Week 3 Worksheet - Solutions

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Power Series

1. 1. Calculate $$\left|{\displaystyle \frac{2^{n+1}}{n+1}}/{\displaystyle \frac{2^n}{n}}\right|={\displaystyle \frac{2n}{n+1}}\to 2$$ as $n\to \mathrm{\infty }$ so the radius of convergence is $\frac{1}{2}$.

   2. Calculate $$\left|{\displaystyle \frac{(n+1)!}{n!}}\right|=n+1\to \mathrm{\infty }$$ as $n\to \mathrm{\infty }$ so the radius of convergence is $0$.

   3. Calculate $$\left|{\displaystyle \frac{(n+1{)}^p}{n^p}}\right|={\left({\displaystyle \frac{n+1}{n}}\right)}^p\to 1$$ as $n\to \mathrm{\infty }$ so the radius of convergence is $1$.

2. Calculate $$\left|{\displaystyle \frac{(-1{)}^{n+1}}{((n+1)!{)}^{2}2(n+1)}}/{\displaystyle \frac{(-1{)}^n}{(n!{)}^{2}2n}}\right|={\displaystyle \frac{n}{n+1}}\cdot {\displaystyle \frac{1}{(n+1{)}^2}}\to 0$$ as $n\to \mathrm{\infty }$ so $R=\mathrm{\infty }$.

3. If $R\ne S$ the radius of convergence will be $min\{R,S\}$. Indeed, if $|z|<min\{R,S\}$ then both power series converge, so their sum does as well, but if $min\{R,S\}<|z|<max\{S,R\}$ the sum cannot converge as then both series would.

   If $R=S$ then the sum can have any radius of convergence $T\ge R$. Indeed, suppose $b_n=-a_n$ and fix a power series $$\sum _{n=0}^{\mathrm{\infty }}{c}_n{z}^n$$ with radius of convergence $T\ge R$. Then $$\sum _{n=0}^{\mathrm{\infty }}(a_n+c_n)z^n\sum _{n=0}^{\mathrm{\infty }}(b_n+c_n)$$ both have radius of convergence $R$ by the previous paragraph, but their sum has radius of convergence $T$.

Holomorphic Functions

4. 1. We calculate $f(z+h)=z^2+2zh+h^2+z+h$ so that $${\displaystyle \frac{f(z+h)-f(z)}{h}}={\displaystyle \frac{2zh+h^2+h}{h}}=2z+h+1$$ and $$\underset{h\to 0}{lim}{\displaystyle \frac{f(z+h)-f(z)}{h}}=\underset{h\to 0}{lim}2z+h+1=2z+1$$ for all $z\in \mathbb{C}$.

   2. We calculate $$\begin{array}{rl}{\displaystyle \frac{f(z+h)-f(z)}{h}}& ={\displaystyle \frac{{\displaystyle \frac{1}{z+h}}-{\displaystyle \frac{1}{z}}}{h}}={\displaystyle \frac{{\displaystyle \frac{-h}{(z+h)z}}}{h}}={\displaystyle \frac{-1}{(z+h)z}}\end{array}$$ and $$\underset{h\to 0}{lim}{\displaystyle \frac{f(z+h)-f(z)}{h}}=\underset{h\to 0}{lim}{\displaystyle \frac{-1}{(z+h)z}}=-{\displaystyle \frac{1}{z^2}}$$ for all $z\in \mathbb{C}\setminus \{0\}$.

5. 1. From the definition $$\underset{h\to 0}{lim}{\displaystyle \frac{f(h)-f(0)}{h}}={\displaystyle \frac{|h{|}^2}{h}}=\underset{h\to 0}{lim}{\displaystyle \frac{h\bar{h}}{h}}=\underset{h\to 0}{lim}\bar{h}=0$$ so the function has a derivative of zero at the origin.

