Answer on Question #50250– Math - Complex Analysis
let ∑ Z n \sum Zn ∑ Z n 0 < Arg ( Z n ) < ( P i / 2 ) 0 < \operatorname{Arg}(Zn) < (\mathrm{Pi} / 2) 0 < Arg ( Z n ) < ( Pi /2 )
Show that
∑ ( Re ( Z n ) . Im ( Z n ) ) \sum (\operatorname{Re}(Zn). \operatorname{Im}(Zn)) ∑ ( Re ( Z n ) . Im ( Z n ))
Solution:
∑ k = 1 ∞ z k = X + i ⋅ Y \sum_{k=1}^{\infty} z_k = X + i \cdot Y ∑ k = 1 ∞ z k = X + i ⋅ Y z n = x n + i y n z_n = x_n + iy_n z n = x n + i y n ∣ X + i ⋅ Y ∣ = X 2 + Y 2 < ∞ |X + i \cdot Y| = \sqrt{X^2 + Y^2} < \infty ∣ X + i ⋅ Y ∣ = X 2 + Y 2 < ∞
If 0 < arg ( Z n ) < π / 2 0 < \arg (Z_n) < \pi / 2 0 < arg ( Z n ) < π /2 x n , y n > 0 x_n, y_n > 0 x n , y n > 0
∑ k = 1 ∞ z k = ∑ k = 1 ∞ ( x k + i y k ) = ∑ k = 1 ∞ x k + i ∑ k = 1 ∞ y k = X + i ⋅ Y \sum_{k=1}^{\infty} z_k = \sum_{k=1}^{\infty} (x_k + iy_k) = \sum_{k=1}^{\infty} x_k + i \sum_{k=1}^{\infty} y_k = X + i \cdot Y k = 1 ∑ ∞ z k = k = 1 ∑ ∞ ( x k + i y k ) = k = 1 ∑ ∞ x k + i k = 1 ∑ ∞ y k = X + i ⋅ Y ∑ k = 1 ∞ Re [ z k ] = ∑ k = 1 ∞ x k = X < X 2 + Y 2 < ∞ , ∑ k = 1 ∞ Im [ z k ] = ∑ k = 1 ∞ y k = Y < X 2 + Y 2 < ∞ \sum_{k=1}^{\infty} \operatorname{Re}[z_k] = \sum_{k=1}^{\infty} x_k = X < \sqrt{X^2 + Y^2} < \infty, \quad \sum_{k=1}^{\infty} \operatorname{Im}[z_k] = \sum_{k=1}^{\infty} y_k = Y < \sqrt{X^2 + Y^2} < \infty k = 1 ∑ ∞ Re [ z k ] = k = 1 ∑ ∞ x k = X < X 2 + Y 2 < ∞ , k = 1 ∑ ∞ Im [ z k ] = k = 1 ∑ ∞ y k = Y < X 2 + Y 2 < ∞ ∣ ∑ x k y n ∣ ≤ ∑ ∣ x k y n ∣ ≤ ∑ k = 1 ∞ ∣ x k ∣ ⋅ ∑ n = 1 ∞ ∣ y n ∣ \left| \sum x_k y_n \right| \leq \sum |x_k y_n| \leq \sum_{k=1}^{\infty} |x_k| \cdot \sum_{n=1}^{\infty} |y_n| ∣ ∣ ∑ x k y n ∣ ∣ ≤ ∑ ∣ x k y n ∣ ≤ k = 1 ∑ ∞ ∣ x k ∣ ⋅ n = 1 ∑ ∞ ∣ y n ∣
As x n , y n > 0 ⇒ ∑ ∣ x k y n ∣ = ∑ x k y n x_n, y_n > 0 \Rightarrow \sum |x_k y_n| = \sum x_k y_n x n , y n > 0 ⇒ ∑ ∣ x k y n ∣ = ∑ x k y n ∑ k = 1 ∞ ∣ x k ∣ = ∑ k = 1 ∞ x k = X \sum_{k=1}^{\infty} |x_k| = \sum_{k=1}^{\infty} x_k = X ∑ k = 1 ∞ ∣ x k ∣ = ∑ k = 1 ∞ x k = X ∑ n = 1 ∞ ∣ y n ∣ = ∑ n = 1 ∞ y n = Y \sum_{n=1}^{\infty} |y_n| = \sum_{n=1}^{\infty} y_n = Y ∑ n = 1 ∞ ∣ y n ∣ = ∑ n = 1 ∞ y n = Y
∑ x k y n < X ⋅ Y < X 2 + Y 2 < ∞ \sum x_k y_n < X \cdot Y < X^2 + Y^2 < \infty ∑ x k y n < X ⋅ Y < X 2 + Y 2 < ∞
Answer: ∑ Re [ z k ] Im [ z n ] \sum \operatorname{Re}[z_k] \operatorname{Im}[z_n] ∑ Re [ z k ] Im [ z n ]
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#339914 on Dec 2023
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