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For complex differentiability at a point, the derivative must have the same value along real and imaginary increments. Along real increments it equals $u_x+i{v}_x$, while along imaginary increments it equals $(u_y+i{v}_y)/i=v_y-i{u}_y$. Equating real and imaginary parts gives the Cauchy-Riemann equationsFor the given real part,Integrating the Cauchy-Riemann equations gives a harmonic conjugateThis open set is open and connected; $F$ is holomorphic there with derivative $-z^{-2}$. The origin must be excluded because the real part is undefined there. On this connected open set, any other harmonic conjugate differs by a real constant: the Cauchy-Riemann equations make both partial derivatives of the difference zero.

Let $v_1,v_2\ne 0$ be tangent vectors to the two smooth curves at $p$. complex differentiability at a point givesWhen $f^{\prime }(p)\ne 0$, its real derivative is multiplication by a nonzero complex number, which is a rotation followed by a positive scaling. It therefore preserves the angle between $v_1$ and $v_2$, so $f$ is conformal at $p$. The condition is essential: $f(z)=z^2$ has $f^{\prime }(0)=0$ and sends rays making angle $\pi /4$ at zero to rays making angle $\pi /2$.

For $J(z)=z+z^{-1}$,If $z,w$ both satisfy $|z|>1$, or both satisfy $0<|z|<1$, the alternative $zw=1$ is impossible unless the points lie on the omitted unit circle. Hence $J$ is one-to-one on each region. Also $J^{\prime }(z)=1-z^{-2}$ has no zero there, so the restrictions are conformal. Solving $z^2-wz+1=0$ shows that $w\in [-2,2]$ exactly when both roots lie on the unit circle; otherwise one root is inside and the other outside. Thus each region has image

Taking the reciprocal of the interior restriction and filling its removable value at zero givesThe reciprocal sends $\mathbb{C}\setminus [-2,2]$ ontoand $g(0)=0$. Therefore $g$ is the required one-to-one conformal map from the unit disc.