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Find the Cauchy-Riemann equations in polar form. In other words, if $f(re^{i\theta}) = u(r,\theta)+iv(r, \theta)$, then find the relations for the partial derivatives of $u$ and $v$ with respect to $f$ and $\theta$ if $f$ is complex differentiable.

Hint: the following version of the chain rule for partial derivatives may be useful: $$\frac{\partial u}{\partial r}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial r}$$

Further, Use the polar version of the Cauchy-Riemann equations to verify that $f(z)=z^n$ is analytic in $\mathbb{C}$ for any $n \in \mathbb{N}$, and that the principal logarithm is analytic in $\mathbb{C} \setminus \{ z \leq 0\}$.

I'm pretty unsure of where to start on this one. Thanks for your help!

Nosrati
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RhythmInk
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2 Answers2

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Hint: As $z = r e^{i\theta} = r \cos \theta + i \,r \sin \theta $. Since $x (r,\theta) = r \cos \theta$ and $y(r,\theta) = r \sin \theta$ we have (by the Chain Rule)

$$ u_r = u_x \cos \theta + u_y \sin \theta$$

and $$u_{\theta} = -u_x r\sin \theta + u_yr \cos \theta$$

Analogously you may find equations to $v_r$ and $v_{\theta}$.

Now, if $f$ is differentiable at $z$ then by CR equations you have

$$u_x = v_y \,\,\,\text{and} \,\,\,\ u_y= - v_x$$

Think you can take it from here?

Aaron Maroja
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C-R relations are

$$ u_x = v_y ; \, u_y = - v_x$$

In a manner of reasoning with partial differentials operation

$$ x \rightarrow r , y \rightarrow r d\theta $$

$$ u_r = \frac1r v_\theta ;\; \frac{u_\theta}{r} = - v_r.\; $$

Narasimham
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