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Find the value of $$\sum^{10}_{k=1}\left (\sin\left (\frac{2k\pi}{11} \right )+i\cos\left (\frac{2k\pi}{11}\right ) \right)$$

My approach:

Since $\cos\theta + i\sin\theta = e^{i\theta}$, we can write the given equation as:

$$\begin{align*} &i \left \{\sum^{10}_{k=1} \left (\cos\frac{2k\pi}{11} -i\sin\frac{2k\pi}{11} \right ) \right \}\\ = &i \left \{\sum^{10}_{k=1}\left (e^{-i\frac{2k\pi}{11}} \right ) \right \} \tag{i} \end{align*}$$

Solving the index part only which is $$\begin{align*} -i\frac{2k\pi}{11} &= -i\frac{2\pi}{11}(1+2+3+\cdots+10) \quad (\text{putting the values of } k)\\ &= -i\frac{2\pi}{11}( 55) \quad \left(\text{By applying sum of first $n$ natural numbers} = \frac{n(n+1)}{2}\right )\\ &=-i10\pi \end{align*}$$

Putting this value in $(\text{i})$ we get:

$e^{-i10\pi} = i\cos10\pi = i.$

But the answer is $-i$. Please suggest where I went wrong… Thanks..

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    $\begingroup$ Why are you adding the exponents? That would be fine if you were multiplying the terms, but its a summation. $\endgroup$ – Joseph G. Jun 6 '13 at 2:37
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    $\begingroup$ Note that $\sum_{k=0}^{10} (e^{-i\frac{2k\pi}{11}})$ is zero. Thus the sum without term $k=0$ should be $-1$. $\endgroup$ – hardmath Jun 6 '13 at 2:41
  • $\begingroup$ Use this formula $ \sum_{k=1}^{n}\rm e^{a k} ={\frac {{{\rm e}^{a ( n+1) }}-{{\rm e}^{a}}}{{{\rm e}^{a}}-1}}.$ $\endgroup$ – Mhenni Benghorbal Jun 6 '13 at 4:49
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I am not sure where your steps came from. You are right that you are summing

$$i \sum_{k=1}^{10} e^{-i 2 \pi k/11}$$

This is a geometric series and has value

$$i\frac{e^{-i 2 \pi/11}-e^{-i 2 \pi}}{1-e^{-i 2 \pi/11}} = i\frac{e^{-i 2 \pi/11}-1}{1-e^{-i 2 \pi/11}}= -i$$

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  • $\begingroup$ Geometric progression with a = $e^{\frac{-i2\pi}{11}}$ and common ratio = $e^{\frac{-i2\pi}{11}}$ Therefore, using sum of n terms of G.P $= \frac{a-r^{n}}{1-r} = [ \frac{e^{\frac{-i2\pi}{11}}\{ \frac{1-e^{-i2\pi}{11}}{1-e^{-i2\pi}{11}} = -i \frac{e^{\frac{-i2\pi}{11}}$ $\endgroup$ – sachin Jun 6 '13 at 3:06
  • $\begingroup$ sorry for previous comment, I jut want to ask sum of G.P = $\frac{a(1-r^n)}{1-r}$ where a is first term and r is common ratio, where both of these are same here, i.e. $e^{\frac{-i2\pi}{11}}$ $\endgroup$ – sachin Jun 6 '13 at 3:15
  • $\begingroup$ Ron Gordon Nice solution....... $\endgroup$ – juantheron Dec 15 '13 at 11:04
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You went wrong when you turned the sum into a product (which you'd have to do in order to sum the indices like that).

Instead, multiply (i) by $1=e^{i2\pi}$ and reindex to get $$i\sum_{k=1}^{10}e^{i2\pi\left(1-\frac{k}{11}\right)}=i\sum_{k=1}^{10}e^{\frac{i2\pi}{11}},$$ then substitute $\omega=e^{\frac{i2\pi}{11}}$ to get $$i\sum_{k=1}^{10}\omega^k.$$ All that's left is to use the fact that $$\sum_{k=0}^{10}\omega^k=0.$$ (Why is this true?)

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