proof of conformal mapping theorem

Let $D\\subset\\mathbb{C}$ be a domain, and let $f\\colon D\\to\\mathbb{C}$ be ananalytic function. By identifying the complex plane $\\mathbb{C}$ with $\\mathbb{R}^{2}$ , we can view $f$ as a function from $\\mathbb{R}^{2}$ toitself:

\\tilde{f}(x,y):=(\\Re f(x+iy),\\Im f(x+iy))=(u(x,y),v(x,y))

with $u$ and $v$ real functions. The Jacobian matrix of $\\tilde{f}$ is

J(x,y)=\\frac{\\partial(u,v)}{\\partial(x,y)}=\\begin{pmatrix}u_{x}&u_{y}\\\\v_{x}&v_{y}\\end{pmatrix}.

As an analytic function, $f$ satisfies the Cauchy-Riemann equations,so that $u_{x}=v_{y}$ and $u_{y}=-v_{x}$ . At a fixed point $z=x+iy\\in D$ , wecan therefore define $a=u_{x}(x,y)=v_{y}(x,y)$ and $b=u_{y}(x,y)=-v_{x}(x,y)$ .We write $(a,b)$ in polar coordinates as $(r\\cos\\theta,r\\sin\\theta)$ and get

J(x,y)=\\begin{pmatrix}a&b\\\\-b&a\\end{pmatrix}=r\\begin{pmatrix}\\cos\\theta&\\sin\\theta\\\\-\\sin\\theta&\\cos\\theta\\end{pmatrix}.

Now we consider two smooth curves through $(x,y)$ , which weparametrize by $\\gamma_{1}(t)=(u_{1}(t),v_{1}(t))$ and $\\gamma_{2}(t)=(u_{2}(t),v_{2}(t))$ . We can choose the parametrization suchthat $\\gamma_{1}(0)=\\gamma_{2}(0)=z$ . The images of these curves under $\\tilde{f}$ are $\\tilde{f}\\circ\\gamma_{1}$ and $\\tilde{f}\\circ\\gamma_{2}$ ,respectively, and their derivatives at $t=0$ are

(\\tilde{f}\\circ\\gamma_{1})^{\\prime}(0)=\\frac{\\partial(u,v)}{\\partial(x,y)}(%\\gamma_{1}(0))\\cdot\\frac{{\\rm d}\\gamma_{1}}{{\\rm d}t}(0)=J(x,y)\\begin{pmatrix}%\\frac{{\\rm d}u_{1}}{{\\rm d}t}\\\\\\frac{{\\rm d}v_{1}}{{\\rm d}t}\\end{pmatrix}

and, similarly,

(\\tilde{f}\\circ\\gamma_{2})^{\\prime}(0)=J(x,y)\\begin{pmatrix}\\frac{{\\rm d}u_{2}%}{{\\rm d}t}\\\\\\frac{{\\rm d}v_{2}}{{\\rm d}t}\\end{pmatrix}

by the chain rule. We see that if $f^{\\prime}(z)\eq 0$ , $f$ transforms thetangent vectors to $\\gamma_{1}$ and $\\gamma_{2}$ at $t=0$ (and thereforein $z$ ) by the orthogonal matrix

J/r=\\begin{pmatrix}\\cos\\theta&\\sin\\theta\\\\-\\sin\\theta&\\cos\\theta\\end{pmatrix}

and scales them by a factor of $r$ . In particular, the transformationby an orthogonal matrix implies that the angle between the tangentvectors is preserved. Since the determinant of $J/r$ is 1, thetransformation also preserves orientation (the direction of the anglebetween the tangent vectors). We conclude that $f$ is a conformalmapping at each point where its derivative is nonzero.