Answer on Question #78435 – Math – Complex Analysis
Question
Find the sum of the 5th roots of unity.
Solution
1 = e 2 π i . 1 = e ^ {2 \pi i}. 1 = e 2 πi . 1 5 = e 2 π i 5 = e 2 π i n m , 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. 5 1 = 5 e 2 πi = e n 2 πi m , m = 0 , 1 , 2 , 3 , 4.
Sum of the roots:
S = ∑ m = 0 4 e 2 π i n m = ∑ m = 0 4 ( e 2 π i n ) m = 1 − ( e 2 π i n ) n 1 − e 2 π i n = 1 − e 2 π i 1 − e 2 π i n = 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. S = m = 0 ∑ 4 e n 2 πi m = m = 0 ∑ 4 ( e n 2 πi ) m = 1 − e n 2 πi 1 − ( e n 2 πi ) n = 1 − e n 2 πi 1 − e 2 πi = 0.
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