Express-this-function-in-both-its-Cartesian-and-polar-form-f-z-ze-iz-Help-

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!

Answered by MJS_new last updated on 11/Jan/23

1^{st} possibility

z = a + b i = \sqrt{a^{2} + b^{2}} e^{i \arctan \frac{b}{a}}

\Rightarrow

f (z) = \sqrt{a^{2} + b^{2}} e^{i \arctan \frac{b}{a}} e^{- b + a i} =

= \frac{\sqrt{a^{2} + b^{2}}}{e^{b}} e^{i (a + \arctan \frac{b}{a})} =

= \frac{\sqrt{a^{2} + b^{2}}}{e^{b}} (\cos (a + \arctan \frac{b}{a}) + i \sin (a + \arctan \frac{b}{a}))

2^{nd} possibility

z = r e^{i θ} = r \cos θ + i r \sin θ

\Rightarrow

f (z) = r e^{i θ} e^{- r \sin θ + i r \cos θ} =

= \frac{r}{e^{r \sin θ}} e^{i (θ + r \cos θ)} =

= \frac{r}{e^{r \sin θ}} (\cos (θ + r \cos θ) + i \sin (θ + r \cos θ))

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 ?

Commented by MJS_new last updated on 11/Jan/23

sorry but I do not understand : what do you

mean with “ ordinary θ^{″} ?

Commented by MJS_new last updated on 19/Jan/23

well, simply insert the value of θ

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