If z=a+bi then f(z)=f(a+bi)=$e^{a+bi}=e^{a}e^{bi}$. For each of the following complex number find...

by Nixie777   Last Updated January 16, 2018 13:20 PM

If z=a+bi then f(z)=f(a+bi)=$e^{a+bi}=e^{a}e^{bi}$. For each of the following complex number $w_{i}$ find a $z_{i}$ so that $f(z_{i})=w_{i}$. If no $z_{i}$ exists explain why not.

a) $w_{1}=e^{2}$

b) $w_{2}=0$

c) $w_{3}=-3i$

I don't need all of them to be computed. I would just like to see an example of one of these because I'm a little lost. I know $e^a$ is just the normal computation and for $e^{bi}$ we need to use Euler's formula, but I'm kinda confused with the Euler formula. Thank you for the help.



Answers 1


Let us take $$w_4=1-i = \sqrt{2}\left[\frac{1-i}{\sqrt 2}\right]$$ $$=\sqrt 2 e^{-i\frac{\pi}4} = e^{\color{red}{(\ln 2)/2 }}e^{i\color{green}{\left(-\frac{\pi}4\right)}}$$ $$=e^{\color{red} {a_4}}e^{i\color{green} {b_4}}$$

After EDIT:

Take for example $$w_3=-3i = 3e^{-\frac{\pi}2i}$$ $$=e^{\color{red}{\ln 3} }e^{\color{green}{-\frac{\pi}2} i}$$ $$=e^{\color{red}{a_3}}e^{i\color{green}{b_3}}$$

Rohan
Rohan
January 16, 2018 12:56 PM

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