Week 3 Worksheet - Solutions

Home | Assessment | Notes | Worksheets | Blackboard

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 $T$.

Differentiation of Power Series

4. We multiple the right-hand side by $A-B$ to get $$\begin{array}{rl}& A(A^{n-1}+A^{n-2}{B}^1+\cdots +A^1{B}^{n-2}+B^{n-1})\\ & -B(A^{n-1}+A^{n-2}{B}^1+\cdots +A^1{B}^{n-2}+B^{n-1})\\ =& (A^n+A^{n-1}{B}^1+\cdots +A^2{B}^{n-2}+A^1{B}^{n-1})\\ & -(A^{n-1}{B}^1+A^{n-2}{B}^2+\cdots +A^1{B}^{n-1}+B^n)\\ =& A^n-B^n\end{array}$$ as desired.

5. 1. We can differentiate the power series for $1/(1-z)$ inside $\mathsf{B}(0,1)$ term-by-term. This gives $$\frac{1}{(1-z{)}^2}=1+2z+3z^2+4z^3+\cdots$$

   2. Replacing $z$ with $z^2$ in the geometric series gives $$\frac{1}{1-z^2}=1+z^2+z^4+z^6+\cdots$$ on $\mathsf{B}(0,1)$ because $|z{|}^2=|z||z|<1$ whenever $|z|<1$.

   3. Differentiating our answer to (b) on $\mathsf{B}(0,1)$ gives $${\displaystyle \frac{2z}{(1-z^2{)}^2}}=2z+4z^3+6z^5+\cdots$$ for all $|z|<1$. \qedhere

The Exponential Function and Trigonometric Functions

6. Fix $w=u+iv\ne 0$. We need to produce $z$ with $\exp (z)=w$. From $$\exp (x+iy)=e^x(\cos (y)+i\sin (y))$$ we should take $x=\ln (|w|)$ and $y=\mathsf{Arg}(w)$.

7. 1. Fix $z\in \mathbb{C}$. We calculate $$\begin{array}{rl}\exp (z+2\pi in)& =\exp (x+i(y+2\pi n))\\ & =e^x(\cos (y+2\pi n)+i\sin (y+2\pi n))\\ & =e^x(\cos (y)+i\sin (y))\\ & =\exp (z)\end{array}$$ because $\sin$ and $\cos$ have $2\pi n$ as a period for any $n\in \mathbb{Z}$.

   2. Suppose $\tau$ is a period of $\exp$. Then $\exp (z+\tau )=\exp (z)$ for all $z\in \mathbb{C}$. In particular $$\exp (z)=\exp (z+\tau )=\exp (z)\exp (\tau )$$ so $\exp (\tau )=1$. Write $\tau =a+ib$. Then $$1=e^a(\cos (b)+i\sin (b))$$ which forces $|1|=e^a$, $\cos (b)=1$ and $\sin (b)=0$. We must have $a=0$ and $b=2\pi n$ so $\exp$ has no other periods.

8. Write $$\exp (x+iy)=e^x(\cos (y)+i\sin (y))=e^x\cos (y)+i{e}^x\sin (y)$$ so $u(x,y)=e^x\cos (y)$ and $v(x,y)=e^x\sin (y)$. Then $$\begin{array}{rl}({\partial }_{1}u)(x,y)=e^x\cos (y)& ({\partial }_{2}u)(x,y)=-e^x\sin (y)\\ ({\partial }_{1}v)(x,y)=e^x\sin (y)& ({\partial }_{2}v)(x,y)=e^x\cos (y)\end{array}$$ and the Cauchy-Riemann equations are seen to be true for all $(x,y)$.

9. If $\exp (x+iy)$ is real then $e^x\sin (y)=0$ which is the same as $y=\pi n$ for some $n\in \mathbb{Z}$. If it is imaginary then $e^x\cos (y)=0$ which is the same as $y=\pi n+\frac{\pi }{2}$.

