Question

How can I solve tough questions related to cube roots of 1 in complex numbers ?

Answer 1

Can you please elaborate the question a little?

Answer 2

In this question
(in the attachment)
I dont know where to begin

Answer 3

try writing it as a summation, see if it makes any more sense that way...

Answer 4

the asker isn't here

Answer 5

so you want to know how to solve it? ok...

Answer 6

yup

Answer 7

\[\sum_{n=1}^{\infty} (n+1)(n+\frac{1}{w})(n+\frac{1}{w^2})\]

Answer 8

I don't quite understand the part about w and w^2 both being non-real cube roots of 1...

Answer 9

the nonreal cube roots of 1 are cis(120) and cis(240)

Answer 10

(cis(120))^2 = cis(240) and (cis(240))^2 = cis(480) = cis(120)

Answer 11

so...
\[\sum_{n=1}^{\infty} (n+1)(n+\frac{1}{-.5+i(.5 \sqrt{3})})(n+\frac{1}{-.5-i(0.5 \sqrt{3})})\]
\[\sum_{n=1}^{\infty} (n+1)(n+\frac{1}{-.5+i(.5 \sqrt{3})})(n-\frac{1}{.5+i(0.5 \sqrt{3})})\]
simplify this down, then see what falls out?

Answer 12

really? do you really think it should go on that way?OMG

Answer 13

looks a bit messy, but unless there's a stunning flash of insight I haven't had, that looks like the only way out

Answer 14

then again, it might be easier to simplify if left in terms of w and w^2...

Answer 15

ah, yes, that's much better. I plugged (n+1)(n+1/w)(n+1/w^2) into wolfram alpha.
it gave me (1+n)(1+wn)(1+w^2n) divided by w^3. I did miss a stunning flash of insight. w^3=1

Answer 16

still messy, but at least no fractions

Answer 17

expand (1+n)(1+wn)(1+w^2n)...
n^3+(1+w+w^2)n^2+(1+w+w^2)n+1

Answer 18

@Loser66 can you see or do you know what 1+w+w^2 is equal to?

Answer 19

the sum of any number of points equally spaced around the unit circle is zero.

Answer 20

oops, -1+1= 0

Answer 21

:) we all make mistakes sometimes

Answer 22

I have been making one since the beginning of the problem in how I write the summations

Answer 23

so now we have \[\sum_{k=1}^{n} k^3+1=\sum_{k=1}^{n}k^3+\sum_{k=1}^{n}1\]

Answer 24

and from here you can see to the end

Answer 25

sorry, friend, above my head, I don't understand a word. hihihi...

Answer 26

what part don't you understand?

Answer 27

I did as wolframalpha does, make the same denominator, take w out of the sumation and don't know how to do next.

Answer 28

but I have no time to stay. I have class now. Thanks for spending time with me.

Answer 29

oh. there is a formula to find the sum of the first n cubes.
have fun in class

Answer 30

nice to see

Answer 31

the sum of the first n cubes is (n(n+1)/2)^2
∑k=1nk3 is essentially the sum of the first n cubes

Answer 32

@Tushi do you see what I did there?

Answer 33

Yes and the answer is (in the attachment)
Thanks for helping me.
You have come pretty close to the answer (just a +4n term is missing)
How can we get the answer ?

Answer 34

simplify (n(n+1)/2)^2 +n

Answer 35

do you see it?

Answer 36

Yes ... Thanks!!!
But the formula for sum of first n cube roots is (n(n+1)/2)^2, where did you get the extra n from ?

Answer 37

\[\sum_{k=1}^{n} k^3+1=\sum_{k=1}^{n}k^3+\sum_{k=1}^{n}1\]

Answer 38

\[\sum_{k=1}^{n}1\] equals n

Answer 39

Now I understand. Thanks a lot!

Answer 40

you're welcome
this was actually a fun problem...

Answer 41

I find these summation problems quite difficult and confusing ... Can you help me a bit? You seem to be a pro in summation.

Answer 42

sure

Answer 43

What are the basic formulas for summation? Our school textbooks assume we know all summation formulae without giving them anywhere in the book and use them frequently. I always get lost after the sigma sign.

Answer 44

see https://en.wikipedia.org/wiki/Summation#Some_summations_of_polynomial_expressions

Answer 45

Thank you... Do I have memorize all of them ?

Answer 46

you'll get it instinctively with time and practice

Answer 47

If you have to answer questions like this one without reference to books , it would appear so.

on the other hand, if you have access to google, then no.

Answer 48

do you want to practice a bit here?

Answer 49

That would be nice !! Where can I get the questions from ?

Answer 50

i'll make them up

Answer 51

\[\sum_{1}^{7}2n+1\]

Answer 52

Now i am lost

Answer 53

ok, first try to split it up into two sums of just one term

Answer 54

7 Summation 1 2n + 7 Summation 1 1
Is that correct ? Where is the summation button on the keyboard ?

Answer 55

oh, it's in the equation button below
vvv down there

Answer 56

yes, you split it up correctly

Answer 57

so now, do you know what to do with the sum from 1 to 7 of 1?

Answer 58

Do I add 1+1+1+1... 7 times to get 7?

Answer 59

YES!

Answer 60

now what do you do with the other one?

Answer 61

No idea :(

Answer 62

it's simple enough you could add it up manually
or you could move the 2 outside the summation
\[\sum_{1}^{7}2n=2\sum_{1}^{7}n\]

Answer 63

do you know the formula for the sum of the first n numbers?

Answer 64

n(n+1)/2

Answer 65

yes

Answer 66

do you see why you can move the constant term (2) outside the summation?

Answer 67

and can you find the final answer now?

Answer 68

How can we move 2 outside the summation? Dont we have to add it 7 times?

Answer 69

the sum from 1 to 7 of 2n is 2+4+6+8...14
we can factor this... 2(1+2+3...7)

Answer 70

we can take it out because it's always the same

Answer 71

sorry, it's my bedtime now. I live in China.

Answer 72

Oh sorry ... I forgot my basic algebra.
So we get the answer
2 x n(n+1)/2 = n(n+1)
We put n=1, add it to n=2, add it to n=3, ... add it to n=7.
Will we get the answer then?

Answer 73

when you replaced the summation with the (n(n+1))/2 you now only have to put in the highest n

Answer 74

so just put in n=7
and don't forget the 7 we had earlier

Answer 75

goodnight

Answer 76

And we get 7 x 8 = 56.
Thanks for helping me and goodnight!

Answer 77

you got this

Answer 78

(just don't forget the 7 we had earlier...)

Answer 79

What do we do with that 7? just put it for n, isn't it?