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Question Number 184731 by Mastermind last updated on 11/Jan/23

Express this function in both its Cartesian and polar form f(z) = ze^(iz) . Help!

$$\mathrm{Express}\:\mathrm{this}\:\mathrm{function}\:\mathrm{in}\:\mathrm{both}\:\mathrm{its} \\ $$$$\mathrm{Cartesian}\:\mathrm{and}\:\mathrm{polar}\:\mathrm{form} \\ $$$$\mathrm{f}\left(\mathrm{z}\right)\:=\:\mathrm{ze}^{\mathrm{iz}} . \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$

Answered by MJS_new last updated on 11/Jan/23

1^(st) possibility z=a+bi=(√(a^2 +b^2 ))e^(i arctan (b/a)) ⇒ f(z)=(√(a^2 +b^2 ))e^(i arctan (b/a)) e^(−b+ai) = =((√(a^2 +b^2 ))/e^b )e^(i(a+arctan (b/a))) = =((√(a^2 +b^2 ))/e^b )(cos (a+arctan (b/a)) +i sin (a+arctan (b/a))) 2^(nd) possibility z=re^(iθ) =rcos θ +irsin θ ⇒ f(z)=re^(iθ) e^(−rsin θ +ircos θ) = =(r/e^(rsin θ) )e^(i(θ+rcos θ)) = =(r/e^(rsin θ) )(cos (θ+rcos θ) +i sin (θ+rcos θ))

$$\mathrm{1}^{\mathrm{st}} \:\mathrm{possibility} \\ $$$${z}={a}+{b}\mathrm{i}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\mathrm{e}^{\mathrm{i}\:\mathrm{arctan}\:\frac{{b}}{{a}}} \\ $$$$\Rightarrow \\ $$$${f}\left({z}\right)=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\mathrm{e}^{\mathrm{i}\:\mathrm{arctan}\:\frac{{b}}{{a}}} \mathrm{e}^{−{b}+{a}\mathrm{i}} = \\ $$$$=\frac{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{\mathrm{e}^{{b}} }\mathrm{e}^{\mathrm{i}\left({a}+\mathrm{arctan}\:\frac{{b}}{{a}}\right)} = \\ $$$$=\frac{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{\mathrm{e}^{{b}} }\left(\mathrm{cos}\:\left({a}+\mathrm{arctan}\:\frac{{b}}{{a}}\right)\:+\mathrm{i}\:\mathrm{sin}\:\left({a}+\mathrm{arctan}\:\frac{{b}}{{a}}\right)\right) \\ $$$$ \\ $$$$\mathrm{2}^{\mathrm{nd}} \:\mathrm{possibility} \\ $$$${z}={r}\mathrm{e}^{\mathrm{i}\theta} ={r}\mathrm{cos}\:\theta\:+\mathrm{i}{r}\mathrm{sin}\:\theta \\ $$$$\Rightarrow \\ $$$${f}\left({z}\right)={r}\mathrm{e}^{\mathrm{i}\theta} \mathrm{e}^{−{r}\mathrm{sin}\:\theta\:+\mathrm{i}{r}\mathrm{cos}\:\theta} = \\ $$$$=\frac{{r}}{\mathrm{e}^{{r}\mathrm{sin}\:\theta} }\mathrm{e}^{\mathrm{i}\left(\theta+{r}\mathrm{cos}\:\theta\right)} = \\ $$$$=\frac{{r}}{\mathrm{e}^{{r}\mathrm{sin}\:\theta} }\left(\mathrm{cos}\:\left(\theta+{r}\mathrm{cos}\:\theta\right)\:+\mathrm{i}\:\mathrm{sin}\:\left(\theta+{r}\mathrm{cos}\:\theta\right)\right) \\ $$

Commented by Mastermind last updated on 11/Jan/23

I′m just seeing your solution Ok now, Assuming we were given the value of θ. how we want to do ordinary θ ?

$$\mathrm{I}'\mathrm{m}\:\mathrm{just}\:\mathrm{seeing}\:\mathrm{your}\:\mathrm{solution} \\ $$$$\mathrm{Ok}\:\mathrm{now},\:\mathrm{Assuming}\:\mathrm{we}\:\mathrm{were}\:\mathrm{given}\: \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\theta. \\ $$$$\mathrm{how}\:\mathrm{we}\:\mathrm{want}\:\mathrm{to}\:\mathrm{do}\:\mathrm{ordinary}\:\theta\:? \\ $$

Commented by MJS_new last updated on 11/Jan/23

sorry but I do not understand: what do you mean with “ordinary θ”?

$$\mathrm{sorry}\:\mathrm{but}\:\mathrm{I}\:\mathrm{do}\:\mathrm{not}\:\mathrm{understand}:\:\mathrm{what}\:\mathrm{do}\:\mathrm{you} \\ $$$$\mathrm{mean}\:\mathrm{with}\:``\mathrm{ordinary}\:\theta''? \\ $$

Commented by MJS_new last updated on 19/Jan/23

well, simply insert the value of θ

$$\mathrm{well},\:\mathrm{simply}\:\mathrm{insert}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\theta \\ $$

Commented by Mastermind last updated on 19/Jan/23

I meant if the value of θ given, how are going to do it in the answer you got

$$\mathrm{I}\:\mathrm{meant}\:\mathrm{if}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\theta\:\mathrm{given},\:\mathrm{how}\:\mathrm{are} \\ $$$$\mathrm{going}\:\mathrm{to}\:\mathrm{do}\:\mathrm{it}\:\mathrm{in}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{you}\:\mathrm{got} \\ $$

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