anonymous
  • anonymous
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!
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@jim_thompson5910 @ganeshie8 @Nnesha @wio @abb0t @zepdrix @Whitemonsterbunny17 @mathmate @jagr2713 @iki @Mehek14 @ikram002p
anonymous
  • anonymous
@ikram002p
anonymous
  • anonymous
multiply \(72\) by \(5\)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
so if I do that and simplify it is right?
anonymous
  • anonymous
you also have to take \(\left(\frac{1}{2}\right)^5\)
anonymous
  • anonymous
if by "simplify" you mean "evaluate the functions" then yes
anonymous
  • anonymous
yup, that is what I meant
anonymous
  • anonymous
@satellite73 Can you help me find the complex fifth roots of 5-5sqrt(3)*i?
anonymous
  • anonymous
yeah first write \[5-5\sqrt{3}i\] in trig form do you now how to do that?
anonymous
  • anonymous
r=sqrt((5)^2+(-5sqrt(3)^2)) for r
anonymous
  • anonymous
and -5sqrt(3)/5=tan(theta) for theta
anonymous
  • anonymous
and then I make a rcis(theta)?
anonymous
  • anonymous
@satellite73
anonymous
  • anonymous
yeah \(r=\sqrt{a^2+b^2}\)
anonymous
  • anonymous
from there what do I do?
anonymous
  • anonymous
did you find \(\theta\)?
anonymous
  • anonymous
theta=-pi/3 Sorry for being a little late. My browser froze :(
anonymous
  • anonymous
r=10
anonymous
  • anonymous
yeah looks like you got it
anonymous
  • anonymous
From there how do I fin the complex fifth roots?
anonymous
  • anonymous
find*
anonymous
  • anonymous
divide the angle by 5
anonymous
  • anonymous
-pi/15
anonymous
  • anonymous
yeah
anonymous
  • anonymous
do I get the 5th root of r?
anonymous
  • anonymous
just say it \[\sqrt[5]{10}\]
anonymous
  • anonymous
you can't really evaluate any of this, just write it in trig form
anonymous
  • anonymous
Is there more to it?
anonymous
  • anonymous
It asks me for the complex fifth roots
anonymous
  • anonymous
there are five of them
anonymous
  • anonymous
go around the circle again, then repeat
anonymous
  • anonymous
so I add 2pik to the angle?
anonymous
  • anonymous
right, the original angle then divide by 5 again
anonymous
  • anonymous
or else you can divide the circle in to 5 equal parts, with \[-\frac{\pi}{15}\]as one of them
anonymous
  • anonymous
each time I add +2pi, should I divide r by 5, or is that a one time thing?
anonymous
  • anonymous
add \(2\pi\) to \(-\frac{\pi}{3}\) then divide that one by \(5\)
anonymous
  • anonymous
then repeat
anonymous
  • anonymous
oh, does r change at all?
anonymous
  • anonymous
no it is going to be \(\sqrt[5]{10}\) each time
anonymous
  • anonymous
okay
anonymous
  • anonymous
oh wait, i think your \(r\) is wrong, check it again
anonymous
  • anonymous
So would the first one be \[\sqrt[5]{10}(cis(\pi/3))?\]
anonymous
  • anonymous
oh, okay
anonymous
  • anonymous
Are you sure, I got 10 again
anonymous
  • anonymous
\[\sqrt{5^2+(5\sqrt{3})^2}\]
anonymous
  • anonymous
\[\sqrt{25+75}=10\]
anonymous
  • anonymous
you are right sorry
anonymous
  • anonymous
it's okay, you're the one helping me :) Would the first answer be \[\sqrt[5]{10}(cis(\pi/3))\]
anonymous
  • anonymous
?
anonymous
  • anonymous
@satellite73
anonymous
  • anonymous
it's fine he is still viewing the chat so just give him a minute or to
anonymous
  • anonymous
oh sorry was away for a minute
anonymous
  • anonymous
kool
anonymous
  • anonymous
it's okay
anonymous
  • anonymous
ok you found \(r=10,\theta =-\frac{\pi}{3}\)
anonymous
  • anonymous
Is the thing I wrote earlier correct?
anonymous
  • anonymous
so \[5-5\sqrt{3}i=10\left(\cos(-\frac{\pi}{3})+i\sin(-\frac{\pi}{3})\right)\]
anonymous
  • anonymous
you want the five fifth roots
anonymous
  • anonymous
\[\sqrt[5]{10}(cis(\pi/3))?\]
anonymous
  • anonymous
one is \[\sqrt[5]{10}\left(\cos(-\frac{\pi}{15})+i\sin(-\frac{\pi}{15})\right)\]
anonymous
  • anonymous
then add \(2\pi\) to \(-\frac{\pi}{3}\) to get \(\frac{5\pi}{3}\)
anonymous
  • anonymous
oh yeah , what you said
anonymous
  • anonymous
divide by 5 and the next one is \[\sqrt[5]{10}\left(\cos(\frac{\pi}{3})+i\sin(\frac{\pi}{3})\right)\]
anonymous
  • anonymous
without adding 2pi?
anonymous
  • anonymous
not sure what you mean
anonymous
  • anonymous
you divided without adding 2pik where k=1
anonymous
  • anonymous
to theta
anonymous
  • 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
  • anonymous
So I can +4pi and divide by 5 for another one?
anonymous
  • 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
  • anonymous
yeah
anonymous
  • anonymous
or add \(2\pi\) to \(\frac{5\pi}{3}\) same thing
anonymous
  • 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
  • anonymous
oh kk
anonymous
  • anonymous
and 2pi+11pi/15
anonymous
  • anonymous
lather, rinse, repeat
anonymous
  • anonymous
oh no careful!
anonymous
  • anonymous
don't add \(2\pi\) to \(\frac{11\pi}{15}\) add it to \(\frac{11\pi}{3}\)
anonymous
  • anonymous
oh right, my bad
anonymous
  • 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
  • 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
  • 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
  • anonymous
yup, so I just continue doing what we were just doing?
anonymous
  • anonymous
yes
anonymous
  • anonymous
So the theta of the roots would be -pi/15, pi/3, 11pi/15, 17pi/15, and 23pi/15?
anonymous
  • anonymous
@satellite73

Looking for something else?

Not the answer you are looking for? Search for more explanations.