Practice final exam solutions

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

Problem 1. Write ${(1+i\sqrt{3})}^3$ in Cartesian and polar coordinates.

The above problem is directed towards Objective 1 (complex numbers, algebra, and functions).

Solution. In polar coordinates, we find the modulus of $1+i\sqrt{3}$ is $r=\sqrt{1^2+{(\sqrt{3})}^2}=\sqrt{4}=2$, and the argument is $\theta =\arctan (\sqrt{3}/1)=\pi /3$. Thus $1+i\sqrt{3}=2e^{\pi i/3}$. Cubing this, we get ${(2e^{\pi i/3})}^3=8e^{\pi i}$. This is the polar form. In Cartesian coordinates, $8e^{\pi i}=8(\cos \pi +i\sin \pi )=-8$.

Problem 2. Find the branch points of the function $f(z)=\sqrt{z^2+2z}$. What are the corresponding phase factors at each branch point?

The above problem is directed towards Objective 1 (complex numbers, algebra, and functions).

Solution. We can rewrite the polynomial inside the square root as $z(z+2)$. The branch points correspond to the zeros of the polynomial, which are $z_1=0$ and $z_2=-2$. Since the function behaves locally like ${(z-z_k)}^{1/2}$ near each root $z_k$, transporting around either branch point once results in multiplying by a phase factor of $e^{2\pi i(1/2)}=e^{\pi i}=-1$.

Problem 3. Use the Cauchy–Riemann equations to determine the set of all points where $f(x+iy)=x^2+i{y}^2$ is complex differentiable. On what domain if any is $f$ analytic?

The above problem is directed towards Objective 2 (analytic functions and the Cauchy–Riemann equations).

Solution. Here $u(x,y)=x^2$ and $v(x,y)=y^2$. The partial derivatives are $u_x=2x,u_y=0$ and $v_x=0,v_y=2y$. The Cauchy–Riemann equations require $u_x=v_y$ and $u_y=-v_x$. The second equation $0=0$ is always satisfied, and the first equation requires $2x=2y$, or $x=y$. Since the partial derivatives are continuous everywhere, $f$ is complex differentiable precisely on the line $y=x$. This contains no domains (indeed it contains no nonempty open sets), so $f$ is nowhere analytic.

Problem 4. Show that if $f(z)$ is analytic on a domain $D$, and its modulus $|f(z)|$ is constant on $D$, then $f(z)$ must be constant.

The above problem is directed towards Objective 2 (analytic functions and the Cauchy–Riemann equations).

Solution. Let $|f(z)|=c$. If $c=0$, then $f(z)=0$ everywhere and is trivially constant. If $c\ne 0$, then $u^2+v^2=c^2$. Differentiating with respect to $x$ and $y$ gives $2u{u}_x+2v{v}_x=0$ and $2u{u}_y+2v{v}_y=0$. Using the Cauchy-Riemann equations $u_y=-v_x$ and $v_y=u_x$ in the second equation gives $-2u{v}_x+2v{u}_x=0$. Thus, we have the system of equations

$$u{u}_x+v{v}_x=0,$$

$$v{u}_x-u{v}_x=0.$$

Multiplying the first by $u$, the second by $v$, and adding them yields $(u^2+v^2)u_x=0$. Since $u^2+v^2=c^2\ne 0$, we must have $u_x=0$. If we instead multiplied the first equation by $v$ and the second by $u$ and subtracted, we would get $(v^2+u^2)v_x=c^2{v}_x=0$, so similarly $v_x=0$. Therefore by the Cauchy–Riemann equations $u_y$ and $v_y$ also vanish, so all the partial derivatives of $f$ vanish hence $f$ is constant.

Problem 5. Verify that $u(x,y)=e^x\cos y$ is harmonic on ${ℝ}^2$, and find a harmonic conjugate $v(x,y)$ for it.

The above problem is directed towards Objective 3 (harmonic functions).

Solution. We check the Laplacian of $u$. The partials are $u_x=e^x\cos y$, $u_{xx}=e^x\cos y$, and $u_y=-e^x\sin y$, $u_{yy}=-e^x\cos y$. Thus $\mathrm{\Delta }u=u_{xx}+u_{yy}=e^x\cos y-e^x\cos y=0$, so $u$ is harmonic everywhere. To find a harmonic conjugate $v$, we use the Cauchy-Riemann equations. We must have $v_y=u_x=e^x\cos y$. Integrating with respect to $y$ gives $v(x,y)=e^x\sin y+h(x)$. We also require $v_x=-u_y=e^x\sin y$. Differentiating our expression for $v$ with respect to $x$ yields $v_x=e^x\sin y+h^{\prime }(x)$. Comparing the two, we have $h^{\prime }(x)=0$, so $h(x)=C$ for some real constant $C$. A harmonic conjugate is thus $v(x,y)=e^x\sin y$.

