Thanks for helping!
I need to simplify this equation to a+bi form.
(1/2(cos(72 degrees)+isin(72 degrees))^5
To solve I should do 1/2^5 and multiply 72 degrees by 5 right? Thanks for helping!

- anonymous

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- anonymous

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- anonymous

@ikram002p

- anonymous

multiply \(72\) by \(5\)

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## More answers

- anonymous

so if I do that and simplify it is right?

- anonymous

you also have to take \(\left(\frac{1}{2}\right)^5\)

- anonymous

if by "simplify" you mean "evaluate the functions" then yes

- anonymous

yup, that is what I meant

- anonymous

@satellite73 Can you help me find the complex fifth roots of 5-5sqrt(3)*i?

- anonymous

yeah first write
\[5-5\sqrt{3}i\] in trig form
do you now how to do that?

- anonymous

r=sqrt((5)^2+(-5sqrt(3)^2)) for r

- anonymous

and -5sqrt(3)/5=tan(theta) for theta

- anonymous

and then I make a rcis(theta)?

- anonymous

@satellite73

- anonymous

yeah \(r=\sqrt{a^2+b^2}\)

- anonymous

from there what do I do?

- anonymous

did you find \(\theta\)?

- anonymous

theta=-pi/3 Sorry for being a little late. My browser froze :(

- anonymous

r=10

- anonymous

yeah looks like you got it

- anonymous

From there how do I fin the complex fifth roots?

- anonymous

find*

- anonymous

divide the angle by 5

- anonymous

-pi/15

- anonymous

yeah

- anonymous

do I get the 5th root of r?

- anonymous

just say it
\[\sqrt[5]{10}\]

- anonymous

you can't really evaluate any of this, just write it in trig form

- anonymous

Is there more to it?

- anonymous

It asks me for the complex fifth roots

- anonymous

there are five of them

- anonymous

go around the circle again, then repeat

- anonymous

so I add 2pik to the angle?

- anonymous

right, the original angle
then divide by 5 again

- anonymous

or else you can divide the circle in to 5 equal parts, with
\[-\frac{\pi}{15}\]as one of them

- anonymous

each time I add +2pi, should I divide r by 5, or is that a one time thing?

- anonymous

add \(2\pi\) to \(-\frac{\pi}{3}\) then divide that one by \(5\)

- anonymous

then repeat

- anonymous

oh, does r change at all?

- anonymous

no it is going to be \(\sqrt[5]{10}\) each time

- anonymous

okay

- anonymous

oh wait, i think your \(r\) is wrong, check it again

- anonymous

So would the first one be \[\sqrt[5]{10}(cis(\pi/3))?\]

- anonymous

oh, okay

- anonymous

Are you sure, I got 10 again

- anonymous

\[\sqrt{5^2+(5\sqrt{3})^2}\]

- anonymous

\[\sqrt{25+75}=10\]

- anonymous

you are right sorry

- anonymous

it's okay, you're the one helping me :)
Would the first answer be \[\sqrt[5]{10}(cis(\pi/3))\]

- anonymous

?

- anonymous

@satellite73

- anonymous

it's fine he is still viewing the chat so just give him a minute or to

- anonymous

oh sorry was away for a minute

- anonymous

kool

- anonymous

it's okay

- anonymous

ok you found \(r=10,\theta =-\frac{\pi}{3}\)

- anonymous

Is the thing I wrote earlier correct?

- anonymous

so
\[5-5\sqrt{3}i=10\left(\cos(-\frac{\pi}{3})+i\sin(-\frac{\pi}{3})\right)\]

- anonymous

you want the five fifth roots

- anonymous

\[\sqrt[5]{10}(cis(\pi/3))?\]

- anonymous

one is
\[\sqrt[5]{10}\left(\cos(-\frac{\pi}{15})+i\sin(-\frac{\pi}{15})\right)\]

- anonymous

then add \(2\pi\) to \(-\frac{\pi}{3}\) to get \(\frac{5\pi}{3}\)

- anonymous

oh yeah , what you said

- anonymous

divide by 5 and the next one is
\[\sqrt[5]{10}\left(\cos(\frac{\pi}{3})+i\sin(\frac{\pi}{3})\right)\]

- anonymous

without adding 2pi?

- anonymous

not sure what you mean

- anonymous

you divided without adding 2pik where k=1

- anonymous

to theta

- anonymous

we now have two fifth roots (3 more to go) one is \[\sqrt[5]{10}\left(\cos(-\frac{\pi}{15})+i\sin(-\frac{\pi}{15})\right)\] and the other is \[\sqrt[5]{10}\left(\cos(\frac{\pi}{3})+i\sin(\frac{\pi}{3})\right)\]

- anonymous

So I can +4pi and divide by 5 for another one?

- anonymous

oh no we did add \(2\pi\) to the original angle of \(-\frac{\pi}{3}\) to get \(\frac{5\pi}{3}\) then we divided that one by 5 to get \(\frac{\pi}{3}\)

- anonymous

yeah

- anonymous

or add \(2\pi\) to \(\frac{5\pi}{3}\) same thing

- anonymous

when you add you get \(\frac{11\pi}{3}\) divide that by 5 and get \(\frac{11\pi}{15}\) for the third root

- anonymous

oh kk

- anonymous

and 2pi+11pi/15

- anonymous

lather, rinse, repeat

- anonymous

oh no careful!

- anonymous

don't add \(2\pi\) to \(\frac{11\pi}{15}\) add it to \(\frac{11\pi}{3}\)

- anonymous

oh right, my bad

- anonymous

hold on a second, it is not a mystery why you are doing this
both sine and cosine are periodic with period \(2\pi\)
that means the angle is NOT unique

- anonymous

so all the original numbers are the same
\[5-5\sqrt{3}i=10(\cos(-\frac{\pi}{3})+i\sin(-\frac{\pi}{3}))=10(\cos(\frac{5\pi}{3})+i\sin(\frac{5\pi}{3}))=...\] then are all equal

- anonymous

you are just expressing the original number in trig form with different values of \(\theta\)
then you divide each of them by 5 to get the different fifth roots

- anonymous

yup, so I just continue doing what we were just doing?

- anonymous

yes

- anonymous

So the theta of the roots would be -pi/15, pi/3, 11pi/15, 17pi/15, and 23pi/15?

- anonymous

@satellite73

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