- Thread starter Vanni
- Start date

- Joined
- Jun 18, 2007

- Messages
- 18,354

Is your function (whose cube-root you are seeking):a cubic root of (-1+i)4√2 is re^{ia}with

1) r=2, a=11π/12

2) r=2√2, a=19π/12

3) r=2√2, a=π/4

4) r=2, a=3π/4

I think they are all wrong

\(\displaystyle \displaystyle{(-1 + i) * 4 * \sqrt{2} }\)

i.e. you want to evaluate:

\(\displaystyle \displaystyle{[(-1 + i) * 4 * \sqrt{2}]^{\frac{1}{3}} }\)

or something else?

- Joined
- Nov 12, 2017

- Messages
- 3,858

Possibly you have forgotten that there area cubic root of (-1+i)4√2 is re^{ia}with

1) r=2, a=11π/12

2) r=2√2, a=19π/12

3) r=2√2, a=π/4

4) r=2, a=3π/4

I think they are all wrong

The first answer I come up with is similar to one of the choices, but not identical. However, if I look at the

the result of my calculations is r = 2 and a = π/4, , I don't understand where I went wrongPossibly you have forgotten that there arethreecube roots, and they are asking aboutoneof them, not all or the principal root.

The first answer I come up with is similar to one of the choices, but not identical. However, if I look at theothertwo roots, one of those is in the list.

yesIs your function (whose cube-root you are seeking):

\(\displaystyle \displaystyle{(-1 + i) * 4 * \sqrt{2} }\)

i.e. you want to evaluate:

\(\displaystyle \displaystyle{[(-1 + i) * 4 * \sqrt{2}]^{\frac{1}{3}} }\)

or something else?

\(\displaystyle \displaystyle{[(-1 + i) * 4 * \sqrt{2}]^{\frac{1}{3}} }\)

this is the one I used for my calculations

- Joined
- Nov 12, 2017

- Messages
- 3,858

You didn't go wrong ... except in stopping here.the result of my calculations isr = 2 and a = π/4, , I don't understand where I went wrong

Your answer is what I got first. But, as I said, that is only

Do you know how to find the other roots? You can multiply by each of the roots of unity; or you can just represent the radicand using an alternative angle (a coterminal angle).

Or, you can just cube each of the options and see which works. Note that only two of them have the correct modulus, so that's all you have to try.

- Joined
- Jan 29, 2005

- Messages
- 8,241

That number can be written as\(\displaystyle z= -\sqrt8+\sqrt8 i \). So the polar form is \(\displaystyle z=4\exp\left(\frac{-\pi i}{4}\right)\)yes

\(\displaystyle \displaystyle{[(-1 + i) * 4 * \sqrt{2}]^{\frac{1}{3}} }\)

this is the one I used for my calculations

Use \(\displaystyle \rho=\exp\left(\frac{2\pi i}{3}\right)\) and \(\displaystyle \eta=\sqrt[3]{4}\exp\left(\frac{-\pi i}{12}\right)\) now the three cube roots of \(\displaystyle z\) are \(\displaystyle \eta\cdot \rho^k,~k=0,1,2\)

So, r=2 and a=11π/12 is correct?You didn't go wrong ... except in stopping here.

Your answer is what I got first. But, as I said, that is onlyoneof three cube roots, and they asked which of the options isacube root, not which isthe oneyou find first.

Do you know how to find the other roots? You can multiply by each of the roots of unity; or you can just represent the radicand using an alternative angle (a coterminal angle).

Or, you can just cube each of the options and see which works. Note that only two of them have the correct modulus, so that's all you have to try.

- Joined
- Jun 18, 2007

- Messages
- 18,354

\(\displaystyle \displaystyle{[(-1 + i) * 4 * \sqrt{2}]^{\frac{1}{3}} }\)yes

\(\displaystyle \displaystyle{[(-1 + i) * 4 * \sqrt{2}]^{\frac{1}{3}} }\)

this is the one I used for my calculations

= \(\displaystyle \displaystyle{[(cos(3\frac{\pi}{4} \pm 2n\pi) + i sin(3\frac{\pi}{4} \pm 2n\pi) * 8]^{\frac{1}{3}} }\)

= \(\displaystyle \displaystyle{2 * e^{[(cos(\frac{\pi}{4} \pm 2n\pi/3) + i sin(\frac{\pi}{4} \pm 2n\pi/3)]} }\)

- Joined
- Nov 12, 2017

- Messages
- 3,858

Yes, a could be 3π/12, 11π/12, or 19π/12, corresponding to n=0, 1, 2 in \(\displaystyle {2 e^{\frac{\pi}{4} + \frac{2n\pi}{3}} }\).So, r=2 and a=11π/12 is correct?

.

- Joined
- Jan 29, 2005

- Messages
- 8,241

EDITThat number can be written as\(\displaystyle z= -\sqrt8+\sqrt8 i \). So the polar form is \(\displaystyle z=4\exp\left(\frac{-\pi i}{4}\right)\)

Use \(\displaystyle \rho=\exp\left(\frac{2\pi i}{3}\right)\) and \(\displaystyle \eta=\sqrt[3]{4}\exp\left(\frac{-\pi i}{12}\right)\) now the three cube roots of \(\displaystyle z\) are \(\displaystyle \eta\cdot \rho^k,~k=0,1,2\)

I missread it should be polar form is \(\displaystyle z=4\exp\left(\frac{3\pi i}{4}\right)\)

Use \(\displaystyle \rho=\exp\left(\frac{2\pi i}{3}\right)\) and \(\displaystyle \eta=\sqrt[3]{4}\exp\left(\frac{3\pi i}{12}\right)\) now the three cube roots of \(\displaystyle z\) are \(\displaystyle \eta\cdot \rho^k,~k=0,1,2\)