   2. Fix $z\in \mathbb{C}$. The limit $$\begin{array}{rl}\underset{h\to 0}{lim}{\displaystyle \frac{f(z+h)-f(z)}{h}}& =\underset{h\to 0}{lim}{\displaystyle \frac{(z+h)\overline{(z+h)}-z\bar{z}}{h}}\\ & =\underset{h\to 0}{lim}{\displaystyle \frac{z\bar{h}+\bar{z}h+|h{|}^2}{h}}\end{array}$$ does not exist because if $h=t$ is real we get $$\underset{t\to 0}{lim}{\displaystyle \frac{zt+\bar{z}t+t^2}{t}}=z+\bar{z}$$ whereas if $h=it$ is imaginary we get $$\underset{t\to 0}{lim}{\displaystyle \frac{zit-\bar{z}it+t^2}{it}}=\bar{z}-z$$ so $f$ is not differentiable at any non-zero $z$. It is therefore not holomorphic on any domain.

Cauchy-Riemann Equations

6. From the definition $$\begin{array}{rl}({\partial }_{1}h)(x,y)& =\underset{t\to 0}{lim}{\displaystyle \frac{h(x+t,y)-h(x,y)}{t}}\\ & =\underset{t\to 0}{lim}{\displaystyle \frac{2(x+t)y-2xy}{t}}=\underset{t\to 0}{lim}2y=2y\end{array}$$ and $$\begin{array}{rl}({\partial }_{2}h)(x,y)& =\underset{t\to 0}{lim}{\displaystyle \frac{h(x,y+t)-h(x,y)}{t}}\\ & =\underset{t\to 0}{lim}{\displaystyle \frac{2x(y+t)-2xy}{t}}=\underset{t\to 0}{lim}2x=2x\end{array}$$ as expected from rules of differentiation.

7. From $$\frac{1}{z}=\frac{x-iy}{x^2+y^2}=\frac{x}{x^2+y^2}+i\frac{(-y)}{x^2+y^2}$$ we get $u(x,y)$ and $v(x,y)$. The partial derivatives are $$\begin{array}{rl}({\partial }_{1}u)(x,y)& ={\displaystyle \frac{(x^2+y^2)-2x(x)}{(x^2+y^2{)}^2}}={\displaystyle \frac{y^2-x^2}{(x^2+y^2{)}^2}}\\ ({\partial }_{2}u)(x,y)& ={\displaystyle \frac{-2xy}{(x^2+y^2{)}^2}}\end{array}$$ and $$\begin{array}{rl}({\partial }_{1}v)(x,y)& ={\displaystyle \frac{2xy}{(x^2+y^2{)}^2}}\\ ({\partial }_{2}u)(x,y)& ={\displaystyle \frac{-(x^2+y^2)+2y(y)}{(x^2+y^2{)}^2}}={\displaystyle \frac{y^2-x^2}{(x^2+y^2{)}^2}}\end{array}$$ so the Cauchy-Riemann equations are satisfied.

8. For this function $u(x,y)=\sqrt{x^2+y^2}$ and $v(x,y)=0$. Thus $$({\partial }_{1}u)(x,y)={\displaystyle \frac{x}{\sqrt{x^2+y^2}}}({\partial }_{2}u)(x,y)={\displaystyle \frac{y}{\sqrt{x^2+y^2}}}$$ and $({\partial }_{1}v)(x,y)=({\partial }_{2}v)(x,y)=0$ for all $(x,y)\ne (0,0)$. The Cauchy- Riemann equations are therefore not satisfied at any non-zero complex number.

   Since the partial derivatives of $u$ do not even exist at $(0,0)$ the function $f$ is certainly not differentiable there.

9. We calculate $$\begin{array}{rl}({\partial }_{1}u)(x,y)& =3x^2-3y^2\\ ({\partial }_{2}u)(x,y)& =-6xy\\ ({\partial }_{1}v)(x,y)& =6xy\\ ({\partial }_{2}v)(x,y)& =3x^2-3y^2\end{array}$$ and see that the equations are satisfied for all $(x,y)$.

