Practice midterm 2 solutions

As usual, be sure to include your method, and remember to write your name.

Problem 1. Let $\gamma$ be the circle of radius $3$ centered at the origin. Compute

$${\int }_{\gamma }\frac{z-1}{z^2}𝑑z$$

via parametrization.

The above problem is directed towards Objective 4 (complex integration).

Solution. Writing $z=\gamma (t)=3e^{it}$ for , this is

$${\int }_0^{2\pi }\frac{3e^{it}-1}{9e^{2it}}\cdot 3i{e}^{it}𝑑t=i{\int }_0^{2\pi }\frac{3e^{it}-1}{3e^{it}}𝑑t=i{\int }_0^{2\pi }(1-e^{-it}/3)𝑑t=2\pi i+{\frac{1}{3}{e}^{-it}|}_0^{2\pi }=2\pi i.$$

Problem 2. Consider the path $\gamma$ given by the straight line segment from $-i$ to $i+1$.

Compute

$${\int }_{\gamma }z{e}^{z^2}𝑑z.$$

The above problem is directed towards Objective 4 (complex integration).

Solution. Since $f(z)=z{e}^{z^2}$ is an analytic function and is defined on all of $ℂ$, which is a star-shaped domain, we can use the fundamental theorem of calculus. Setting $u=z^2$, we have $z{e}^{z^2}dz=\frac{1}{2}{e}^{u}du$ with antiderivative $\frac{1}{2}{e}^u=\frac{1}{2}{e}^{z^2}$, so the integral is

$${\frac{1}{2}{e}^{z^2}|}_{-i}^{i+1}=\frac{1}{2}(e^{{(i+1)}^2}-e^{{(-i)}^2}).$$

This is an acceptable answer, but one could simplify further: ${(i+1)}^2=2i$ and ${(-i)}^2=-1$, so this is $\frac{1}{2}(e^{2i}-e^{-1})$. (If you like, you could write this in terms of sine and cosine, but it isn’t really any simpler.)

Problem 3. Let $R$ be the rectangle with corners at $-2-i$, $-2+i$, $2-i$, $2+i$:

Compute

$${\int }_{\partial R}\frac{z^2+1}{z^4-z^3}𝑑z.$$

Hint/timesaver: you may find the formulas $g^{\prime }(z)=1-\frac{2}{{(z-1)}^2}$ and $g^{\prime \prime }(z)=\frac{4}{{(z-1)}^3}$ useful, where $g(z)=\frac{z^2+1}{z-1}$.

The above problem is directed towards Objective 5 (Cauchy’s integral theorem and formula).

Solution. Let $f(z)=\frac{z^2+1}{z^3}$, $g(z)=\frac{z^2+1}{z-1}$, and $h(z)=\frac{z^2+1}{z^4-z^3}=\frac{f(z)}{z-1}=\frac{g(z)}{z^3}$. Note that $f$ is analytic on $R\setminus \{0\}$ and $g$ is analytic on $R\setminus \{1\}$. Letting $D_0$ and $D_1$ be disks of small radius $r$ around $0$ and $1$ respectively and $D=R\setminus D_0\setminus D_1$, we have $\partial D=\partial R-\partial D_0-\partial D_1$ (i.e. the union of the boundaries with the negative orientations on the second terms), and since $h$ is analytic on $D$ we have

$$0={\int }_{\partial D}h(z)𝑑z={\int }_{\partial R}h(z)𝑑z-{\int }_{\partial D_0}h(z)𝑑z-{\int }_{\partial D_1}h(z)𝑑z,$$

so

$${\int }_{\partial R}h(z)𝑑z={\int }_{\partial D_0}h(z)𝑑z+{\int }_{\partial D_1}h(z)𝑑z.$$

On $D_0$, we have $h(z)=\frac{g(z)}{z^3}$ and $g(z)$ is analytic, so by Cauchy’s formula for derivatives

$${\int }_{\partial D_0}\frac{g(z)}{z^3}𝑑z=\frac{2\pi i}{2!}{g}^{(2)}(0)=\pi i\cdot \frac{4}{{(-1)}^3}=-4\pi i.$$

Similarly, on $D_1$ we have $h(z)=\frac{f(z)}{z-1}$ and $f$ analytic, so by Cauchy’s formula

$${\int }_{\partial D_1}\frac{f(z)}{z-1}𝑑z=2\pi if(1)=4\pi i.$$

Hence the overall integral is $0$.

Problem 4. Let $f:D\to ℂ$ be a smooth function. For $z_0\in D$, let $g(z)=f(z)-f(z_0)$. Show that

$${\int }_{\partial D}g(z)𝑑z={\int }_{\partial D}f(z)𝑑z.$$

The above problem is directed towards Objective 5 (Cauchy’s integral theorem and formula).

Solution. Since $z\mapsto f(z_0)$ is an analytic function (since it is constant), by Cauchy’s theorem

$${\int }_{\partial D}g(z)𝑑z={\int }_{\partial D}f(z)𝑑z-{\int }_{\partial D}f(z_0)𝑑z={\int }_{\partial D}f(z)𝑑z.$$

Problem 5. Suppose that $f:ℂ\to ℂ$ is an analytic function such that $|f(z)|\le {|z|}^n$ for all $z\in ℂ$. Conclude that $|f^{(n)}(0)|\le n!$.

The above problem is directed towards Objective 6 (consequences of Cauchy’s theorem and formula).

Solution. Since $f$ is analytic on $ℂ$, it is analytic on any disk of radius $R$ centered at the origin; on its boundary the bound is then ${|z|}^n=R^n$. By the Cauchy estimates it follows that $|f^{(n)}(0)|\le \frac{n!}{R^n}{R}^n=n!$.

Problem 6. Let $D$ be the disk of radius $1$ centered at the origin. Verify Pompeiu’s formula for $f(z)={|z|}^2$ at $z_0=0$.

The above problem is directed towards Objective 6 (consequences of Cauchy’s theorem and formula).

Solution. First, note ${|z|}^2=z\overline{z}$, so $\frac{\partial f}{\partial \overline{z}}=z$. Therefore Pompeiu’s formula reads

$${|0|}^2=0=\frac{1}{2\pi i}{\int }_{\partial D}\frac{{|z|}^2}{z}𝑑z-\frac{1}{\pi }{\int }_{D}z\cdot \frac{1}{z}𝑑x𝑑y.$$

The second integral is just the area of $D$, which is $\pi$, so the second term is $-\frac{1}{\pi }\cdot \pi =-1$. On $\partial D$ the absolute value ${|z|}^2$ is just $1$, so the first integral is

$${\int }_{\partial D}\frac{1}{z}𝑑z,$$

which we’ve seen by a number of methods is $2\pi i$. Hence the first term is $\frac{1}{2\pi i}\cdot 2\pi i=1$. Combining these together, the right-hand side is $1-1=0$, matching the left-hand side.