Practice midterm 3 solutions

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

Problem 1. Find the Taylor expansion of $f(z)=e^{2z}-e^{-2}$ centered at $z_0=-1$. What is its radius of convergence?

The above problem is directed towards Objective 7 (power series).

Solution. The $n$th derivative of $f$ for $n\ge 1$ is $2^n{e}^{2z}$, so evaluating at $-1$ gives $2^n{e}^{-2}$; evaluating $f$ at $-1$ gives $0$, so the Taylor series is

$$f(z)=\sum _{n=1}^{\mathrm{\infty }}\frac{2^n{e}^{-2}}{n!}{(z+1)}^n.$$

Alternatively we could observe that

$$e^{2z}=e^{-2}{e}^{2(z+1)}=\sum _{n=0}^{\mathrm{\infty }}\frac{2^n{e}^{-2}}{n!}{(z+1)}^n,$$

and subtracting $e^{-2}$ kills off the constant term. Finally, the radius of convergence is infinite, either since that of $e^z$ is and we’re substituting $2z$, so holds for all $z$; or since $f$ is entire, so the distance to the nearest singularity is infinite; or by the ratio test, the radius of convergence is

$$\underset{n\to \mathrm{\infty }}{lim}\left|\frac{2^n{e}^{-2}/n!}{2^{n+1}{e}^{-2}/(n+1)!}\right|=\underset{n\to \mathrm{\infty }}{lim}\left|\frac{n+1}{2}\right|=\mathrm{\infty }.$$

Problem 2. Find the Taylor expansion of $f(z)=\frac{z^2}{z^3-1}$ at infinity.

The above problem is directed towards Objective 7 (power series).

Solution. Let $w=\frac{1}{z}$. Then $f(z)=\frac{z^2}{z^3-1}=\frac{w^{-2}}{w^{-3}-1}=\frac{w}{1-w^3}$, which by the geometric series expansion is

$$f(z)=\sum _{n=0}^{\mathrm{\infty }}{w}^{3n+1}=\sum _{n=0}^{\mathrm{\infty }}{z}^{-3n-1}=z^{-1}+z^{-4}+z^{-7}+\mathrm{\cdots }.$$

We could also write this as

$$\sum _{n=0}^{\mathrm{\infty }}{a}_n{z}^{-n}$$

where $a_n=1$ if $n=3m+1$ for some integer $n$, and $a_n=0$ otherwise; either form is fine.

Problem 3. Compute the Laurent expansion of $f(z)=\frac{1}{(z+1)(z+2)}$ on the annulus .

The above problem is directed towards Objective 8 (Laurent series).

Solution. Writing $f(z)=\frac{1}{z+1}\cdot \frac{1}{z+2}$, we use partial fractions to write it as $\frac{A}{z+1}+\frac{B}{z+2}$ for some constants $A,B$: we have $\frac{A}{z+1}+\frac{B}{z+2}=\frac{A(z+2)+B(z+1)}{(z+1)(z+2)}=\frac{1}{(z+1)(z+2)}$, so $2A+B=1$ and $A+B=0$, hence $A=1$ and $B=-1$; and we can check that indeed $\frac{1}{z+1}-\frac{1}{z+2}=\frac{1}{(z+1)(z+2)}$. Since $\frac{1}{z+1}$ is analytic on $|z|>1$ (since its only singularity is at $z=-1$) and $-\frac{1}{z+2}$ is analytic on (since its only singularity is at $z=-2$), we can take $f_0(z)=-\frac{1}{z+2}$ and $f_1(z)=\frac{1}{z+1}$, and find their Taylor expansions at $0$ and at infinity respectively. By the geometric series formula,

$$-\frac{1}{z+2}=-\frac{1}{2}\cdot \frac{1}{1+z/2}=-\frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}{(-z/2)}^n=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^{n+1}{2}^{-n-1}{z}^n.$$

Meanwhile

$$\frac{1}{z+1}=\frac{1}{1/(1/z)+1}=\frac{1/z}{1+1/z}=\sum _{n=1}^{\mathrm{\infty }}{(-1)}^{n+1}{z}^{-n}.$$

Combining the two, we get

$$f(z)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_n{z}^n$$

where $a_n={(-1)}^{n+1}{2}^{-n-1}$ for $n\ge 0$ and $a_n={(-1)}^{n+1}$ for .

Note that the first power series converges for , i.e. , and the second converges for , i.e. $|z|>1$, so the overall Laurent series converges on the desired annulus.

Problem 4. Find and classify the isolated singularities of $f(z)=\frac{\cos (1/z)}{z^4-1}$. For any poles, find their order.

The above problem is directed towards Objective 8 (Laurent series).

Solution. The singularities are at $z=0$ for $\cos (1/z)$ and wherever $z^4=1$, i.e. $z=1,i,-1,-i$. Factoring $z^4-1=(z^2-1)(z^2+1)=(z+1)(z-1)(z+i)(z-i)$, since $\cos (1/z)$ is nonzero at each of these points these are all simple poles.

