Question

Hi, I am new to this website :) I'm currently taking an online trig class on De Moivre's theorem and I don't understand it at all! The question is: Write each expression in the standard form for a complex number, a+bi.
1. [3cos(27))+isin(27)]^5
2. [2(cos(40))+isin(40)]^6
any info regarding this question would be extremely helpful! Thanks!

Answer 1

De Moivre's theorem says that if you have

z = r*[ cos(theta) + i*sin(theta) ]

then

z^n = r^n*[ cos(n*theta) + i*sin(n*theta) ]

Answer 2

so that means

[3(cos(27)+isin(27)]^5

3^5*[cos(5*27)+isin(5*27)]

243*[cos(135)+isin(135)]

I'll let you finish up. Use the unit circle from here and simplify

Answer 3

Oh!! That makes so much more sense now, thank you! :)

Answer 4

you're welcome

Answer 5

I have one more question regarding the same there, how do I find the complex cube roots of a function?

Answer 6

Same *theorem*

Answer 7

what's the question

Answer 8

Find the complex cube roots of 8(cos(4pi/5) + isin(4pi/5))

Answer 9

well one cube root (there are 3 total) is found by taking the cube root of 8 to get 2

then you divide each argument by 3, so one cube root is

2*[cos(4pi/15) + i*sin(4pi/15)]

Answer 10

actually, i found a better formula

Answer 11

if

z = r*[ cos(theta) + i*sin(theta) ]

then

z^(1/n) = r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]


the notation z^(1/n) means the nth root of z

the term k is a constant that ranges from 0 to n

Answer 12

so for instance, in this case

r = 8
theta = 4pi/5
n = 3

then you'll go through 3 values of k: k = 0, k = 1, k = 2

sorry I meant to say that k ranges from 0 to n-1

Answer 13

oh and replace 360 with 2pi if you're using radians

Answer 14

Ok so the one part I don't get is how you got 3 exactly? Not to sound dumb lol I understood everything else you said

Answer 15

Oh it is because n is the root right! Sorry

Answer 16

yes exactly, cube roots will have n = 3

square roots will use n = 2

etc etc

Answer 17

Right, so I will have 3 different solutions all together right? For each root from 0-3?

Answer 18

0-2 actually, i made that mistake myself lol

Answer 19

so the first cube root will have you use

r = 8
theta = 4pi/5
n = 3
k = 0

the second cube root will use the same values but k will be different

r = 8
theta = 4pi/5
n = 3
k = 1

the last cube root will only change k as well

r = 8
theta = 4pi/5
n = 3
k = 2

Answer 20

And just to clarify that I am solving these correctly, should my answer be just one number or a function or something? Sorry for all these questions lol

Answer 21

its ok, questions are good

Answer 22

with cube roots, you'll have 3 answers total

Answer 23

it's analogous to saying that square roots produce two answers

Answer 24

ex:

if x^2 = 25 then x = -5 or x = 5

Answer 25

turns out that if x^3 = 216, then x = 6, x = -3-3i*sqrt(3), x = -3+3i*sqrt(3)

Answer 26

this is because the fundamental theorem of algebra states that

if f(x) is a polynomial function of degree n, then f(x) will have n complex roots