0

Multiplying and dividing by $-i^3$

$$\frac {1}{-i^3} \sum \left[e^{i \frac{2k\pi}{11}}\right]$$

It can be seen as $w+w^2+w^3....w^{10}$, where $w$ is a complex root of the complex equation. Also $$1+w+w^2+w^3....w^{10}=0$$

Since sum is till 10, the final value would be -1

$$\frac{-1}{-i^3}=\frac{1}{-i}=\frac{i}{-i^2}=i$$

But the answer given is $-i$. What is going wrong?

Aditya
  • 6,191

3 Answers3

1

Your error seems to be that you are assuming

$$\sin\left(2\pi k\over11\right)+i\cos\left(2\pi k\over 11\right)={-1\over i^3}\left(\cos\left(2\pi k\over11\right)+i\sin\left(2\pi k\over11\right) \right)$$

You can see that this is wrong if you multiply both sides by $i^3$ and note that the real part on the left becomes $i^4\cos(2\pi k/11)=\cos(2\pi k/11)$, which is not equal to the real part on the right, $-\cos(2\pi k/11)$.

Rather than multiplying and dividing by a power of $i$, it's easier to simply factor the $i$ out front:

$$\begin{align} \sum_{k=1}^{10}\left(\sin\left(2\pi k\over11\right)+i\cos\left(2\pi k\over11\right)\right) &=i\sum_{k=1}^{10}\left(\cos\left(2\pi k\over11\right)-i\sin\left(2\pi k\over11\right)\right)\\ &=i\sum_{k=1}^{10}e^{-2\pi ik/11}\\ &=i\left(\sum_{k=0}^{10}e^{-2\pi ik/11}-1 \right)\\ &=i(0-1)\\ &=-i \end{align}$$

Barry Cipra
  • 79,832
1

Let $a$ be $\exp(2\pi i/11)$. Then we have: $$ \begin{aligned} 0&=\frac {1-a^{11}}{1-a}=1+a+a^2+\dots+a^{10}\ ,\text{ so }\\ -1&=a+a^2+\dots+a^{10}\ . \end{aligned} $$ Now multiply with $i$.

dan_fulea
  • 32,856
  • Simply factoring put $i$ gives a negative sign to $\sin$ which is undesirable, even though it’s fixable. Using $-i_3$ gives a positive sign – Aditya Oct 29 '19 at 11:49
  • @Aditya: Your question does not define $w$, so i defined an $a$ that is either $w$ or $w^{-1}$. (The difference being the fact that we become the sum of the sines values with plus or with minus, in direct or reversed order, before or after we divide or multiply with the first or the third power of $i$. But the sine values do not contribute to the final sum, so let us look at the other part.) Now we go to the line: Since sum is till 10, the final value would be -1 as stated in the OP. It is unclear which sum is till $10$, so i suppose it is your $w+\dots+w^{10}$ or my $a+\dots +a^{10}$... – dan_fulea Oct 29 '19 at 12:37
  • That’s true. I got it thanks! – Aditya Oct 29 '19 at 12:39
  • @Aditya ... but it may be also the other more complicated sum from the OP, the one $$\frac {1}{-i^3} \sum \left[e^{i \frac{2k\pi}{11}}\right]$$ (we do not need the right brackets, and there is no need to place such a complicated factor in front of it.) Please always make clear which object is the subject in your sentences. It is else hard to understand your problem, your way to think, and your comments.... – dan_fulea Oct 29 '19 at 12:44
  • @Aditya So - for instance - it is hard to figure out your comment Simply factoring put i gives a negative sign to sin which is undesirable :: Simply factoring what, where, gives a negative sign for which sine value... (Note that i have to write an answer for the community. I was assuming the sign was lost when "moving" the term $1$ "to the other side" with a negative one, and/or when "multiplication with $i$" was done as "multiplication / division with $i^3$... The calculus was extracted in such a way, that operations and signs are transparent.) – dan_fulea Oct 29 '19 at 12:45
  • @Aditya As a final comment, please try to present (for the own safety) clear and clean simple arguments and computations, it is the way that worked for me. My comments are not offensive, i only try politely to show that the communication part in mathematics is more complicated than the mathematics itself - people just thing differently. My path is to join the flow, think as the other mathematicians - up to some research projects, where i need to go crazy... All the best! – dan_fulea Oct 29 '19 at 12:48
  • It was simple calculation problem, I misread the question. I Ought to delete my comment be because it’s no longer relevant. I was wrong, and my comment is wrong. My comment was trying to prove my wrong assumption. Regardless, I don’t see how my question was ‘hard’ to understand. I feel my language was concise enough to make my point. Although if it helps (both me and the community), I will bear your suggestions in my mind. – Aditya Oct 29 '19 at 12:49
0

Use Intuition behind euler's formula

$$\sum_{k=1}^{10}e^{i (\pi/2-2\pi k/11)}=-e^{i\pi/2}+e^{i\pi/2}\cdot\dfrac{e^{-2\pi i}-1}{\cdots}=?$$

lab bhattacharjee
  • 274,582