- Show that in polar coordinates, the Cauchy-Riemann equations take the form
\[ {\frac {\partial u}{\partial r}} = {\frac {1}{r}} {\frac {\partial v}{\partial \theta}} \quad {\mathrm{and}} \quad {\frac {1}{r}} {\frac {\partial u}{\partial \theta}} = - {\frac {\partial v}{\partial r}}. \]
Use these equations to show that the logarithm function defined by
\[ \log z = \log r + i \theta \quad \mathrm{where} z = r e ^ {i \theta} \mathrm{with} - \pi < \theta < \pi \]
is holomorphic in the region \(r > 0\) and \(- \pi < \theta < \pi\)
Solution
Let \(z=x+iy=re^{i\theta}\), so \[ x=r\cos\theta, \qquad y=r\sin\theta. \] For \(f=u+iv\), the ordinary chain rule gives \[ u_r=u_x\cos\theta+u_y\sin\theta, \qquad u_\theta=-r u_x\sin\theta+r u_y\cos\theta, \] and similarly for \(v\). If the Cartesian Cauchy–Riemann equations \(u_x=v_y\) and \(u_y=-v_x\) hold, then \[ \frac1r v_\theta =-v_x\sin\theta+v_y\cos\theta =u_y\sin\theta+u_x\cos\theta =u_r, \] and \[ v_r=v_x\cos\theta+v_y\sin\theta =-u_y\cos\theta+u_x\sin\theta =-\frac1r u_\theta. \] Thus \[ u_r=\frac1r v_\theta, \qquad \frac1r u_\theta=-v_r. \] Conversely, reversing the calculation gives the Cartesian Cauchy–Riemann equations, so these are the polar form for \(r>0\).
For the principal logarithm, \[ u(r,\theta)=\log r, \qquad v(r,\theta)=\theta. \] Hence \[ u_r=\frac1r, \quad u_\theta=0, \quad v_r=0, \quad v_\theta=1, \] so the polar Cauchy–Riemann equations hold throughout \[ r>0,\qquad -\pi<\theta<\pi. \] These partial derivatives are continuous there, and \((r,\theta)\mapsto re^{i\theta}\) is a \(C^1\) diffeomorphism onto the slit plane, so \(u\) and \(v\) are \(C^1\) functions of \((x,y)\) satisfying the Cartesian Cauchy–Riemann equations. A \(C^1\) function satisfying the Cauchy–Riemann equations is holomorphic. Therefore the principal branch \(\log z=\log r+i\theta\) is holomorphic on that region.