Week 6 Worksheet - Solutions

Home | Assessment | Notes | Worksheets | Blackboard

Cauchy's Integral Formula

1. 1. We have $f(z) = |z+1|^2 = (z+1)(\overline{z} + 1) = x^2 + y^2 + 2x + 1$ and calculate \[ (\partial_1 u)(x,y) = 2x + 2 \quad (\partial_2 u)(x,y) = 2y \] as well as $(\partial_1 v)(x,y) = 0 = (\partial_2 v)(x,y)$ for all $x,y$. The Cauchy-Riemann equations are only satisfed at $z = -1$ so $f$ is not holomorphic on any domain whatsoever.
   2. Note that $f(z) = 2 + z + \overline{z}$ when $|z| = 1$. It just so happens that \[ \dfrac{1}{z} = \dfrac{\overline{z}}{|z|^2} = \overline{z} \] when $|z| = 1$ so let's take $g(z) = 2 + z + \dfrac{1}{z}$.
   3. Now \[ \int\limits_\gamma |z+1|^2 \intd z = \int\limits_\gamma 2 + z \intd z + \int\limits_\gamma \dfrac{1}{z} \intd z \] and, since $\gamma$ is closed, the existence of an antiderivative for $2+z$ on $\C$ leaves us with \[ \int\limits_\gamma |z+1|^2 \intd z = 2 \pi i \left( \dfrac{1}{2 \pi i} \int\limits_\gamma \dfrac{1}{z} \intd z \right) \] to be evaluated. The term in brackets is the Cauchy integral formula applied to $f(z) = 1$ and $b = 0$.
2. Since $f$ is entire it is represented by a power series \[ \sum_{n=0}^\infty a_n z^n \] centered at 0 with infinite radius of convergence. Moreover, for every $R > 0$ we have \[ a_n = \dfrac{1}{2\pi i} \int\limits_{\gamma_R} \dfrac{f(z)}{z^{n+1}} \intd z \] where $\gamma_R(t) = Re^{it}$ on $[0,2\pi]$. Thus \[ a_n = \dfrac{1}{2\pi} \int\limits_0^{2\pi} \dfrac{f(Re^{it})}{(Re^{it})^n} \intd t \] and the estimation lemma, together with the hypothesis, gives \[ |a_n| \le \dfrac{1}{2\pi} \cdot 2 \pi \cdot \dfrac{K R^k}{R^n} = K \dfrac{R^k}{R^n} \] which, for $n > k$, converges to zero as $R \to \infty$. Thus all but finitely many of the coefficients $a_n$ are equal to zero and $f$ is a polynomial.

