= Solution
Let the regular curves be $z_j(t)$ with $z_j(0)=z_0$ and nonzero tangent vectors $a_j=z_j'(0)$. By the <chain rule>, their transformed tangents are $f'(z_0)a_j$. Multiplication by the nonzero complex number $f'(z_0)=\rho e^{i\alpha}$ scales both tangent lengths by $\rho$ and rotates both directions by $\alpha$. In particular,
$$
\arg\frac{f'(z_0)a_2}{f'(z_0)a_1}=\arg\frac{a_2}{a_1}.
$$
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> is
$$
\boxed{f(z)=-\frac1z.}
$$
If $z=re^{i\theta}$ 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 $f'(z)=1/z^2\ne0$ there.
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