Proving a sum that contains complex numbers

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homad2000
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Homework Statement


show that:

Equation: sum from n=0 to infinity of e^(i d l) divided by 2^l equals negative 2 divided by (-2 + e^(i d)), approximately negative 2 divided by (-2 + 2.71828^d).



I tried changing the form to the sin and cos, but I couldn't complete it..

Any hints?
 
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  • Equation: sum from i=0 to infinity of e^(i d l) / 2^l equals minus 2 over (-2 + e^(i d)), approximately minus 2 over (-2 + 2.71828^(i d)).
    MSP12919hf2d39f2ch8fdd000012d5597g3dceh4bf.gif
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That's a geometric sequence. The sum of any geometric series is
[tex]\sum_{n=0}^\infty r^n= \frac{1}{1- r}[/tex]

Here, your r is [itex]e^{id}/2[/itex].