Question

Find the cube roots of 27(cos 330° + i sin 330°).

Answer 1

can anyone explain this to me?

Answer 2

Use De moivres Theorem

Answer 3

i've seen it but i don't understand how to use it in this context

Answer 4

just get the cube root of 27 then the angles divide it by 3

Answer 5

3(cos 110 + i sin 110)
3(cos 230 + i sin 230)
3(cos 350 + i sin 350)
so would these be the right answer?

Answer 6

Note that we will be using De'Moivre's theorem which for any complex number \(\bf z=a+bi\) states:\[\bf z^n=r^n[\cos(n \theta)+isin( n \theta)]=r^n e^{i n \theta}\]Where \(\bf r=|z|\), i.e. \(\bf r\) is the modulus of \(\bf z\). And \(\bf \theta\) is the angle that the complex number \(\bf z\) makes with the x-axis, i.e. the angle that the point \(\bf (a,b)\) makes on the Cartesian plane with the x-axis.

Now we are already given that:\[\bf z^n=27[\cos(330)+isin(330)]\]However, we don't know what \(\bf n\) is. From what it looks like however, we can say that \(\bf n=3\) would work and so we can re-write as the following:\[\bf \implies z^3=3^3[\cos(3(110))+isin(3(110))]\]Taking the third-root of both sides implies that now we are basically finding \(\bf (z^3)^{1/3}=z\) which implies that now \(\bf n=1\). So where we see a \(\bf n=3\), we replace it with \(\bf n=1\):\[\bf \implies z^1=3^1[\cos(1(110)+isin(1(110))] \implies z = 3[\cos(110)+isin(110)]\]And we are done.

Answer 7

@rasoolni

Answer 8

Do you understand?

Answer 9

nice @genius12 it helps a lot

Answer 10

ok so I understand until the implication that n=1. I got 3(cos 110 + i sin 110) earlier, but are there other roots?

Answer 11

Nope. For De'Moivre's theorem to work, \(\bf n\) must be an integer. Here we can take the cube root of given complex number and obtain it using de'moivre's theorem because the the new 'n' is 1, which is an integer. If we took a square root instead of a cube-root let's say, then the new would be 3/2, which is not an integer. Hence if we want to use de'moivre's theorem, then the cube root is the only one we can get for this particular question.

@rasoolni

Answer 12

Do you get it now? @rasoolni

Answer 13

ok @genius12 I understand but i'm a little confused about writher or not there is more than one cube root of 27(cos 330° + i sin 330°)

Answer 14

@rasoolni There is only one cube-root and that is the one I stated already. Why would there be more?

Answer 15

@rasoolni

Answer 16

The theorem gives us a single cube root and that is and will be the only cube root. There won't be more. @rasoolni

Answer 17

ok thank you

Answer 18

@genius12..not sure but i think there should be more than one cube root..
for example,,there are 3 cube roots of 1..
and so if we have to find cube roots of z=1,,then there should be 3 cube roots and not one
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