Question

Let \(f(z) = \displaystyle\sum_{n=1}^{\infty} 5^{-n}\cos(nz)\). On which one of the following domains, does \(f(z)\) represent an analytic function?

• A. \(\{z\in\mathbb{C} : |\operatorname{Im} z| < \ln 5\}\)
• B. \(\{z\in\mathbb{C} : |\operatorname{Im} z| > \ln 5\}\)
• C. \(\{z\in\mathbb{C} : |\operatorname{Re} z| < \ln 5\}\)
• D. \(\{z\in\mathbb{C} : |\operatorname{Re} z| > \ln 5\}\)

Hint:

Split \(\cos(nz)\) into \(\dfrac{e^{inz}+e^{-inz}}{2}\) and check when both resulting geometric series converge.

Answer 1

The Correct Option is A

Step 1: Rewrite the cosine terms using complex exponentials.
Write \(z=x+iy\), so \(\operatorname{Im} z = y\). Using \(\cos(nz) = \dfrac{e^{inz}+e^{-inz}}{2}\), the series becomes
\[ f(z) = \frac{1}{2}\sum_{n=1}^{\infty}\left(\frac{e^{iz}}{5}\right)^{n} + \frac{1}{2}\sum_{n=1}^{\infty}\left(\frac{e^{-iz}}{5}\right)^{n} \] which is a sum of two geometric series in the ratios \(w_1=\dfrac{e^{iz}}{5}\) and \(w_2=\dfrac{e^{-iz}}{5}\).

Step 2: Find the modulus of each ratio.
Since \(z=x+iy\), \(e^{iz}=e^{ix-y}=e^{-y}e^{ix}\), so \(|e^{iz}|=e^{-y}\), and \(e^{-iz}=e^{-ix+y}=e^{y}e^{-ix}\), so \(|e^{-iz}|=e^{y}\).
So \(|w_1| = \dfrac{e^{-y}}{5}\) and \(|w_2|=\dfrac{e^{y}}{5}\).

Step 3: Apply the geometric series convergence condition.
A geometric series \(\sum w^n\) converges, and sums to an analytic function, exactly when \(|w|<1\). Both \(|w_1|<1\) and \(|w_2|<1\) are needed together, since \(f(z)\) is the sum of both series.
\(|w_1|<1\) means \(e^{-y}<5\), that is \(y>-\ln 5\).
\(|w_2|<1\) means \(e^{y}<5\), that is \(y<\ln 5\).
Together, \(-\ln 5<y<\ln 5\), which is exactly \(|\operatorname{Im} z|<\ln 5\).

Step 4: Check the wrong options.
Option (B) reverses the inequality and describes the region where the series actually diverges, since there \(e^{y}\ge 5\) or \(e^{-y}\ge 5\), so it is wrong.
Options (C) and (D) depend on \(\operatorname{Re} z=x\), but the modulus of \(e^{\pm iz}\) does not depend on \(x\) at all, only on \(y\), so the convergence of this series cannot depend on \(\operatorname{Re} z\); both are wrong.

Final Answer:
The series converges absolutely, and \(f(z)\) is analytic, exactly on the strip \(|\operatorname{Im} z|<\ln 5\). \[ \boxed{\{z\in\mathbb{C} : |\operatorname{Im} z| < \ln 5\}} \]