Find Re(α+α^2+α^3+α^4+α^5): Solve Trig Series

[utkarshakash]

Homework Statement

If $\large α = e^{i\frac{8∏}{11}}$, then find $Re(α+α^{2}+α^{3}+α^{4}+α^{5})$

Homework Equations

The Attempt at a Solution

$\large e^{i\frac{8∏}{11}}+e^{i\frac{16∏}{11}}...+e^{i\frac{40∏}{11}}$

$cos \frac{8∏}{11}+isin \frac{8∏}{11}...$

Since I am interested only in real part so now I have to find the value of

$cosθ+cos2θ...cos5θ$

where $θ= \frac{8∏}{11}$

I think some trigonometry must be applied since it seems to me sum of a trigonometrical series.

 

[ehild]

Homework Statement

If $\large α = e^{i\frac{8∏}{11}}$, then find $Re(α+α^{2}+α^{3}+α^{4}+α^{5})$

Homework Equations

The Attempt at a Solution

$\large e^{i\frac{8∏}{11}}+e^{i\frac{16∏}{11}}...+e^{i\frac{40∏}{11}}$

$cos \frac{8∏}{11}+isin \frac{8∏}{11}...$

Since I am interested only in real part so now I have to find the value of

$cosθ+cos2θ...cos5θ$

where $θ= \frac{8∏}{11}$

I think some trigonometry must be applied since it seems to me sum of a trigonometrical series.

Is not

$α+α^{2}+α^{3}+α^{4}+α^{5}$

a geometric series?

hild

 

[Ray Vickson]

Homework Statement

If $\large α = e^{i\frac{8∏}{11}}$, then find $Re(α+α^{2}+α^{3}+α^{4}+α^{5})$

Homework Equations

The Attempt at a Solution

$\large e^{i\frac{8∏}{11}}+e^{i\frac{16∏}{11}}...+e^{i\frac{40∏}{11}}$

$cos \frac{8∏}{11}+isin \frac{8∏}{11}...$

Since I am interested only in real part so now I have to find the value of

$cosθ+cos2θ...cos5θ$

where $θ= \frac{8∏}{11}$

I think some trigonometry must be applied since it seems to me sum of a trigonometrical series.

You have found the real part; it is a sum of 5 terms. What is wrong with that answer?

RGV

 

[utkarshakash]

You have found the real part; it is a sum of 5 terms. What is wrong with that answer?

RGV

Hey I have found the answer but not completely. I have to find the value of cosθ+cos2θ...
which I don't know how to solve

 

[Mentallic]

Hey I have found the answer but not completely. I have to find the value of cosθ+cos2θ...
which I don't know how to solve

Well as people have already mentioned, the answer IS
$\cos(8\pi/11)+\cos(16\pi/11)+...+\cos(40\pi/11)$
but that's messy, and this question has been cleverly constructed so that there is a nice answer.

$$\alpha+\alpha^2+...+\alpha^5$$
$$=1+\alpha+\alpha^2+...+\alpha^5-1$$

$$=\frac{1-\alpha^6}{1-\alpha}-1$$

Now, notice that

$$\alpha^6=e^{48\pi i/11}=e^{4\pi i/11}=\left(e^{8\pi i/11}\right)^{1/2}=\alpha^{1/2}$$

So we can now turn the expression into

$$=\frac{1-\alpha^{1/2}}{1-\alpha}-1$$

$$=\frac{1-\alpha^{1/2}}{(1-\alpha^{1/2})(1+\alpha^{1/2})}-1$$

I'm sure you can finish it off from here