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Question Number 180295 by Mastermind last updated on 10/Nov/22

$$\mathrm{Express}\mathrm{these}\mathrm{both}\mathrm{Cartesian}\mathrm{and}$$$$\mathrm{polar}\mathrm{form}$$$$\left(1\right)\mathrm{f}\left(\mathrm{z}\right)={3\mathrm{z}}^2-2\mathrm{z}+\frac{1}{\mathrm{z}}$$$$\left(2\right)\mathrm{f}\left(\mathrm{z}\right)=\mathrm{z}+\frac{1}{\mathrm{z}}$$$$$$$$$$$$\mathrm{Thanks}$$

Commented by Mastermind last updated on 10/Nov/22

$$\mathrm{Express}\mathrm{these}\mathrm{both}\mathrm{in}\mathrm{Cartesian}(\mathrm{z}=\mathrm{x}+\mathrm{iy})$$$$\mathrm{and}\mathrm{polar}\mathrm{form}(\mathrm{z}={\mathrm{re}}^{\mathrm{i}\theta })$$$$\left(1\right)\mathrm{f}\left(\mathrm{z}\right)={3\mathrm{z}}^2-2\mathrm{z}+\frac{1}{\mathrm{z}}$$$$\left(2\right)\mathrm{f}\left(\mathrm{z}\right)=\mathrm{z}+\frac{1}{\mathrm{z}}$$$$$$$$$$$$\mathrm{Thanks}$$

Commented by Frix last updated on 10/Nov/22

$$z=x+\mathrm{i}y$$$$z=r{\mathrm{e}}^{\mathrm{i}\theta }=r\cos \theta +\mathrm{i}r\sin \theta$$$$\Rightarrow \mathrm{you}\mathrm{only}\mathrm{have}\mathrm{to}\mathrm{do}\mathrm{it}\mathrm{once}$$

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