Problem 6. Let $u(x,y)$ be a harmonic function on a domain $D$. Prove that the function $g(z)=u_x(x,y)-i{u}_y(x,y)$ is an analytic function on $D$.

The above problem is directed towards Objective 3 (harmonic functions).

Solution. Let $g(z)=U(x,y)+iV(x,y)$, where $U=u_x$ and $V=-u_y$. To show $g$ is analytic, we must verify the Cauchy-Riemann equations for $U$ and $V$, which are $U_x=V_y$ and $U_y=-V_x$. Checking the first: $U_x={(u_x)}_x=u_{xx}$ and $V_y={(-u_y)}_y=-u_{yy}$. Because $u$ is harmonic, $u_{xx}+u_{yy}=0$, which means $u_{xx}=-u_{yy}$, so $U_x=V_y$. Checking the second: $U_y=u_{xy}$ and $-V_x=-{(-u_y)}_x=u_{yx}$. Since partial derivatives commute, these are equal. Since $u$ is harmonic, all partial derivatives of $U$ and $V$ are continuous. Thus $g(z)$ is analytic.

Problem 7. Let $\gamma$ be the upper half of the unit circle $|z|=1$, oriented from $1$ to $-1$. Compute

$${\int }_{\gamma }\overline{z}𝑑z.$$

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

Solution. We can parameterize $\gamma$ by $z(t)=e^{it}$ for $0\le t\le \pi$. Then $dz=i{e}^{it}dt$. The integral becomes:

$${\int }_0^{\pi }\overline{e^{it}}\cdot i{e}^{it}𝑑t={\int }_0^{\pi }{e}^{-it}i{e}^{it}𝑑t={\int }_0^{\pi }i𝑑t=\pi i.$$

Problem 8. Let $\gamma$ be the straight line segment from $0$ to $i\sqrt{\pi }$. Compute

$${\int }_{\gamma }z\cos (z^2)𝑑z.$$

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

Solution. The integrand $f(z)=z\cos (z^2)$ is entire and has an analytic antiderivative on $ℂ$, namely $F(z)=\frac{1}{2}\sin (z^2)$. By the fundamental theorem of calculus, the integral depends only on the endpoints:

$${\int }_{\gamma }z\cos (z^2)𝑑z={\frac{1}{2}\sin (z^2)|}_0^{i\sqrt{\pi }}=\frac{1}{2}\sin ({(i\sqrt{\pi })}^2)-\frac{1}{2}\sin (0)=\frac{1}{2}\sin (-\pi )-0=0.$$

Problem 9. Let $C$ be the circle $|z|=1$. Compute

$${\int }_C\frac{e^z}{z(z-2)}𝑑z.$$

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

Solution. The integrand has isolated singularities at $z=0$ and $z=2$. Only $z=0$ lies inside the contour $C$. Let $f(z)=\frac{e^z}{z-2}$, which is analytic on the closed unit disk. By Cauchy’s integral formula,

$${\int }_C\frac{f(z)}{z}𝑑z=2\pi if(0)=2\pi i\left(\frac{e^0}{0-2}\right)=2\pi i\left(-\frac{1}{2}\right)=-\pi i.$$

Problem 10. Let $f$ be analytic inside and on a simple closed contour $\gamma$. Show that for any $z_0$ strictly inside $\gamma$,

$${\int }_{\gamma }\frac{f^{\prime }(z)}{z-z_0}𝑑z={\int }_{\gamma }\frac{f(z)}{{(z-z_0)}^2}𝑑z.$$

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

Solution. Since $f(z)$ is analytic, its derivative $f^{\prime }(z)$ is also analytic. By Cauchy’s integral formula applied to the function $f^{\prime }(z)$, the left side evaluates to $2\pi i{f}^{\prime }(z_0)$. Applying the Cauchy integral formula for derivatives to the function $f(z)$ on the right side, we also get $\frac{2\pi i}{1!}{f}^{\prime }(z_0)=2\pi i{f}^{\prime }(z_0)$. Since both integrals evaluate to $2\pi i{f}^{\prime }(z_0)$, they are equal.

