{n-1}z^{1}h^{n-1}+h^n\big)-z^n}{h}$$$$=\lim_{h\to 0}\bigg(\binom{n}{1}z^{n-1}+\binom{n}{2}z^{n-2}h+...+\binom{n}{n-1}z^{1}h^{n-2}+h^{n-1}\bigg)$$$$=\binom{n}{1}z^{n-1}+\binom{n}{2}z^{n-2}0+...+\binom{n}{1}z^{1}0^{n-2}+0^{n-1}$$$$=nz^{n-1}.$$

Answer 2

The binomial theorem says that

$$(a+b)^n=a^n+na^{n-1}b+\binom n2a^{n-2}b^2+\binom n3 a^{n-3}b^3+\dots+nab^{n-1}+b^n.$$

Use that with $a=x$ and $b=iy$ to get $z^n$.

Also note that $i^{4k}=1, i^{4k+1}=i, i^{4k+2}=-1, $ and $i^{4k+3}=-i$.