Question

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!

Answer 1

@jim_thompson5910 @ganeshie8 @Nnesha @wio @abb0t @zepdrix @Whitemonsterbunny17 @mathmate @jagr2713 @iki
@Mehek14 @ikram002p

Answer 2

@ikram002p

Answer 3

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

Answer 4

so if I do that and simplify it is right?

Answer 5

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

Answer 6

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

Answer 7

yup, that is what I meant

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

and then I make a rcis(theta)?

Answer 13

@satellite73

Answer 14

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

Answer 15

from there what do I do?

Answer 16

did you find \(\theta\)?

Answer 17

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

Answer 18

r=10

Answer 19

yeah looks like you got it

Answer 20

From there how do I fin the complex fifth roots?

Answer 21

find*

Answer 22

divide the angle by 5

Answer 23

-pi/15

Answer 24

yeah

Answer 25

do I get the 5th root of r?

Answer 26

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

Answer 27

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

Answer 28

Is there more to it?

Answer 29

It asks me for the complex fifth roots

Answer 30

there are five of them

Answer 31

go around the circle again, then repeat

Answer 32

so I add 2pik to the angle?

Answer 33

right, the original angle
then divide by 5 again

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

then repeat

Answer 38

oh, does r change at all?

Answer 39

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

Answer 40

okay

Answer 41

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

Answer 42

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

Answer 43

oh, okay

Answer 44

Are you sure, I got 10 again

Answer 45

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

Answer 46

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

Answer 47

you are right sorry

Answer 48

it's okay, you're the one helping me :)

Would the first answer be \[\sqrt[5]{10}(cis(\pi/3))\]

Answer 49

?

Answer 50

@satellite73

Answer 51

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

Answer 52

oh sorry was away for a minute

Answer 53

kool

Answer 54

it's okay

Answer 55

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

Answer 56

Is the thing I wrote earlier correct?

Answer 57

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

Answer 58

you want the five fifth roots

Answer 59

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

Answer 60

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

Answer 61

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

Answer 62

oh yeah , what you said

Answer 63

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

Answer 64

without adding 2pi?

Answer 65

not sure what you mean

Answer 66

you divided without adding 2pik where k=1

Answer 67

to theta

Answer 68

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

Answer 69

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

Answer 70

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 \(\frac{\pi}{3}\)

Answer 71

yeah

Answer 72

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

Answer 73

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

Answer 74

oh kk

Answer 75

and 2pi+11pi/15

Answer 76

lather, rinse, repeat

Answer 77

oh no careful!

Answer 78

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

Answer 79

oh right, my bad

Answer 80

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

Answer 81

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

Answer 82

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

Answer 83

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

Answer 84

yes

Answer 85

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

Answer 86

@satellite73