   From $$u(x,y)+iv(x,y)=x^3+3i{x}^{2}y-3x{y}^2-y^3$$ we see that $$f(z)=z^3=(x+iy{)}^3$$ has the above real and imaginary parts.

10. We write $${\displaystyle \frac{1}{(x+iy{)}^4}}={\displaystyle \frac{(x-iy{)}^4}{(x^2+y^2{)}^4}}={\displaystyle \frac{x^4-4i{x}^{3}y-6x^2{y}^2+4ix{y}^3+y^4}{(x^2+y^2{)}^4}}$$ so that $u$ and $v$ are the real and imaginary parts of $f(z)=\frac{1}{z^4}$ on $\mathbb{C}\setminus \{0\}$. Since $f$ is holomorphic on that domain, the Cauchy-Riemann equations are satisfied for all $z$ therein.

11. We calculate $$\begin{array}{rl}{\partial }_1({\partial }_{1}u)+{\partial }_2({\partial }_{2}u)& ={\partial }_1({\partial }_{2}v)+{\partial }_2(-{\partial }_{1}v)\\ & ={\partial }_1({\partial }_{2}v)-{\partial }_2({\partial }_{1}v)=0\end{array}$$ by Clairaut's theorem.

12. From an earlier problem we have $$u(x,y)=x^3-3x{y}^{2}v(x,y)=3x^{2}y-y^3$$ and can calculate that $$\begin{array}{rl}{\partial }_1({\partial }_{1}u)+{\partial }_2({\partial }_{2}u)& ={\partial }_1(3x^2-3y^2)+{\partial }_2(-6xy)=6x-6x=0\\ {\partial }_1({\partial }_{1}v)+{\partial }_2({\partial }_{2}v)& ={\partial }_1(3x^2-3y^2)+{\partial }_2(-6xy)=6x-6x=0\end{array}$$ as desired.

13. We have $$\begin{array}{rl}({\partial }_{1}u)(x,y)& =5x^4-30x^2{y}^2+5y^4\\ ({\partial }_{2}u)(x,y)& =-20x^{3}y+20x{y}^3\end{array}$$ so $$\begin{array}{rl}({\partial }_{1}v)(x,y)& =20x^{3}y-20x{y}^3\\ ({\partial }_{2}v)(x,y)& =5x^4-30x^2{y}^2+5y^4\end{array}$$ from the Cauchy-Riemann equations. Partially integrating gives $$5x^{4}y-10x^2{y}^3+y^5+h(x)=v(x,y)=5x^{4}y-10x^2{y}^3+g(y)$$ which will be satisfied if e.g. $h(x)=0$ and $g(y)=y^5$. Then $$f(x+iy)=(x^5-10x^3{y}^2+5x{y}^4)+i(5x^{4}y-10x^2{y}^3+y^5)=(x+iy{)}^5$$ and $f(z)=z^5$.

14. We have $$\begin{array}{rl}({\partial }_{1}u)(x,y)& =3x^2-k{y}^2+12y-12\\ ({\partial }_{2}u)(x,y)& =-2kxy+12x\end{array}$$ so $$\begin{array}{rl}({\partial }_{1}v)(x,y)& =2kxy-12x\\ ({\partial }_{2}v)(x,y)& =3x^2-k{y}^2+12y-12\end{array}$$ from the Cauchy-Riemann equations. Partially integrating gives $$3x^{2}y-\frac{k}{3}{y}^3+6y^2-12y+h(x)=v(x,y)=k{x}^{2}y-6x^2+g(y)$$ and therefore $$3x^{2}y=k{x}^{2}y$$ if $h(x)=-6x^2$ and $g(y)=-\frac{k}{3}{y}^3+6y^2-12y$. Therefore, the only value of $k$ for which $u$ could possibly be the real part of a holomorphic function is $3$. When $k=3$ we have $$v(x,y)=3x^{2}y-6x^2-y^3+6y^2-12y$$ and $u,v$ are the real and imaginary parts of $f(z)=z^3+6i{z}^2-12z$.

