complex number

Vanni

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a cubic root of (-1+i)4√2 is reia 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
 

Subhotosh Khan

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a cubic root of (-1+i)4√2 is reia 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
Is 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?
 

Dr.Peterson

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Nov 12, 2017
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a cubic root of (-1+i)4√2 is reia 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
Possibly you have forgotten that there are three cube roots, and they are asking about one of 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 the other two roots, one of those is in the list.
 

Vanni

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Possibly you have forgotten that there are three cube roots, and they are asking about one of 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 the other two roots, one of those is in the list.
the result of my calculations is r = 2 and a = π/4, , I don't understand where I went wrong
 

Vanni

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Is 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?
yes
\(\displaystyle \displaystyle{[(-1 + i) * 4 * \sqrt{2}]^{\frac{1}{3}} }\)
this is the one I used for my calculations
 

Dr.Peterson

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

Your answer is what I got first. But, as I said, that is only one of three cube roots, and they asked which of the options is a cube root, not which is the one you 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.
 

pka

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yes
\(\displaystyle \displaystyle{[(-1 + i) * 4 * \sqrt{2}]^{\frac{1}{3}} }\)
this is the one I used for my calculations
That number can be written as\(\displaystyle z= -\sqrt8+\sqrt8 i \). So the polar form is \(\displaystyle z=4\exp\left(\frac{-\pi i}{4}\right)\)
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 \(\displaystyle \eta\cdot \rho^k,~k=0,1,2\)
 

Vanni

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You didn't go wrong ... except in stopping here.

Your answer is what I got first. But, as I said, that is only one of three cube roots, and they asked which of the options is a cube root, not which is the one you 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.
So, r=2 and a=11π/12 is correct?
 

Subhotosh Khan

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yes
\(\displaystyle \displaystyle{[(-1 + i) * 4 * \sqrt{2}]^{\frac{1}{3}} }\)
this is the one I used for my calculations
\(\displaystyle \displaystyle{[(-1 + i) * 4 * \sqrt{2}]^{\frac{1}{3}} }\)

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

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

Dr.Peterson

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

.
 

pka

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8,313
That number can be written as\(\displaystyle z= -\sqrt8+\sqrt8 i \). So the polar form is \(\displaystyle z=4\exp\left(\frac{-\pi i}{4}\right)\)
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 \(\displaystyle \eta\cdot \rho^k,~k=0,1,2\)
EDIT
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 \(\displaystyle \eta\cdot \rho^k,~k=0,1,2\)
 
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