Cauchy-Riemann equations in polar form.

Question

Show that in polar coordinates, the Cauchy-Riemann equations take the form $\dfrac{\partial u}{\partial r} = \dfrac{1}r \dfrac{\partial v}{\partial \theta}$ and $\dfrac{1}r \dfrac{\partial u}{\partial \theta} = −\dfrac{\partial v}{\partial r}$.

Use these equations to show that the logarithm function defined by $\log z = \log r + i\theta$ where $z=re^{i\theta}$ with $-\pi<\theta<\pi$ is holomorphic in the region $r > 0$ and $-\pi<\theta<\pi$.

What I have so far:

Cauchy-Riemann Equations: Let $f(z)$ = $u(x, y)$ +$iv(x, y)$ be a function on an open domain with continuous partial derivatives in the underlying real variables. Then f is differentiable at $z = x+iy$ if and only if $\frac{∂u}{∂ x}(x, y)$ = $\frac{∂ v}{∂ y}(x, y)$ and $\frac{∂u}{∂ y}(x, y)$ = −$\frac{∂ v}{∂ x}(x, y)$. So we have $f'(z)= \frac{∂u}{∂ x}(z) +i \frac{∂ v}{∂ x}(z)$. Let $f(z)$ = $f(re^{iθ})$= $u(r,θ)$ +$iv(r,θ)$ be a function on an open domain that does not contain zero and with continuous partial derivatives in the underlying real variables. Then f is differentiable at $z$ = $re^{iθ}$ if and only if $r \frac{∂u}{∂r}=\frac{∂ v}{∂θ}$ and $\frac{∂u}{∂θ}$ = $−r \frac{∂v}{∂ r}$.

Sorry, if this is not very good. I just decided to start learning complex analysis today...

Answer 1

Proof of Polar C.R Let $f=u+iv$ be analytic, then the usual Cauchy-Riemann equations are satisfied \begin{equation} \frac{\partial u}{\partial x} =\frac{\partial v}{\partial y} \ \ \ \ \ \text{and} \ \ \ \ \frac{\partial u}{\partial y} =-\frac{\partial v}{\partial x} \ \ \ \ \ \ \ (C.R.E) \end{equation} Since $z=x+iy=r(\cos\theta + i \sin\theta)$, then $x(r, \theta)=r\cos\theta$ and $y(r,\theta)=r\sin\theta$. By the chain rule: \begin{align*} \frac{\partial u}{\partial r} & = \frac{\partial u}{\partial x} \cos\theta+ \frac{\partial u}{\partial y} \sin\theta \\ & \overset{(C.R.E)}{=} \frac{1}{r} \left( \frac{\partial v}{\partial y} r\cos\theta - \frac{\partial v}{\partial x} r\sin\theta\right) =\frac{1}{r} \left( \frac{\partial v}{\partial \theta}\right) \end{align*} and again, by the chain rule: \begin{align*} \frac{\partial v}{\partial r} & = \frac{\partial v}{\partial x} \cos\theta+ \frac{\partial v}{\partial y} \sin\theta \\ & \overset{(C.R.E)}{=} \frac{-1}{r} \left( \frac{\partial u}{\partial y} r\cos\theta - \frac{\partial u}{\partial x} r\sin\theta\right) =\frac{-1}{r} \left( \frac{\partial u}{\partial \theta}\right) \end{align*} So indeed $$ \left( \frac{\partial u}{\partial r}\right) = \frac{1}{r} \left( \frac{\partial v}{\partial \theta}\right) \ \ \ \ \ \text{and} \ \ \ \ \left(\frac{\partial v}{\partial r} \right) = \frac{-1}{r} \left( \frac{\partial u}{\partial \theta}\right) \ \ \ \ \ \ \ \ \ \blacksquare $$

Logarithm Example $log(z)=\ln(r)+i \theta$ with $z=re^{i\theta}$, $r>0$ and $-\pi<\theta<\pi$. Then $$ u(r, \theta)=\ln(r) \ \ \ \text{ and } \ \ \ v(r, \theta) =\theta $$ and $$ \left( \frac{\partial u}{\partial r}\right) =\frac{1}{r}= \frac{1}{r} \cdot 1 = \frac{1}{r} \cdot \left( \frac{\partial v}{\partial \theta}\right) \ \ \ \ \ \text{and } \ \ \ \ \left(\frac{\partial v}{\partial r} \right) = 0 = \frac{-1}{r}\cdot 0 = \frac{-1}{r} \left( \frac{\partial u}{\partial \theta}\right) $$ So indeed, $log(z)$ is analytic.

Answer 2

One way to derive CR equations in polar form is to find $u_r$, $u_{\theta}$, $v_r$, $v_{\theta}$ in terms of $u_x$, $u_y$, $v_x$, $v_y$ and $\sin{\theta}$, $\cos{\theta}$, $r$. Then plug in this information in the polar form of equations and verify that $LHS = RHS$ (by using the cartesian form of equations).

Another way is to find $u_x$, $u_y$, $v_x$, $v_y$ in terms of $u_r$, $u_{\theta}$, $v_r$, $v_{\theta}$ and $\sin{\theta}$, $\cos{\theta}$, $r$. Then plug in this information in the cartesian form of equations and derive the polar form through algebraic manipulations. This goes as follows:

Let $f(z) = U(x,y) + iV(x,y) = u(r,\theta) + iv(r,\theta)$. (Equating real and imaginary, we have $U(x,y) = u(r,\theta)$ and $V(x,y) = v(r,\theta)$. Also, $r = \sqrt{x^2 + y^2}$ and $\theta = \arctan{\frac{y}{x}}$).

$f$ is analytic at $z$ iff $U_x = V_y$ and $U_y = -V_x$.

Using chain rule,

\begin{eqnarray} U_x &=& u_r \cdot r_x + u_{\theta} \cdot \theta_x % &=& u_r \cos{\theta} - u_{\theta} \frac{\sin{\theta}}{r} \tag{1} \\ U_y &=& u_r \cdot r_y + u_{\theta} \cdot \theta_y % &=& u_r \sin{\theta} + u_{\theta} \frac{\cos{\theta}}{r} \tag{2} \\ V_x &=& v_r \cdot r_x + v_{\theta} \cdot \theta_x % &=& v_r \cos{\theta} - v_{\theta} \frac{\sin{\theta}}{r} \tag{3} \\ V_y &=& v_r \cdot r_y + v_{\theta} \cdot \theta_y % &=& v_r \sin{\theta} + v_{\theta} \frac{\cos{\theta}}{r} \tag{4} \end{eqnarray}

Now $U_x = V_y$ gives

\begin{eqnarray} u_r \cos{\theta} - u_{\theta} \frac{\sin{\theta}}{r} &=& % v_r \sin{\theta} + v_{\theta} \frac{\cos{\theta}}{r} \label{a} \tag{5} \end{eqnarray}

And $U_y = -V_x$ gives

\begin{eqnarray} u_r \sin{\theta} + u_{\theta} \frac{\cos{\theta}}{r} &=& % - v_r \cos{\theta} + v_{\theta} \frac{\sin{\theta}}{r} \label{b} \tag{6} \end{eqnarray}

Thus, $\cos{\theta} \cdot (\ref{a}) + \sin{\theta} \cdot (\ref{b})$ gives $u_r = \frac{1}{r} v_{\theta}$, and $\sin{\theta} \cdot (\ref{a}) - \cos{\theta} \cdot (\ref{b})$ gives $- \frac{1}{r} u_{\theta} = v_r$.