Question

How do you find three complex cubed roots of 6????

Answer 1

Well a complex number is "i" so we're doing with that right?

Answer 2

i believe so, i am just using this to study for my final and this was a question on the study guide that idk how to do

Answer 3

you could write 6 in polar form
\[ 6 = 6 e^{i 2 \pi} \]
now raise to the 1/3 power:
\[ 6^{\frac{1}{3} } e^{i \frac{2 \pi}{3} } \]
to get the other roots notice you could start with
\[ 6 = 6 e^{i 4 \pi} \]
and do the same thing
also
\[ 6 = 6 e^{i 6 \pi} \]

doing it again will start repeating the roots

Answer 4

See
http://tutorial.math.lamar.edu/Extras/ComplexPrimer/Roots.aspx
for more background

Answer 5

so would the roots be for example 6=6e^i6π or no? @phi

Answer 6

to the 1/3 power

Answer 7

the idea is to write 6 as
\[ 6 e^{i 2 \pi k} \] where k is an integer
we can use euler's formula to show this is
\[ 6( \cos (2 \pi k) + i \sin(2 \pi k)) \]
for k=0 this is
\[ 6( \cos (0) + i \sin(0)) = 6(1+0) = 6 \]
for k= 1 we still will get 6 (because 2 pi is the same angle as 0 )
same for k=2, and so on...

now raise 6 to the 1/3 power. you get
\[ 6^{\frac{1}{3}} e^{i \frac{2 \pi k}{3}} \]
for k=0,1,2 (or even negative integer k's)

Answer 8

you might think there are an infinite number of roots because k goes forever, but k=0 gives the same answer as k=3, and k=1 gives the same answer as k=4, and so on....
you get 3 distinct roots

Answer 9

we start with 6 written as
\[ \left( 6 e^{i 2 \pi k}\right) , \text{ k is an integer }\]
and raise it the to 1/3 power
\[ \left( 6 e^{i 2 \pi k}\right)^{\frac{1}{3}} \]
which can be written as
\[ 6^{\frac{1}{3}} e^{i \frac{2 \pi k}{3}} \]

Answer 10

you can write the roots in rectangular form
\[ 6^{\frac{1}{3}} \left(\cos(0) + i \sin(0)\right)\]
\[ 6^{\frac{1}{3}} \left(\cos(120º) + i \sin(120º)\right)\]
\[ 6^{\frac{1}{3}} \left(\cos(240º) + i \sin(240º)\right)\]
notice we add 120 degrees to get to the next root. also, the next root 240+120= 360 gets us back to 0 degrees

Answer 11

so for k=2 i would do 6(cos(2π(2))+isin(2π(2))) --> 6(cos(4π)+isin(4π)) -->
6((1) + i(-2*10^-13))
I feel like this is wrong.. @phi

Answer 12

it looks like you used a calculator that 10^-13 is a tiny number, and should be exactly 0
you just found that 6 can be written as
6*( cos (4 pi) + i sin( 4 pi) )
see
http://www.wolframalpha.com/input/?i=6*%28+cos+%284+pi%29+%2B+i+sin%28+4+pi%29+%29

to find the cube root (i.e. raised to the 1/3 power), you have to raise 6 to the 1/3 power

\[ \left( 6 e^{i 2 \pi k}\right)^{\frac{1}{3}} = 6^{\frac{1}{3}} e^{i \frac{2 \pi k}{3}} \]
what do you get when k= 0 ?

Answer 13

6^.333 (e^i(0/3)
6^.333 (e^i(0)
6^.333 (e^0)
6^.333 (1)
6^.333
1.817
@phi

Answer 14

yes,
now what do you get for k=1 ?

Answer 15

6^.333 (e^i(2pi(1)/3)
6^.333 (e^i(2pi/3)
6^.333 (e^i(2.09)
I am not positive what to do with the i(2.09) but on my calculator i did imaginary(2.09) = 0
6^.333 (e^0)
6^.333 (1)
6^.333
1.817
@phi

Answer 16

you have to use euler's formula
http://en.wikipedia.org/wiki/Euler's_formula

because your calculator does not do complex numbers (I don't think)

be sure to be in radian mode (or change the radians to degrees)

2 pi/3 in degrees is 2 pi /3 * 180/pi = 120 degrees

Answer 17

6^.333 (e^i(2pi(1)/3)
6^.333 (e^i(2pi/3)
cos^i(2pi/3)=cos(2pi/3)+isin(2pi/3)
cos^i(2pi/3)=-.5+i(.866)
6^.333 (e^cos^i(2pi/3)=-.5+i(.866))
am i doing this right? if so what next?

Answer 18

6^.333 (e^i(2pi(1)/3)
6^.333 (e^i(2pi/3)
you should say (not cos^i(2pi/3))
e^i(2pi/3)=cos(2pi/3)+isin(2pi/3)

that means you replace e^i(2pi/3) with the stuff on the right to get
6^.333 ( cos(2pi/3)+isin(2pi/3) )

you should memorize cos(120) is -1/2 and sin(120) is sqrt(3)/2
so you get (remember 6^(1/3) is about 1.817
\[ 1.817\left( - \frac{1}{2} + \frac{\sqrt{3}}{2} i \right) \]
or
\[ 0.909 ( -1 + \sqrt{3} i) \]
or (to be exact)
\[ \frac{\sqrt[3]{6}}{2}\left( -1 + \sqrt{3} \ i\right) \]

Answer 19

or roughly,
-0.909 + 1.574 i

Answer 20

thanks! and now k=2

1.87 (e^i(2π(2)/3))
1.87 (e^i(4π/3))
e^i(4π/3)=cos(4π/3+isin(4π/3)
e^i(4π/3)= -.5 + -sqrt3/2 i
1.87 (-.5 + -sqrt3/2 i)
0.909(−1-√3i)
3sqrt/2 (−1-√3i)

@phi

Answer 21

if in the last line 3 sqrt/2 is a typo for
\[ \frac{\sqrt[3]{6} }{2} \]
then yes, looks good.

notice that complex numbers are *complex conjugates*
(roots of real numbers always come in complex conjugate pairs)

Answer 22

yeah thats what i meant and thanks so mucH!