# Thread: Complex no.

1. ## Complex no.

If $\alpha = e^{i\frac{8\pi}{11}}$ . Then find $Re(\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5)=$

Where $Re =$ Real part of $\alpha$

2. Originally Posted by jacks
If $\alpha = e^{i\frac{8\pi}{11}}$ . Then find $Re(\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5)=$

Where $Re =$ Real part of $\alpha$
This seems pretty straight forward. What part are you having trouble with?
Evaluating the sum?
Finding the real part?

3. As you need the real part of these then start with the fact that $e^{i\theta} = \cos \theta + i\sin \theta$

4. Actually you should end with that fact. Do the exponentiation first while it's in exponential form.

5. Here I Have Calculate $Re(\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5)$ and Getting

$\Leftrightarrow cos(\frac{8\pi}{11})+cos(\frac{16\pi}{11})........ ............cos(\frac{40\pi}{11})$

Now How can I calculate that sum.....

6. You can't get exact values, so you can leave it like that.

7. Originally Posted by jacks
Here I Have Calculate $Re(\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5)$ and Getting

$\Leftrightarrow cos(\frac{8\pi}{11})+cos(\frac{16\pi}{11})........ ............cos(\frac{40\pi}{11})$

Now How can I calculate that sum.....
How did you get that?

Edit: I now see!

8. Originally Posted by jacks
Here I Have Calculate $Re(\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5)$ and Getting

$\Leftrightarrow cos(\frac{8\pi}{11})+cos(\frac{16\pi}{11})........ ............cos(\frac{40\pi}{11})$

Now How can I calculate that sum.....
Let us denote $\beta=8\pi/11$ , we can use the following method:

$\alpha+\alpha^2+\ldots+\alpha^5=\dfrac{\alpha(\alp ha^5-1)}{\alpha -1}=\dfrac{e^{i\beta}(e^{5\beta i}-1)}{e^{\beta i} -1}=$

$e^{i\beta}\cdot \dfrac{e^{-i\beta/2}}{e^{i\beta/2}-e^{-i\beta/2}}\cdot \dfrac{1}{e^{-i5\beta/2}}\cdot (e^{i5\beta/2}-e^{-i5\beta/2})=$

$e^{3i\beta}\cdot \dfrac{1}{2i\sin (\beta/2)}\cdot (2i\sin (5\beta/2))=(\cos 3\beta+i\sin 3\beta)\cdot \dfrac{\sin (5\beta/2)}{\sin (\beta/2)}=\ldots$

The rest is easy.

Fernando Revilla

9. Originally Posted by jacks
Here I Have Calculate $Re(\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5)$ and Getting

$\Leftrightarrow \cos(\frac{8\pi}{11})+\cos(\frac{16\pi}{11})...... ..............\cos(\frac{40\pi}{11})$

Now How can I calculate that sum.....
You can still do quite a bit of simplifying.

Use some identities: $\cos(-\theta)=\cos(\theta)$ $,\ \cos(\pi\pm\theta)=-\cos(\theta)$ $,\ \cos(2{\pi}n\pm\theta)=\cos(\theta)$

$\cos({{8\pi}\over{11}})=\cos(\pi-{{3\pi}\over{11}})=-\cos({{3\pi}\over{11}})$

$\cos({{16\pi}\over{11}})=\cos(\pi+{{5\pi}\over{11} })=-\cos({{5\pi}\over{11}})$

$\cos({{24\pi}\over{11}})=\cos(2\pi+{{2\pi}\over{11 }})=\cos({{2\pi}\over{11}})$

$\cos({{32\pi}\over{11}})=\cos({{10\pi}\over{11}})= \cos(\pi-{{\pi}\over{11}})=-\cos({{\pi}\over{11}})$

$\cos({{40\pi}\over{11}})=\cos(4\pi-{{4\pi}\over{11}})=\cos({{4\pi}\over{11}})$

This result is quite interesting.

Now look at the sum to product identity: $\displaystyle \cos(\varphi)+\cos(\theta)=2\cos\!\left({{\varphi+ \theta}\over{2}}\right)\cos\!\left({{\varphi-\theta}\over{2}}\right)\,.$ This also can give an interesting result.