At $z=0$, we have

$$\cos (1/z)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{(2n)!}{z}^{-2n},$$

which has infinitely many negative terms; since $z^4-1$ is analytic and nonzero at $z=0$, dividing by it does not change the type of singularity, so we have an essential singularity at $z=0$.

Since we’ve had less time with the residue theorem, I’ve included an extra practice problem on it below (so three instead of two). However, only two will appear on the actual exam.

Problem 5. Let $D$ be the rectangle in $ℂ$ with corners at $i-1$, $-i-1$, $4+i$, and $4-i$. Using the residue theorem, compute

$${\int }_{\partial D}\frac{z}{\sin z}𝑑z.$$

The above problem is directed towards Objective 9 (the residue theorem).

Solution. The singularities of $\frac{z}{\sin z}$ are at zeros of $\sin z$, i.e. at $z=\pi n$ for $n\in \mathbb{Z}$. Since ${\sin }^{\prime }(\pi n)=\cos (\pi n)\ne 0$, these are simple zeros, hence at worst simple poles of $\frac{z}{\sin z}$. The only zeros in $D$ are at $0$, which is removable since both numerator and denominator have a zero of order $1$, and at $\pi$, where since it is a simple pole we get residue $\frac{\pi }{\cos \pi }=-\pi$. Therefore by the residue theorem the integral is $-\pi \cdot 2\pi i=-2{\pi }^{2}i$.

Problem 6. Using the residue theorem, compute

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{x^4+1}𝑑x.$$

The above problem is directed towards Objective 9 (the residue theorem).

Solution. Let $D$ be the upper half-disk of radius $N$, with boundary $[-N,N]$ together with the arc $\gamma$ from $N$ to $-N$ of radius $N$. The integral of $\frac{1}{z^4+1}$ over $\gamma$ will vanish in the limit by the ML bound: in absolute value the integrand is bounded by something of order $\frac{1}{N^4}$ and the arc is of length $\pi N$, so the product goes to $0$. Therefore it’s enough to evaluate the integral over $\partial D$ and take the limit. By the residue theorem, it suffices to find the residues at singularities in $D$, i.e. zeros of $z^4+1$, which are at $e^{\pi i/4},e^{3\pi i/4},e^{5\pi i/4},e^{7\pi i/4}$, of which the only two in $D$ are $e^{\pi i/4}$ and $e^{3\pi i/4}$, at each of which $\frac{1}{z^4+1}$ has a simple pole with residue $\frac{1}{4z_0^3}$, which at these points gives $\frac{1}{4e^{3\pi i/4}}=-\frac{1}{4}{e}^{\pi i/4}=-\frac{\sqrt{2}}{8}(1+i)$ and $\frac{1}{4e^{9\pi i/4}}=\frac{1}{4}{e}^{-\pi i/4}=\frac{\sqrt{2}}{8}(1-i)$. Adding these together gives $-\frac{\sqrt{2}}{4}i$, so by the residue theorem the integral is $-2\pi i\cdot \frac{\sqrt{2}}{4}i=\frac{\pi \sqrt{2}}{2}=\frac{\pi }{\sqrt{2}}$.

Problem 7. Using the residue theorem, compute

$${\int }_{-\pi }^{\pi }\frac{1}{2+\sin \theta }𝑑\theta .$$

The above problem is directed towards Objective 9 (the residue theorem).

Solution. Writing $z=e^{i\theta }$, so $d\theta =\frac{dz}{iz}$ as in class and $\sin \theta =\frac{1}{2i}(z-1/z)$, this is

${\displaystyle {\int }_{|z|=1}}{\displaystyle \frac{1}{2+\frac{1}{2i}(z-1/z)}}\cdot {\displaystyle \frac{1}{iz}}𝑑z$ $=2{\displaystyle {\int }_{|z|=1}}{\displaystyle \frac{1}{4iz+z^2-1}}𝑑z.$

To apply the residue theorem, we need to find the singularities of the integrand, for which we need the zeros of $z^2+4iz-1$. A nice way to do this is to write $y=iz$, so $z^2+4iz-1=-y^2+4y-1$, so it suffices to find the zeros of $y^2-4y+1$, which are at $y=2\pm \sqrt{3}$ by the quadratic formula. Therefore the zeros of the original polynomial in $z$ are at $z=-iy=-i(2\pm \sqrt{3})$. Since $-i(2+\sqrt{3})$ is not in the unit disk but $-i(2-\sqrt{3})$ is, we only need the residue at the latter, which is ${\frac{1}{2z+4i}|}_{z=-i(2-\sqrt{3})}=\frac{1}{2i\sqrt{3}}$, and we have a leading factor of $2$, so the overall integral is $2\cdot \frac{2\pi i}{2i\sqrt{3}}=\frac{2\pi }{\sqrt{3}}$.