Taylor's Theorem

3. We can differentiate power series. By induction \[ f^{(k)}(z) = \sum_{n=k}^\infty a_n n(n-1) \cdots (n-(k-1)) (z-b)^{n-k} \] and plugging in 0 gives $f^{(k)}(b) = a_k k (k-1) \cdots 2 \cdot 1 = k! a_k$.
4. 1. First we verify that \[ (\sin(z))^2 = \left( \dfrac{\exp(iz) - \exp(-iz)}{2i} \right)^2 = \dfrac{\exp(2iz) - 2 + \exp(-2iz)}{-4} = \dfrac{1}{2} - \dfrac{\cos(2z)}{2} \] so that \[ (\sin(z))^2 = \frac{1}{2} - \dfrac{1}{2} \sum_{n=0}^\infty \dfrac{(-1)^n}{(2n)!} (2z)^{2n} = z^2 - \dfrac{1}{3} z^4 + \dfrac{2}{45} z^6 - \cdots \] with infinite radius of convergence.
   2. Write \[ \dfrac{1}{2z + 1} = \dfrac{1}{1 - (-2z)} = \sum_{n=0}^\infty (-2z)^n = \sum_{n=0}^\infty (-2)^n z^n \] whenever $|-2z| < 1$ using the geometric series. The radius of convergence is $1/2$.
   3. We can plug $z^2$ into the power series of the exponential function \[ \exp(z^2) = \sum_{n=0}^\infty \dfrac{1}{n!} (z^2)^n = \sum_{n=0}^\infty \dfrac{1}{n!} z^{2n} \] giving a Taylor series with infinite radius of convergence.
5. We know that $\Log(w)$ is holomorphic on $\C \setminus (-\infty,0]$. Thus $f(z) = \Log(1+z)$ is holomorphic on $\C \setminus (-\infty,-1]$. It therefore has a Taylor on $\ball(0,1)$. First $a_0 = f(0) = \Log(1) = 0$. Now \[ f'(z) = (\Log(1+z))' = \dfrac{1}{1+z} \] from the chain rule. But we know \[ f'(z) = \sum_{n=0}^\infty (-1)^n z^n \] on $\ball(0,1)$ from the geometric series. Thus \[ f(z) = \sum_{n=0}^\infty \dfrac{(-1)^n}{n+1} z^{n+1} = \sum_{n=1}^\infty \dfrac{(-1)^{n+1}}{n} z^n = z - \dfrac{1}{2} z^2 + \dfrac{1}{3} z^3 - \dfrac{1}{4} z^4 + \cdots \] on $\ball(0,1)$.
6. We can write \begin{align*} \dfrac{1}{1+z^2} & = \dfrac{1}{2} \left( \dfrac{1}{1-iz} + \dfrac{1}{1 - (-iz)} \right) \\ & = \dfrac{1}{2} \sum_{n=0}^\infty (iz)^n + \dfrac{1}{2} \sum_{n=0}^\infty (-iz)^n \\ & = \sum_{n=0}^\infty \dfrac{i^n + (-i)^n}{2} z^n \\ & = 1 - z^2 + z^4 - z^6 + z^8 - z^{10} + \cdots \end{align*} whenever $|iz| < 1$ and $|-iz| < 1$ using the geometric series. Thus the series is valid on $\ball(0,1)$.
7. Whenever $\ball(b,R) \subset \C \setminus \{c,d\}$ we will have a Taylor series for $f$ on $\ball(b,R)$. The largest ball centered at $b$ and entirely contained within $\C \setminus \{c,d\}$ is the ball of radius $\min \{ |b-c|, |b-d| \}$.

Applications

8. Plug into the quadratic formula \[ \dfrac{-(2+2i) \pm \sqrt{(2+2i)^2 - 4(-2)i}}{2i} = \dfrac{-(2+2i) \pm \sqrt{16i}}{2i} = \dfrac{-(2+2i) \pm 4(\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}})}{2i} \] to get \[ (1+i)i + \sqrt{2}(1+i)i = (1+i)(1+\sqrt{2})i \qquad (1+i)i - \sqrt{2}(1+i)i = (1+i)(1 - \sqrt{2}) i \] as the roots.
9. If $p$ has degree one then \[ p(z) = a_1 z + a_0 = a_1 (z - \tfrac{a_0}{a_1}) \] as desired. We proceed by induction. Fix a polynomial \[ p(z) = a_n z^n + a_{n-1} z^{n-1} + \cdots + a_2 z^2 + a_1 z + a_0 \] of degree $n$. By the fundamental theorem of algebra $p(c_n) = 0$ for some $c_n \in \C$. Polynomial long division therefore lets us write \[ p(z) = (z-c_n) q(z) + d \] where $q(z)$ is a polynomial of degree $n=1$. Plugging in $z = c_n$ gives $d = 0$ so $p(z) = (z-c_n) q(z)$. By induction \[ q(z) = b (z - c_{n-1}) \cdots (z - c_2)(z - c_1) \] and therefore $p$ has the desired form.
10. Suppose there is $a \in \C$ not equal to any $p(z)$. Then the function $f(z) = 1 / (p(z) - a)$ is defined on all of $\C$ and holomorphic there. Moreover, as in the proof of the fundamental theorem of algebra, the function is bounded. Liouville's theorem then implies $f$ and therefore $p$ is constant. This is a contradiction, as no polynomial of degree at least 1 is constant.