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Question Number 180277 by Mastermind last updated on 09/Nov/22
Express the function f(z)=ze^(iz)  into  cartesian form and separate it into  Real and Imaginary part.    M.m
$$\mathrm{Express}\:\mathrm{the}\:\mathrm{function}\:\mathrm{f}\left(\mathrm{z}\right)=\mathrm{ze}^{\mathrm{iz}} \:\mathrm{into} \\ $$$$\mathrm{cartesian}\:\mathrm{form}\:\mathrm{and}\:\mathrm{separate}\:\mathrm{it}\:\mathrm{into} \\ $$$$\mathrm{Real}\:\mathrm{and}\:\mathrm{Imaginary}\:\mathrm{part}. \\ $$$$ \\ $$$$\mathrm{M}.\mathrm{m} \\ $$
Commented by Frix last updated on 09/Nov/22
That′s what I did before  It seems you don′t know much about this...
$$\mathrm{That}'\mathrm{s}\:\mathrm{what}\:\mathrm{I}\:\mathrm{did}\:\mathrm{before} \\ $$$$\mathrm{It}\:\mathrm{seems}\:\mathrm{you}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{much}\:\mathrm{about}\:\mathrm{this}… \\ $$
Commented by Mastermind last updated on 09/Nov/22
Smile  okay, lets see in polar form...  when z=re^(iθ)
$$\mathrm{Smile} \\ $$$$\mathrm{okay},\:\mathrm{lets}\:\mathrm{see}\:\mathrm{in}\:\mathrm{polar}\:\mathrm{form}… \\ $$$$\mathrm{when}\:\mathrm{z}=\mathrm{re}^{\mathrm{i}\theta} \\ $$
Commented by Frix last updated on 09/Nov/22
f(z)=ze^(iz) =re^(iθ) e^(ire^(iθ) ) =  =re^(iθ) e^(ir(cos θ +i sin θ)) =  =re^(iθ) e^(−rsin θ +ircos θ) =  =re^(−rsin θ +i(θ+rcos θ)) =  =(r/e^(rsin θ) )e^(i(θ+rcos θ)) =  =((rcos (θ+rcos θ))/e^(rsin θ) )+i((rsin (θ+rcos θ))/e^(rsin θ) )
$${f}\left({z}\right)={z}\mathrm{e}^{\mathrm{i}{z}} ={r}\mathrm{e}^{\mathrm{i}\theta} \mathrm{e}^{\mathrm{i}{r}\mathrm{e}^{\mathrm{i}\theta} } = \\ $$$$={r}\mathrm{e}^{\mathrm{i}\theta} \mathrm{e}^{\mathrm{i}{r}\left(\mathrm{cos}\:\theta\:+\mathrm{i}\:\mathrm{sin}\:\theta\right)} = \\ $$$$={r}\mathrm{e}^{\mathrm{i}\theta} \mathrm{e}^{−{r}\mathrm{sin}\:\theta\:+\mathrm{i}{r}\mathrm{cos}\:\theta} = \\ $$$$={r}\mathrm{e}^{−{r}\mathrm{sin}\:\theta\:+\mathrm{i}\left(\theta+{r}\mathrm{cos}\:\theta\right)} = \\ $$$$=\frac{{r}}{\mathrm{e}^{{r}\mathrm{sin}\:\theta} }\mathrm{e}^{\mathrm{i}\left(\theta+{r}\mathrm{cos}\:\theta\right)} = \\ $$$$=\frac{{r}\mathrm{cos}\:\left(\theta+{r}\mathrm{cos}\:\theta\right)}{\mathrm{e}^{{r}\mathrm{sin}\:\theta} }+\mathrm{i}\frac{{r}\mathrm{sin}\:\left(\theta+{r}\mathrm{cos}\:\theta\right)}{\mathrm{e}^{{r}\mathrm{sin}\:\theta} } \\ $$
Commented by Mastermind last updated on 09/Nov/22
Thanks man
$$\mathrm{Thanks}\:\mathrm{man} \\ $$

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