Question #78435

Find the sum of the 5th roots of unity.
1

Expert's answer

2018-06-22T14:27:08-0400

Answer on Question #78435 – Math – Complex Analysis

Question

Find the sum of the 5th roots of unity.

Solution

1=e2πi.1 = e ^ {2 \pi i}.15=e2πi5=e2πinm,m=0,1,2,3,4.\sqrt[5]{1} = \sqrt[5]{e ^ {2 \pi i}} = e ^ {\frac {2 \pi i}{n} m}, \quad m = 0, 1, 2, 3, 4.


Sum of the roots:


S=m=04e2πinm=m=04(e2πin)m=1(e2πin)n1e2πin=1e2πi1e2πin=0.S = \sum_ {m = 0} ^ {4} e ^ {\frac {2 \pi i}{n} m} = \sum_ {m = 0} ^ {4} \left(e ^ {\frac {2 \pi i}{n}}\right) ^ {m} = \frac {1 - \left(e ^ {\frac {2 \pi i}{n}}\right) ^ {n}}{1 - e ^ {\frac {2 \pi i}{n}}} = \frac {1 - e ^ {2 \pi i}}{1 - e ^ {\frac {2 \pi i}{n}}} = 0.


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Comments

Assignment Expert
15.07.18, 18:39

Formula exp(2*Pi*i)=1 follows from the Euleur's formula exp(ix)=cos(x)+isin(x), where i*i=-1, cos(2*Pi)=1, sin(2*Pi)=0. To find the sum of the 5th roots of unity, the formula for the sum of geometric sequence was applied, where r=exp(2*Pi*i/n) is the ratio and the sum starts from m=0 to m=n-1=5-1=4.

Ss
15.07.18, 09:17

Kindly show all the steps so that it will be easy for us.....

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