 Solution

Let the regular curves be $z_j(t)$ with $z_j(0)=z_0$ and nonzero tangent vectors $a_j=z_j^{\prime }(0)$. By the chain rule, their transformed tangents are $f^{\prime }(z_0)a_j$. Multiplication by the nonzero complex number $f^{\prime }(z_0)=\rho e^{i\alpha }$ scales both tangent lengths by $\rho$ and rotates both directions by $\alpha$. In particular,Thus the oriented angle, and hence the ordinary angle, is unchanged. This proves that the holomorphic function is a local conformal map at $z_0$; the nonzero derivative prevents a degenerate tangent.

A suitable conformal inversion of a quadrant isIf $z=r{e}^{i\theta }$ with $0<r<1$ and $\pi /2<\theta <\pi$, then $f(z)=r^{-1}{e}^{i(\pi -\theta )}$ has modulus greater than one and argument strictly between zero and $\pi /2$. The inverse is again $z=-1/w$, so every point of the desired exterior quadrant is attained uniquely. The origin is outside the open source region, and $f^{\prime }(z)=1/z^2\ne 0$ there.