## owlet one year ago Question and my solution below. Which part did I do wrong?

1. owlet

2. owlet

This is a review question and my book only has the solution & answers to odd number questions and this is an even number question. My answer is not even one of the choices lol

3. hartnn

I see no error in your working...

4. owlet

are you sure? maybe the book is wrong. thanks for confirming it :) I'll probably ask my prof about that.

5. hartnn
6. IrishBoy123

you are starting with $\sqrt[3] { -3e^{\frac {\pi } {4} } }$??

7. owlet

why is it -3? r is equal to $$\sqrt{18}$$, so I have multiply it by the nth power which is 1/3 so it will be $$\sqrt[6]{18}$$ since 1/2 times 1/3 is 1/6. De moivres + euler's formula: $$\large z^n=r^ne^{in \theta}$$ sub the values in..it will be: $$\large z^{1/3}=\sqrt[6]{18}e^{i \frac{1}{3}( \frac{5 \pi}{4} + 2 \pi k)}$$

8. IrishBoy123

if we draw it first we have this |dw:1442695851885:dw| which kinda makes it easy as we have $$3 e^{i \frac{5 \pi}{4}}$$

9. owlet

ok. I got that.. but where did the "3" came from?

10. IrishBoy123

sorry we have $$\sqrt{18} e^{i \frac{5 \pi}{4}}$$ which gives $$\large \sqrt[3]{\sqrt{18} \, e^{i(\frac{5 \pi}{4} + 2n\pi)} }$$ $$\large = \sqrt[6]{18} \sqrt[3]{ \, e^{i(\frac{5 \pi}{4} + 2n\pi)} }$$ $$\large =\sqrt[6]{18} \ \, e^{i(\frac{5 \pi}{12} + \frac{2n\pi}{3})}$$ $$\large = \sqrt[6]{18} \ \, e^{i(\frac{5 \pi + 8n \pi}{12}) }$$ $$\large \implies \sqrt[6]{18} \ \, e^{i(\frac{5}{12}) }$$ $$\large \implies \sqrt[6]{18} \ \, e^{i(\frac{13}{12}) }$$ $$\large \implies \sqrt[6]{18} \ \, e^{i(\frac{21}{12}) }$$

11. owlet

we got the same! :D the method is the same, I just used exponents instead of radical. Thanks for clarifying it also. I already sent a message to my prof. I'm just waiting for his response.

12. IrishBoy123

good luck!