(a) Find the reciprocal of 
, working entirely in the Cartesian representation.
(b) Repeat part (a), working in polar form but expressing the final result in Cartesian form.
Solution.
(a)
(b)
Let
where 
and 
.
Then, the reciprocal of 
is
![\begin{aligned} \frac{1}{z} & = \frac{1}{re^{\mathrm{i}\varphi}} \\ & = \frac{1}{r}\, e^{\mathrm{i}(-\varphi )} \\ & = \frac{1}{r}\, \mathrm{cis}(-\varphi )\\ & = \frac{1}{r}\big(\cos (-\varphi)+\mathrm{i}\sin (-\varphi )\big)\\ & = \frac{1}{r}(\cos\varphi -\mathrm{i}\sin\varphi ) \\ & = \frac{1}{\sqrt{x^2+y^2}}\bigg[ \frac{x}{\sqrt{x^2+y^2}} - \mathrm{i}\Big( \frac{y}{\sqrt{x^2+y^2}} \Big) \bigg] \\ & = \bigg( \frac{x}{x^2+y^2} \bigg) - \mathrm{i}\bigg( \frac{y}{x^2+y^2} \bigg) \end{aligned}](https://physicspupil.com/wp-content/ql-cache/quicklatex.com-83d96b05396a5ba91612cfd121bbabad_l3.svg "Rendered by QuickLaTeX.com")
物理子衿
物理子衿
(a) Find the reciprocal of , working entirely in the Cartesian representation.
(b) Repeat part (a), working in polar form but expressing the final result in Cartesian form.
Solution.
(a)
(b)
Let
where and .
Then, the reciprocal of is