Cauchy-Riemann Equations in Polar Form

The Cauchy-Riemann equations in polar form are given as follows: ∂u/∂r = (1/r) ∂v/∂θ and ∂u/∂θ = – r ∂v/∂r if f(z)=u(r,θ)+iv(r,θ) is an analytic function. These C-R equations are very useful to test the analyticity of a complex function that can be expressed in polar co-ordinates easily. In this article, we learn the polar of the Cauchy-Riemann equations with some solved problems.

The Cauchy-Riemann equations are also knowns as C-R equations.

Statement of C-R Equations in Polar Form

Let f(z) = u(r,θ)+iv(r,θ) be a complex function expressed in polar co-ordinates (r, θ). The Cauchy-Riemann equation states that if f(z) is differentiable at z₀, then

$\boxed{\dfrac{\partial u}{\partial r}=\dfrac{1}{r} \dfrac{\partial v}{\partial \theta}, \quad \dfrac{\partial v}{\partial r} = -\dfrac{1}{r}\dfrac{\partial u}{\partial \theta}}$

holds at z₀. That is, the CR equations are given by

$u_r = \dfrac{v_\theta}{r}$ and $v_r = -\dfrac{u_\theta}{r}$.

Related Article: Cauchy-Riemann Equations: Statement, Proof, Questions

Solved Problems

Q1: Show that f(z)=zn satisfies the Cauchy-Riemann equations.

Solution:

If z=x+iy, then it is very difficult to find the real and the imaginary parts of f(z)=(x+iy)ⁿ. But, one can easily find the same if we express it in polar co-ordinates. We have:

In polar co-ordinates, z=re^iθ.

So, f(z)=zⁿ = (re^iθ)ⁿ

f(z) = rneinθ ⇒ f(z) = rn(cosnθ+i sinnθ) ⇒ f(z) = rncosnθ+i rnsinnθ

So the real part and the imaginary part of f(z)=u+iv are given by

u=rⁿcosnθ and v=rⁿsinnθ.

Now,

$\dfrac{\partial u}{\partial r}=nr^{n-1}\cos n\theta, \quad$ $\dfrac{\partial u}{\partial \theta}=-nr^n \sin n\theta$, $\dfrac{\partial v}{\partial r}=nr^{n-1}\sin n\theta, \quad$ $\dfrac{\partial v}{\partial \theta}=nr^n \cos n\theta$.

Thus, ∂u/∂r = (1/r) ∂v/∂θ and ∂u/∂θ = – r ∂v/∂r are satisfied. Therefore, it follows that the function f(z)=zⁿ satisfies the Cauchy-Riemann equations.

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