Question

let ∑ Zn be A convergent series such that 0< Arg (Zn) < (Pi /2)

Show that
∑ ( Re ( Zn). Im (Zn) ) is also Convergent

Answer 1

Answer on Question #50236 – Math - Complex Analysis

let $\sum Zn$ be A convergent series such that $0 < \operatorname{Arg}(Zn) < (\mathrm{Pi} / 2)$

Show that

$\sum (\operatorname{Re}(Zn). \operatorname{Im}(Zn))$ is also Convergent

Solution:

$\sum_{k=1}^{\infty} z_k = X + i \cdot Y$ is the convergent series, where $z_n = x_n + iy_n$. $|X + i \cdot Y| = \sqrt{X^2 + Y^2} < \infty$

If $0 < \arg (Z_n) < \pi / 2$ then $x_n, y_n > 0$.

$$\sum_{k=1}^{\infty} z_k = \sum_{k=1}^{\infty} (x_k + iy_k) = \sum_{k=1}^{\infty} x_k + i \sum_{k=1}^{\infty} y_k = X + i \cdot Y$$ $$\sum_{k=1}^{\infty} \operatorname{Re}[z_k] = \sum_{k=1}^{\infty} x_k = X < \sqrt{X^2 + Y^2} < \infty, \quad \sum_{k=1}^{\infty} \operatorname{Im}[z_k] = \sum_{k=1}^{\infty} y_k = Y < \sqrt{X^2 + Y^2} < \infty$$ $$\left| \sum x_k y_n \right| \leq \sum |x_k y_n| \leq \sum_{k=1}^{\infty} |x_k| \cdot \sum_{n=1}^{\infty} |y_n|$$

As $x_n, y_n > 0 \Rightarrow \sum |x_k y_n| = \sum x_k y_n$, $\sum_{k=1}^{\infty} |x_k| = \sum_{k=1}^{\infty} x_k = X$, $\sum_{n=1}^{\infty} |y_n| = \sum_{n=1}^{\infty} y_n = Y$, then

$$\sum x_k y_n < X \cdot Y < X^2 + Y^2 < \infty$$

Answer: $\sum \operatorname{Re}[z_k] \operatorname{Im}[z_n]$ is also convergent.

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