Question

find all the cube roots of -3+3i in polar form

Answer 1

polar form: |z| = sqrt((-3)^2+(3)^2))
the you find the argument \(\theta = y/x\)

Answer 2

\[\huge -3+3i=2\sqrt{3}cis\frac{3\pi}{4}\]

Answer 3

Polar form:
\( r (cos\theta + isin\theta)\)

Answer 4

i think he/she should solve it themselves :)

Answer 5

With this level of maths, don't think it matters.

Answer 6

People with this calibre of maths, should know that the answer isn't everything. They should know how they got that answer as the most important thing out of a question.

Answer 7

\[\huge \sqrt[3]{-3+3i}=(2\sqrt{3})^{\frac{1}{3}}cis\frac{3\pi}{4}\times \frac{1}{3}\]

Answer 8

azteck i got \[\sqrt{18}cis(3\Pi/4)\]

but then i get confused with the next steps. i agree with working it out it more important than the answer i can get the answer but i cant do the working properly

Answer 9

You know De Moivre's theorem correct?

Answer 10

You work backwards.

Answer 11

nope i dont. i have tried working it out but the notes i have dont show a good explanation of it

Answer 12

When you want to find the cube of something:
\[\huge (rcis\theta)^{3}=r^3cis3\theta\]

Answer 13

Using that example above, what would
\[\huge (rcis)^{\frac{1}{3}}=?\]
Do the exact same thing as what I did but with 1/3.

Answer 14

\[\huge rcis\theta\]

Answer 15

sorry about that, left the theta.

Answer 16

The angle is multiplied by the value of the exponent and r (the modulus) is raised by the same value of the exponent.

Answer 17

\[r1/3cis1/3\]

Answer 18

theta on the end

Answer 19

Correct. SO using what you got so far for your question:
\[\sqrt{18}cis\frac{3\pi}{4}\]
You know r is \[\sqrt{18}\] and theta is\[\frac{3\pi}{4}\]
What would your final result be now?

Answer 20

Because:
You're trying to find:
\[\huge (\sqrt{18}cis\frac{3\pi}{4})^{\frac{1}{3}}\]

Answer 21

\[\sqrt{18} cis (3\pi/4)?\]

Answer 22

Look at my previous post. You're trying to find that.

Answer 23

hangon..

Answer 24

You can draw it out if the equation tool is wasting your time.

Answer 25

ye.. i have no idea :/

Answer 26

THere's a draw tool beside the equation tool. It could save you a tonne of time.