15. If $u$ is constant and the real part of a holomorphic function then the Cauchy-Riemann equations would force $({\partial }_{1}v)(x,y)=0=({\partial }_{2}v)(x,y)$ whence $v$ is constant as well.

16. In the purported situation the Cauchy-Riemann equations become $$\begin{array}{c}{u}^{\prime }(x)=({\partial }_{1}u)(x,y)=({\partial }_{2}v)(x,y)=v^{\prime }(y)\\ 0=({\partial }_{2}u)(x,y)=-({\partial }_{1}v)(x,y)=0\end{array}$$ which tells us that $u^{\prime }$ and $v^{\prime }$ are constant. But then $u(x)=\mu x+b$ and $v(y)=\nu y+c$. Put $\lambda =\mu +i\nu$ and $a=b+ic$.

17. If we apply ${\partial }_1$ to the equation we get $$0=2({\partial }_{1}u)(x,y)+({\partial }_{1}v)(x,y)=2({\partial }_{2}v)(x,y)+({\partial }_{1}v)(x,y)$$ from the Cauchy-Riemann equations. We get $$0=2({\partial }_{2}u)(x,y)+({\partial }_{2}v)(x,y)=-2({\partial }_{1}v)(x,y)+({\partial }_{2}v)(x,y)$$ if we apply instead ${\partial }_2$. Adding twice the first equation to the second gives $5({\partial }_{2}v)(x,y)=0$ for all $(x,y)$. It then follows from the second equation that $({\partial }_{1}v)(x,y)=0$ for all $(x,y)$. The Cauchy-Riemann equations then gives $({\partial }_{1}u)(x,y)=0$ and $({\partial }_{2}u)(x,y)=0$ for all $(x,y)$. We conclude that $u$ and $v$ are constant, so that $f$ is as well.

A Cauchy-Riemann Converse

18. 1. Write $h=a+ib$. We need to verify that the limit $$\underset{h\to 0}{lim}{\displaystyle \frac{f(h)-f(0)}{h}}=\underset{h\to 0}{lim}{\displaystyle \frac{\sqrt{|ab|}}{h}}$$ does not exist. Taking $h=t$ with $t$ real gives the limit $$\underset{t\to 0}{lim}{\displaystyle \frac{\sqrt{0}}{t}}=0$$ while taking $h=t+it$ with $t$ real gives the limit $$\underset{t\to 0}{lim}{\displaystyle \frac{\sqrt{t^2}}{t+it}}={\displaystyle \frac{1}{1+i}}$$ so the complex limit does not exist and $f$ is not differentiable at $0$.

    2. We have $u(x,y)=\sqrt{|xy|}$ and $v(x,y)=0$. The partial derivatives at the origin are $$\begin{array}{rl}({\partial }_{1}u)(0,0)& =\underset{t\to 0}{lim}{\displaystyle \frac{u(t,0)-u(0,0)}{t}}=\underset{t\to 0}{lim}0=0\\ ({\partial }_{2}u)(0,0)& =\underset{t\to 0}{lim}{\displaystyle \frac{u(0,t)-u(0,0)}{t}}=\underset{t\to 0}{lim}0=0\end{array}$$ and $({\partial }_{1}v)(x,y)=0=({\partial }_{2}v)(x,y)$. The Cauchy-Riemann equations are certainly satisfied at the origin. However, for example, the partial derivative ${\partial }_{2}u$ is not defined on any ball centered at the origin, because $$\underset{t\to 0}{lim}{\displaystyle \frac{u(a,t)-u(a,0)}{t}}=\underset{t\to 0}{lim}{\displaystyle \frac{\sqrt{at}}{t}}=\underset{t\to 0}{lim}{\displaystyle \frac{\sqrt{a}}{\sqrt{t}}}$$ does not exist and therefore $({\partial }_{2}u)(a,0)$ does not exist for any $a>0$.