MAT334-2018F > Quiz-5

Q5 TUT 0201

(1/1)

Victor Ivrii:
Use Morera's Theorem to show that the following function is analytic on the indicated domain; find a power-series expansion for the function by using the known power series for the integrand and interchanging the summation and integration.
$$\int_0^1 \frac{dt}{1-tz}\qquad\text{on}\; |z|<1.$$

Junya Zhang:
Morera's Theorem:
If $f$ is a continuous function on a domain $D$, and if $\int_{\gamma} f(z) dz = 0$ for every triangle $\gamma$ that lies, together with its interior, in $D$, then $f$ is analytic on $D$.

Define $f(z) = \int_0^1 \frac{dt}{1-tz}$ and let $D = \{z\in\mathbb{C}:|z|≤1\}$
Let $\gamma$ be a triangle that lies together with its interior in $D$.
Note that $1-tz = 0$ iff $z = \frac{1}{t}$ which implies $|z| = |1/t| ≥ 1$ (since $t\in[0,1]$)
Therefore, we can conclude that $1-tz ≠ 0$ for all $z \in \{z\in\mathbb{C}:|z|<1\}$.
This implies that $f$ is continuous on $\{z\in\mathbb{C}:|z|<1\}$, and for fixed $t\in[0,1]$, $f$ is analytic on and inside $\gamma$.
By Cauchy's theorem, $$\int_\gamma \frac{1}{1-tz} dz = 0$$ for fixed $t\in[0,1]$
Then, $$\int_{\gamma} f(z) dz = \int_{\gamma} \left( \int_0^1 \frac{dt}{1-tz}\right) dz = \int_0^1 \left(\int_{\gamma}\frac{1}{1-tz} dz\right) dt = \int_0^1 0 dt = 0$$
By Morera's Theorem, we can conclude that $\int_0^1 \frac{dt}{1-tz}$ is analytic on the domain $|z|<1$.

$$f(z) = \int_0^1 \frac{dt}{1-tz} = \int_0^1 \sum_{n=0} ^{\infty} (tz)^n dt = \sum_{n=0} ^{\infty} \int_0^1(tz)^n dt = \sum_{n=0} ^{\infty} \left[\frac{z^n\cdot t^{n+1}}{n+1}\right]_{0}^{1} = \sum_{n=0} ^{\infty} \frac{z^n}{n+1}$$