The problem shown in the image asks to find the radius of convergence of the series:

$\sum_{n=1}^{\infty} \frac{(z^n)}{\cos(n)}$ where $z = x + iy$ is a complex number.

Steps to Solve:

To determine the radius of convergence of a complex series of the form $\sum a_n z^n$, we use the formula for the radius of convergence $R$:

$\frac{1}{R} = \limsup_{n \to \infty} \sqrt[n]{|a_n|}$

For the given series, the general term is:

$a_n = \frac{1}{\cos(n)}$

Key Points:

1. $\cos(n)$ oscillates between -1 and 1, meaning that $\frac{1}{\cos(n)}$ is bounded for $n$, but $\cos(n)$ is never zero (except for specific values where $n$ is not a real number).
2. Since $a_n$ does not tend to 0 as $n$ approaches infinity, the series does not converge for any nonzero $z$.

Therefore, the radius of convergence for this series is:

$R = 1$

Hence, the correct answer is:

$\boxed{1}$

Would you like more details or have any questions?

Here are five related questions to deepen understanding:

1. How is the radius of convergence related to the $\limsup$ of coefficients?
2. What is the importance of $\cos(n)$'s behavior in this series?
3. How would the series behave if $\cos(n)$ approached zero?
4. Can the radius of convergence ever be infinite?
5. What role does the imaginary part $iy$ of $z$ play in convergence?

Tip: When analyzing complex series, always check the behavior of the series' coefficients and apply the root test or ratio test to determine convergence.