$\begingroup$Write the following in polar form:$$\frac{1}{z}$$
$$\frac{1}{z}=\frac{1}{x+yi}$$
$$\frac{1}{x+yi}\cdot \frac{x-yi}{x-yi}=\frac{x-yi}{x^2+y^2}=\frac{x-yi}{r^2}$$
Because:
$$x-yi=rcis(-\frac{\pi}{4})$$
We get:
$$\frac{rcis(-\frac{\pi}{4})}{r^2}=\frac{cis(-\frac{\pi}{4})}{r}$$
Is the following correct?
$\endgroup$2 Answers
$\begingroup$It's unclear why $\textrm{cis}\left(-\frac{\pi}{4}\right)$ appears here: Since every nonzero complex number is invertible, this would imply that every complex number has argument $-\frac{\pi}{4}$, which isn't the case.
Hint If we write $z$ in polar form, as $r e^{i\theta}$, how can we write $\frac{1}{z} = \frac{1}{r e^{i \theta}}$ in polar form, i.e., as $\rho e^{i \alpha}$ for appropriate real $\rho, \alpha$?
$\endgroup$3$\begingroup$Hint:
write $\frac{1}{z}=z^{-1}$ for $z=\rho e^{i\theta}$ using the property of exponents $(x^n)^m=x^{nm}$.
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