Problem 11. Suppose $f$ is analytic in the open disk and satisfies $|f(z)|\le 5$ for all $z\in D$. Find an upper bound for $|f^{(3)}(0)|$. (For full credit, give the best possible bound, though you don’t need to prove it is the best one.)

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

Solution. We apply Cauchy’s estimates to the circle $C_R$ centered at $0$ with radius . Since $f$ is analytic in $D$, the Cauchy estimate formula gives

$$|f^{(n)}(0)|\le \frac{n!M_R}{R^n}$$

where $M_R={\max }_{|z|=R}|f(z)|$. Here we are given $n=3$ and that $|f(z)|\le 5$ for all $z\in D$, meaning $M_R\le 5$. Thus, for any , we have

$$|f^{(3)}(0)|\le \frac{3!\cdot 5}{R^3}=\frac{30}{R^3}.$$

Since this inequality holds for all , we can take the limit as $R\to 2$ from below to obtain the tightest bound:

$$|f^{(3)}(0)|\le \underset{R\to 2^{-}}{lim}\frac{30}{R^3}=\frac{30}{8}=\frac{15}{4}.$$

(In fact, this is achieved by $f(z)=\frac{5}{8}{z}^3$: note since , we always have , and $f^{(3)}(0)=\frac{5}{8}\cdot 6=\frac{15}{4}$, so this is the best possible bound.)

Problem 12. Show, without using Taylor series methods such as the identity theorem, that there does not exist an entire function $f:ℂ\to ℂ$ such that for all $|z|>1$, $f(z)=\frac{1}{z}$.

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

Solution. Since $|1/z|=1/|z|$ is bounded on $|z|>1$ and any entire function is bounded on the compact disk $|z|\le 1$, such a function $f$ would be entire and bounded, hence constant by Liouville’s theorem.

Problem 13. Find the Taylor series of $f(z)=\frac{1}{{(1-z)}^2}$ centered at $z_0=0$. What is its radius of convergence?

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

Solution. We start with the geometric series $\frac{1}{1-z}={\sum }_{n=0}^{\mathrm{\infty }}{z}^n$. Since we are inside the radius of convergence, we can differentiate term by term. Differentiating both sides with respect to $z$ yields $\frac{1}{{(1-z)}^2}={\sum }_{n=1}^{\mathrm{\infty }}n{z}^{n-1}={\sum }_{n=0}^{\mathrm{\infty }}(n+1)z^n$. The closest singularity to the center $0$ is at $z=1$, which is at distance $1$, so the radius of convergence is $1$.

Problem 14. Find the Taylor expansion of $f(z)=\frac{1}{1-z}$ at infinity.

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

Solution. Expanding at infinity means expanding in powers of $1/z$, which is equivalent to substituting $w=1/z$ and finding the Taylor series around $w=0$. We have $f(1/w)=\frac{1}{1-1/w}=\frac{w}{w-1}=-w\frac{1}{1-w}$. Expanding this as a geometric series gives $-w{\sum }_{n=0}^{\mathrm{\infty }}{w}^n=-{\sum }_{n=0}^{\mathrm{\infty }}{w}^{n+1}=-{\sum }_{n=1}^{\mathrm{\infty }}{w}^n$. Substituting back $w=1/z$, the expansion is $-{\sum }_{n=1}^{\mathrm{\infty }}\frac{1}{z^n}$.

Problem 15. Find the Laurent expansion of $f(z)=\frac{1}{z^2+1}$ valid in the punctured disk .

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

Solution. We want an expansion in powers of $z-i$. We factor the denominator as $z^2+1=(z-i)(z+i)$. We keep the $\frac{1}{z-i}$ factor and expand $\frac{1}{z+i}$ around $z=i$: $\frac{1}{z+i}=\frac{1}{(z-i)+2i}=\frac{1}{2i}\frac{1}{1+\frac{z-i}{2i}}$. Because , we have , so we can use a geometric series:

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

Multiplying by $\frac{1}{z-i}$ gives the final Laurent series:

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

Problem 16. Find and classify the isolated singularities of $f(z)=\frac{e^z-1}{z^2}$. For any poles, state their order.

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

Solution. The only singularity is at $z=0$. To classify it, we look at the Laurent series expansion. The Taylor series for $e^z-1$ is $z+\frac{z^2}{2!}+\frac{z^3}{3!}+\mathrm{\dots }$. Dividing by $z^2$, we get $f(z)=\frac{1}{z^2}\left(z+\frac{z^2}{2!}+\frac{z^3}{3!}+\mathrm{\dots }\right)=\frac{1}{z}+\frac{1}{2!}+\frac{z}{3!}+\mathrm{\dots }$. Since the lowest power of $z$ with a non-zero coefficient is $-1$, the singularity at $z=0$ is a simple pole (a pole of order $1$).

