Given $f:\mathbb{R}^2\simeq \mathbb{C}\rightarrow \mathbb{R}^2$ with $f(z)=z^n$. Prove that $$deg(f, B(0,1), 0)=n.$$ //Proof// Suppose that $f(z)=u(x,y)+iv(x,y)$, it is easy to see that $f$ is analytic, hence $u,v$ satisfy Cauchy-Riemann equation (i.e. $u_x=v_y, u_y=-v_x$). Hence, we see that $$(1)\hspace{25pt}det(\p f)=|\begin{pmatrix}u_x&u_y ~\\ v_x&v_y\end{pmatrix}| = u_x^2+u_y^2\geq 0.$$ Hence, $det(\p f)$ is $0$ if and only if $u_x=u_y=0$ iff $z=0$. Hence, $z=0$ is a critical point. Thus $0$ is a critical value. By definition of Brouwser degree, we have $$deg(f, B(0,1), 0) = deg(f, B(0,1),1/2).$$ We know $z^n=1/2$ has $n$ solutions and by (1), we see that $deg(f,B(0,1),1/2)=n$ by definition.