Question

Given the complex number, find a cube root of the number. It should be exact and in the form z^1/3 = r(cos(theta) + i sin(theta))

z = 4 + i4*sqrt(3)...

HELP!!!

Answer 1

\[z=4+i4 \sqrt{3}\]

Answer 2

r = the distance form 0+0i to 4+4sqrt(3) i

Answer 3

Need in \[z ^{1/3}=r*\cos(\theta)+i \sin(\theta)\]

Answer 4

Okay...

Answer 5

\[z^n=r^n(cos(nx)+isin(nx))\]

Answer 6

r = sqrt(4^2+(4sqrt(3))^2)

Answer 7

lol or,given z=a+bi\[r=\sqrt{a^2+b^2}\]

Answer 8

why replace z with r?

Answer 9

becuase z is a complex number; r is the modulus of z which is analologulous to the distance from the origin of (a,b)

Answer 10

analagus?

Answer 11

anomoluos?

Answer 12

you are above my knowledge now. i am just taking trig.

Answer 13

this is trig :)

Answer 14

we haven't covered modulus.

Answer 15

i know what it means.

Answer 16

it is remainder.

Answer 17

but i am trying to rework trig now that i am in group theory, but we have done nothing with complex.

Answer 18

so what is z^1/3???

Answer 19

If you recognize the factor of four times the one and square root of three in the coefficients, you can divide by eight to get\[z = 4 + i4\sqrt(3)=8(1/2+i \sqrt{3}/2)=8(\cos(\pi/3)+isin(\pi/3))\]

Answer 20

is that r^1/3

Answer 21

z^(1/3) = r^(1/3) * (cos(x/3) + i sin(x/3))

Answer 22

i dont think pi enters into the picture there

Answer 23

\[z=a+bi\]
\[a=rcos(x);\ bi=irsin(x)\]

Answer 24

\[z=rcos(x)+irsin(x)\]
\[z=r(cos(x)+isin(x))\]
\[z^{1/3}=(r(cos(x)+isin(x)))^{1/3} \]
\[z^{1/3}=r^{1/3}\ (cos(x)+isin(x)))^{1/3} \]

Answer 25

From that, we can get \[z ^{1/3}=2(\cos \theta+isin \theta)\]\[\theta=\pi/9\pm2\pi/3\]

Answer 26

oh yeah, we got an actual angle dont we :)

Answer 27

okay. i am just SO confused with this trig. I am retaking all the stuff from the class 2 years ago and trying to make sense of it. Thanks!

Answer 28

good luck with it :)

Answer 29

Thanks!