Question

Assume that the complex numbers z., n =1, 2, ... are in the right half-plane and that both the series Ž - and = converge. Show that the series Ž. converges. Give a counterexample when the z, are not restricted to the right half-plane.

Answer 1

The series Σz_n converges when the complex numbers z_n, n = 1, 2, … are in the right half-plane and both the series Σ|z_n| and ΣRe(z_n) converge.

Step-by-step explanation:

In order to show that the series Σz_n converges when the complex numbers z_n, n = 1, 2, … are in the right half-plane and both the series Σ|z_n| and ΣRe(z_n) converge, we can use the comparison test. Since the series Σ|z_n| and ΣRe(z_n) converge, we know that both of these series are bounded. Let M be a real number such that |z_n| ≤ M for all n. Then, for any n, we have |Re(z_n)| ≤ |z_n| ≤ M. Therefore, the real part of z_n is also bounded. Now, by the comparison test, since |z_n| and Re(z_n) are bounded, we can conclude that Σz_n converges.

However, if the z_n are not restricted to the right half-plane, convergence of the series Σz_n is not guaranteed. For instance, consider the series Σ(−1)^n/n. This series does not converge even though both Σ|(-1)^n/n| and ΣRe((-1)^n/n) converge. This counterexample illustrates that without the restriction of z_n to the right half-plane, convergence of Σz_n cannot be assured.

In summary, when z_n are in the right half-plane and both Σ|z_n| and ΣRe(z_n) converge, then the series Σz_n also converges due to their boundedness. However, without this restriction on z_n, convergence of Σz_n cannot be guaranteed.