Problem 17. Using the residue theorem, compute

$${\int }_{|z|=2}\frac{1}{z^3(z+4)}𝑑z.$$

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

Solution. The singularities are at $z=0$ (a pole of order 3) and $z=-4$ (a simple pole). Only $z=0$ is inside the contour $|z|=2$. The residue at $0$ is the coefficient of $z^2$ in the Taylor series of $g(z)=\frac{1}{z+4}$ around $0$. We have $g(z)=\frac{1}{4(1+z/4)}=\frac{1}{4}(1-z/4+z^2/16-\mathrm{\dots })=\frac{1}{4}-\frac{z}{16}+\frac{z^2}{64}-\mathrm{\dots }$, so the coefficient of $z^2$ is $1/64$. By the residue theorem, the integral is $2\pi i{\mathrm{Res}}_0(f)=2\pi i\cdot \frac{1}{64}=\frac{\pi i}{32}$.

Problem 18. Using the residue theorem, compute

$${\int }_0^{2\pi }\frac{1}{3+2\cos \theta }𝑑\theta .$$

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

Solution. We substitute $z=e^{i\theta }$, so $dz=izd\theta$ and $\cos \theta =\frac{z+z^{-1}}{2}$. The integral becomes:

$${\int }_{|z|=1}\frac{1}{3+2\frac{z+z^{-1}}{2}}\frac{dz}{iz}=\frac{1}{i}{\int }_{|z|=1}\frac{1}{z(3+z+z^{-1})}𝑑z=\frac{1}{i}{\int }_{|z|=1}\frac{1}{z^2+3z+1}𝑑z.$$

The roots of the denominator are $z=\frac{-3\pm \sqrt{9-4}}{2}=\frac{-3\pm \sqrt{5}}{2}$. The root $z_1=\frac{-3+\sqrt{5}}{2}$ is inside the unit circle, while $z_2=\frac{-3-\sqrt{5}}{2}$ is outside. The residue at $z_1$ is $\frac{1}{2z_1+3}=\frac{1}{-3+\sqrt{5}+3}=\frac{1}{\sqrt{5}}$. The integral over the circle is $2\pi i(\frac{1}{\sqrt{5}})$, so the original integral is $\frac{1}{i}(2\pi i\frac{1}{\sqrt{5}})=\frac{2\pi }{\sqrt{5}}$.

Problem 19. Use Rouché’s theorem to find the number of zeros (with multiplicity) of the polynomial $p(z)=z^5+4z^2+1$ inside the unit disk .

The above problem is directed towards Objective 10 (special topics).

Solution. We split the polynomial $p(z)$ into two parts: let $f(z)=4z^2$ and $g(z)=z^5+1$. We want to compare the magnitudes of $f(z)$ and $g(z)$ on the boundary circle $|z|=1$.

On $|z|=1$, we have

$$|f(z)|=|4z^2|=4{|z|}^2=4{(1)}^2=4.$$

For $g(z)$, by the triangle inequality

$$|g(z)|=|z^5+1|\le {|z|}^5+1=1^5+1=2.$$

Since $4>2$, we have $|f(z)|>|g(z)|$ everywhere on the unit circle $|z|=1$, so by Rouché’s Theorem, $f(z)$ and $f(z)+g(z)=P(z)$ must have the same number of zeros strictly inside the unit circle . The function $f(z)=4z^2$ has a zero of order 2 at the origin, meaning it has exactly $2$ zeros inside . Therefore, $P(z)$ also has exactly $2$ zeros inside the unit disk.

Problem 20. Let $p(z)$ be a polynomial of degree $n\ge 2$. Prove that $p$ cannot be an analytic bijection from $ℂ$ to $ℂ$.

The above problem is directed towards Objective 10 (special topics).

Solution. If $p$ were an analytic bijection, in class we saw as a consequence that $p^{\prime }(z)$ is never zero. (If it were, then $p^{-1}$ would have derivative undefined at some points.) Since $p$ is a polynomial of degree $n\ge 2$, its derivative $p^{\prime }$ is a polynomial of degree $n-1\ge 1$, in particular non-constant; so by the fundamental theorem of algebra it has at least one zero in $ℂ$, so $p$ cannot be an analytic bijection $ℂ\to ℂ$.