Question

fifth roots

Answer 1

find the fifth roots of 243(cos300+isin300)

Answer 2

mmmmm :D
z=243(cos300+isin300)
so u have r = 243
and theta =300
remember for complex
if z=r (cos theta + i cos theta )
then
\(\large z^n=r^n (\cos n \theta + i \cos n \theta )\)

Answer 3

fifth root means n=1/5 :D

Answer 4

wait how did you get that?

Answer 5

:O
ok so its not in ur text book mm
forget it
do u know this one ?
z= r (cos theta + i cos theta ) =r e^i theta

Answer 6

my text book mitaswell be in french. i cant understand any of it

Answer 7

\(\large z = 243(\cos (300)+i\sin (300) ) \)

Answer 8

since 2pi is the period of both sin and cos functions :

\(\large z = 243(\cos (300 + 2k\pi)+i\sin (300+2k\pi ) ) \)

Answer 9

next use the euler definition : \(\large \cos \theta + i \sin \theta = e^{i\theta }\)

Answer 10

\(\large z = 243e^{i (300 + 2k \pi)} \)

Answer 11

now you're ready to take 5th root both sides

Answer 12

\(\large \left(z\right)^{\frac{1}{5}} = \left(243e^{i (300 + 2k \pi)}\right)^{\frac{1}{5}} \)

Answer 13

k = 0,1,2,3,4

Answer 14

why \( 2k\pi \) ?

Answer 15

for infinitly solutions ?

Answer 16

for getting all the 5 roots

Answer 17

\(\large \left(z\right)^{\frac{1}{5}} = \left(243e^{i (300 + 2k \pi)}\right)^{\frac{1}{5}} \)

\( 0\le k \le 4\)

Answer 18

my computer shut down and while i was gone you guys took my problem and exploded it....

Answer 19

haha got options ?

Answer 20

haha xD

Answer 21

nope its a short essay .-.

Answer 22

and nothing you wrote makes a bit of sense to me .-.

Answer 23

im sorry i gtg now , cant continue to explain :D

Answer 24

i thought so

Answer 25

\(\large z = 243(\cos (300)+i\sin (300) )\)

since the period of both sin and cos is 2pi :
\(\large z = 243(\cos (300 + 2k\pi)+i\sin (300+2k\pi ) ) \)

Answer 26

fine with this step ?

Answer 27

sure .-.

Answer 28

good next take fifth root both sides

Answer 29

\(\large z^{\frac{1}{5}} = \left( 243(\cos (300 + 2k\pi)+i\sin (300+2k\pi ) )\right)^{\frac{1}{5}}\)

Answer 30

still fine ?

Answer 31

.-. yes

Answer 32

\(\large z^{\frac{1}{5}} = 243^{\frac{1}{5}}\left( \cos (300 + 2k\pi)+i\sin (300+2k\pi ) \right)^{\frac{1}{5}}\)

Answer 33

still wid me ?

Answer 34

sorry my computer keeps blue screening -.- and yes

Answer 35

@ganeshie8

Answer 36

its okay, next we use this formula
http://en.wikipedia.org/wiki/De_Moivre's_formula

Answer 37

\(\large z^{\frac{1}{5}} = 243^{\frac{1}{5}}\cos (\frac{300 + 2k\pi}{5})+i\sin (\frac{300+2k\pi}{5} ) \)

Answer 38

see if that looks okay..

Answer 39

it looks okay .-. i dont know what you mean

Answer 40

\(\large z^{\frac{1}{5}} = 3 \cos (\frac{300 + 2k\pi}{5})+i\sin (\frac{300+2k\pi}{5} )\)

Answer 41

plugin k = 0, you will get the first 5th root

Answer 42

first fifth root :

\(\large z^{\frac{1}{5}} = 3 \left( \cos (\frac{300 + 0 }{5})+i\sin (\frac{300 + 0}{5} )\right) \\ \large = 3\left( \cos (60) + i\sin (60)\right)\)

Answer 43

plugin k = 1,2,3,4 to get the remaining four fifth roots

Answer 44

so the main part that changes is 60.2, 60.4, 60.6, and 60.8?