(62) Utilize coordenadas polores no plano x=pi cos theta e y=11 operatorname(sen) theta , para obter as condicóes de Cauchy-Rermann ma forma polar.

Para obter as condições de Cauchy-Riemann na forma polar, começamos com as coordenadas polares dadas:<br /><br />\[ x = \pi \cos \theta \]<br />\[ y = 11 \sin \theta \]<br /><br />As condições de Cauchy-Riemann na forma cartesiana são:<br /><br />\[ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \]<br />\[ \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \]<br /><br />Vamos converter essas condições para a forma polar. Primeiro, expressamos \( u \) e \( v \) em termos de \( r \) e \( \theta \):<br /><br />\[ u = u(r, \theta) \]<br />\[ v = v(r, \theta) \]<br /><br />Usamos as relações entre as derivadas parciais nas coordenadas polares:<br /><br />\[ \frac{\partial}{\partial x} = \frac{\partial r}{\partial x} \frac{\partial}{\partial r} + \frac{\partial \theta}{\partial x} \frac{\partial}{\partial \theta} \]<br />\[ \frac{\partial}{\partial y} = \frac{\partial r}{\partial y} \frac{\partial}{\partial r} + \frac{\partial \theta}{\partial y} \frac{\partial}{\partial \theta} \]<br /><br />Sabemos que:<br /><br />\[ r = \sqrt{x^2 + y^2} \]<br />\[ \theta = \tan^{-1}\left(\frac{y}{x}\right) \]<br /><br />Calculamos as derivadas parciais:<br /><br />\[ \frac{\partial r}{\partial x} = \frac{x}{r}, \quad \frac{\partial r}{\partial y} = \frac{y}{r} \]<br />\[ \frac{\partial \theta}{\partial x} = -\frac{y}{r^2}, \quad \frac{\partial \theta}{\partial y} = \frac{x}{r^2} \]<br /><br />Substituímos essas derivadas nas condições de Cauchy-Riemann:<br /><br />\[ \frac{\partial u}{\partial x} = \frac{x}{r} \frac{\partial u}{\partial r} - \frac{y}{r^2} \frac{\partial u}{\partial \theta} \]<br />\[ \frac{\partial u}{\partial y} = \frac{y}{r} \frac{\partial u}{\partial r} + \frac{x}{r^2} \frac{\partial u}{\partial \theta} \]<br />\[ \frac{\partial v}{\partial x} = \frac{x}{r} \frac{\partial v}{\partial r} - \frac{y}{r^2} \frac{\partial v}{\partial \theta} \]<br />\[ \frac{\partial v}{\partial y} = \frac{y}{r} \frac{\partial v}{\partial r} + \frac{x}{r^2} \frac{\partial v}{\partial \theta} \]<br /><br />Aplicando as condições de Cauchy-Riemann:<br /><br />\[ \frac{x}{r} \frac{\partial u}{\partial r} - \frac{y}{r^2} \frac{\partial u}{\partial \theta} = \frac{y}{r} \frac{\partial v}{\partial r} + \frac{x}{r^2} \frac{\partial v}{\partial \theta} \]<br />\[ \frac{y}{r} \frac{\partial u}{\partial r} + \frac{x}{r^2} \frac{\partial u}{\partial \theta} = -\left( \frac{x}{r} \frac{\partial v}{\partial r} - \frac{y}{r^2} \frac{\partial v}{\partial \theta} \right) \]<br /><br />Simplificando, obtemos as condições de Cauchy-Riemann na forma polar:<br /><br />\[ \frac{\partial u}{\partial r} = \frac{1}{r} \frac{\partial v}{\partial \theta} \]<br />\[ \frac{\partial v}{\partial r} = -\frac{1}{r} \frac{\partial u}{\partial \theta} \]