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Question Number 180159 by Mastermind last updated on 08/Nov/22

$$\mathrm{Express}\mathrm{the}\mathrm{function}\mathrm{f}\left(\mathrm{z}\right)={\mathrm{ze}}^{\mathrm{iz}}\mathrm{in}\mathrm{polar}$$$$\mathrm{form}\mathrm{and}\mathrm{separate}\mathrm{it}\mathrm{into}\mathrm{Real}\mathrm{and}$$$$\mathrm{Imaginary}\mathrm{part}.$$$$$$$$\mathrm{M}.\mathrm{m}$$

Answered by Frix last updated on 08/Nov/22

$$z\in \mathbb{C}$$$$z=a+b\mathrm{i}$$$$f\left(z\right)=(a+b\mathrm{i}){\mathrm{e}}^{-b+a\mathrm{i}}=(a+b\mathrm{i}){\mathrm{e}}^{-b}{\mathrm{e}}^{a\mathrm{i}}=\frac{a}{{\mathrm{e}}^b}{\mathrm{e}}^{a\mathrm{i}}+\frac{b\mathrm{i}}{{\mathrm{e}}^b}{\mathrm{e}}^{a\mathrm{i}}=$$$$=\frac{a}{{\mathrm{e}}^b}(\cos a+\mathrm{i}\sin a)+\frac{b\mathrm{i}}{{\mathrm{e}}^b}(\cos a+\mathrm{i}\sin a)=$$$$=\frac{a}{{\mathrm{e}}^b}(\cos a+\mathrm{i}\sin a)+\frac{b}{{\mathrm{e}}^b}(-\sin a+\mathrm{i}\cos a)=$$$$=\frac{a\cos a-b\sin a}{{\mathrm{e}}^b}+\frac{a\sin a+b\cos a}{{\mathrm{e}}^b}\mathrm{i}$$

Commented by Mastermind last updated on 09/Nov/22

$$\mathrm{Thank}\mathrm{you}\mathrm{man}$$

Commented by Mastermind last updated on 09/Nov/22

$$\mathrm{But}\mathrm{in}\mathrm{polar}\mathrm{form}$$

Commented by Frix last updated on 09/Nov/22

$$z{\mathrm{e}}^{\mathrm{i}z}\mathrm{is}\mathrm{the}\mathrm{polar}\mathrm{form}$$$$\mathrm{and}\mathrm{in}\mathrm{polar}\mathrm{form}\mathrm{you}\mathrm{cannot}\mathrm{seperate}\mathrm{the}$$$$\mathrm{real}\mathrm{and}\mathrm{imaginary}\mathrm{parts}$$

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