1) decompose at simple elements U_n = ((n x^(n−1) )/(x^n −1)) 2) calculste ∫


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Question Number 50406 by Abdo msup. last updated on 16/Dec/18

$1) d e c o m p o s e a t s i m p l e e l e m e n t s$ $U_{n} = \frac{n x^{n - 1}}{x^{n} - 1}$ $2) c a l c u l s t e \int_{0}^{2 π} \frac{d t}{x - e^{i t}}$
Commented by Abdo msup. last updated on 25/Dec/18

$l e t p (x) = x^{n} - 1 t h e r o o t s o f p (x) a r e z_{k} = e^{i \frac{2 k π}{n}}$ $w i t h k \in [[0, n - 1]] b u t U_{n} = \sum_{k = 0}^{n - 1} \frac{λ_{k}}{x - z_{k}}$ $λ_{k} = \frac{{n z}_{k}^{n - 1}}{p^{^{'}} (z_{k})} = \frac{n}{z_{k}} \frac{1}{n z_{k}^{n - 1}} = 1 \Rightarrow U_{n} = \sum_{k = 0}^{n - 1} \frac{1}{x - z_{k}}$ $= \sum_{k = 0}^{n - 1} \frac{1}{x - e^{i \frac{2 k π}{n}}}$
Commented by Abdo msup. last updated on 25/Dec/18

$2) c h a n g e m e n t e^{i t} = z g i v e$ $\int_{0}^{2 π} \frac{d t}{x - e^{i t}} = \int_{∣ z ∣= 1} \frac{1}{x - z} \frac{d z}{i z}$ $= \int_{∣ z ∣= 1} \frac{- i d z}{x z - z^{2}} = \int_{∣ z ∣= 1} \frac{i d z}{z^{2} - x z} l e t$ $φ (z) = \frac{i}{z^{2} - x z} = \frac{i}{z (z - x)} s o t h e p o l e s x o f φ a r e 0 a n d$ $x i f ∣ x ∣< 1 \int_{∣ z ∣= 1} φ (z) d z = 2 i π (R e s (φ, 0) + R e s (φ, x)}$ $R e s (φ, 0) = {l i m}_{z \to 0} z φ (z) = - \frac{i}{x}$ $R e s (φ, x) = {l i m}_{z \to x} (z - x) φ (z) = \frac{i}{x} \Rightarrow$ $\int_{∣ z ∣= 1} φ (z) d z = 0$ $i f ∣ x ∣> 1 \int_{∣ z ∣= 1} φ (z) d z = 2 i π R e s (φ, 0)$ $= 2 i π (\frac{- i}{x}) = \frac{2 π}{x} (w e s u p p o s e x \neq 0)$ $i f x = 0 \int_{0}^{2 π} \frac{d t}{x - e^{i t}} = - \int_{0}^{2 π} e^{- i t} d t$ $= - {[- \frac{1}{i} e^{- i t}]}_{0}^{2 π} = \frac{1}{i} (1 - 1) = 0$
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