Question

Which of the following holds for complex variable $z$:
A. Let $C$ be the unit circle $z = e^{i\theta}, (-\pi \le \theta \le \pi)$. Then $\int_C \frac{e^{iz}}{z} dz = 2\pi i$ B. The $\{z \mid 1 < |z| < 2\}$ set is simply connected. C. $\frac{1}{1+z} = \sum_{n=0}^{\infty} z^n$, whenever $|z| < 1$ D. The unit disk $\{z \mid |z| < 1\}$ is simply connected. E. If $f$ is analytic at $z_0$, then $f$ is continuous at $z_0$.
Choose the correct answer from the options given below:

• A. A, B, E Only
• B. A, D, E Only
• C. B, C, E Only
• D. A, C, D Only

Hint:

An annulus $\{r_1 < |z| < r_2\}$ is the classic example of a region that is connected but NOT simply connected!

Answer 1

The Correct Option is B

Step 1 : Concept:
This question tests contour integration via Cauchy's Integral Formula, topological properties of complex domains (simply connected sets), power series expansions, and basic continuity of analytic functions.

Step 2 : Key Formulas and Approach:
1. Cauchy's Integral Formula: If $g(z)$ is analytic inside and on a simple closed contour $C$, and $z_0$ lies inside $C$: \[ \int_C \frac{g(z)}{z - z_0} dz = 2\pi i \, g(z_0) \] 2. Geometric series: $\frac{1}{1 - z} = \sum_{n=0}^\infty z^n$ for $|z| < 1$, and $\frac{1}{1 + z} = \sum_{n=0}^\infty (-1)^n z^n$.
3. A domain $D$ is simply connected if every simple closed curve in $D$ encloses only points in $D$ (i.e., it has no "holes").

Step 3 : Step-by-step Explanation:

• Statement A:
Let $g(z) = e^{iz}$, which is entire. The pole is at $z_0 = 0$, which lies inside the unit circle $C$.
By Cauchy's Integral Formula: \[ \int_C \frac{e^{iz}}{z} dz = 2\pi i \, g(0) = 2\pi i \, e^{0} = 2\pi i \] Hence, Statement A is correct.

• Statement B:
The set $\{z \mid 1 < |z| < 2\}$ represents an annulus (ring-shaped region), which contains a hole centered at $z = 0$. Hence, it is multiply connected, not simply connected. Statement B is incorrect.

• Statement C:
For $|z| < 1$, $\frac{1}{1+z} = \sum_{n=0}^{\infty} (-1)^n z^n = 1 - z + z^2 - z^3 + \dots$.
The expression $\sum_{n=0}^\infty z^n$ represents $\frac{1}{1-z}$, not $\frac{1}{1+z}$. Statement C is incorrect.

• Statement D:
The open unit disk $\{z \mid |z| < 1\}$ is a convex set with no holes. Any closed loop can be continuously shrunk to a point. Thus, it is simply connected. Statement D is correct.

• Statement E:
Analyticity at $z_0$ implies differentiability in a neighborhood of $z_0$. Since differentiability implies continuity, $f$ must be continuous at $z_0$. Statement E is correct.

Step 4 : Final Answer:
Statements A, D, and E are correct. Therefore, option (B) is the correct answer.