10. If $\exp (z)=-1$ then $e^x\cos (y)=-1$ and $e^x\sin (y)=0$ so $x=0$ and $y=2\pi n+\pi$ for some $n\in \mathbb{Z}$.

    If $\exp (z)=1+i$ then $e^x\cos (y)=1$ and $e^x\sin (y)=1$. We have $e^x=|1+i|=\sqrt{2}$ so $x=\frac{1}{2}\ln (2)$. From $\sin (y)=\frac{1}{\sqrt{2}}=\cos (y)$ we get $y=2\pi n+\frac{\pi }{4}$.

11. $$\cos (i)={\displaystyle \frac{\exp (i^2)+\exp (-i^2)}{2}}={\displaystyle \frac{\frac{1}{e}+e}{2}}={\displaystyle \frac{1+e^2}{2e}}$$ $$\sin (i)={\displaystyle \frac{\exp (i^2)-\exp (-i^2}{2i}}={\displaystyle \frac{\frac{1}{e}-e}{2i}}={\displaystyle \frac{1-e^2}{2ei}}$$

12. No because the odd (respectively even) coefficients are zero: the quotient $a_{n+1}/a_n$ is not defined for all $n\in \mathbb{N}$.

13. Calculate $$\begin{array}{rl}& \cos (z)\cos (w)-\sin (z)\sin (w)\\ =& {\displaystyle \frac{\exp (iz)+\exp (-iz)}{2}}\cdot {\displaystyle \frac{\exp (iw)+\exp (-iw)}{2}}\\ & -{\displaystyle \frac{\exp (iz)-\exp (-iz)}{2i}}\cdot {\displaystyle \frac{\exp (iw)-\exp (-iw)}{2i}}\\ =& {\displaystyle \frac{\exp (i(z+w))+\exp (i(z-w))+\exp (i(-z+w))+\exp (i(-z-w))}{4}}\\ & +{\displaystyle \frac{\exp (i(z+w))-\exp (i(z-w))-\exp (i(-z+w))+\exp (i(-z-w))}{4}}\\ =& {\displaystyle \frac{\exp (i(z+w))+\exp (-i(z+w))}{2}}=\cos (z+w)\end{array}$$ and $$\begin{array}{rl}& \sin (z)\cos (w)+\sin (w)\cos (z)\\ =& {\displaystyle \frac{\exp (iz)-\exp (-iz)}{2i}}\cdot {\displaystyle \frac{\exp (iw)+\exp (-iw)}{2}}\\ & +{\displaystyle \frac{\exp (iw)-\exp (-iw)}{2i}}\cdot {\displaystyle \frac{\exp (iz)+\exp (-iz)}{2}}\\ =& {\displaystyle \frac{\exp (i(z+w))+\exp (i(z-w))-\exp (i(-z+w))-\exp (i(-z-w))}{4i}}\\ & +{\displaystyle \frac{\exp (i(w+z))+\exp (i(w-z))-\exp (i(-w+z))-\exp (i(-w-z))}{4i}}\\ =& {\displaystyle \frac{\exp (i(w+z))-\exp (i(-w-z))}{2i}}=\sin (w+z)\end{array}$$

14. Calculate $$\begin{array}{rl}\cosh (z)+\sinh (z)& ={\displaystyle \frac{\exp (z)+\exp (-z)}{2}}+{\displaystyle \frac{\exp (z)-\exp (-z)}{2}}\\ & =\exp (z)\end{array}$$

15. Calculate $$\begin{array}{rl}& (\cosh (z){)}^2-(\sinh (z{)}^2\\ =& {\left({\displaystyle \frac{\exp (z)+\exp (-z)}{2}}\right)}^2-{\left({\displaystyle \frac{\exp (z)-\exp (-z)}{2}}\right)}^2\\ =& {\displaystyle \frac{\exp (2z)+2+\exp (-2z)}{4}}-{\displaystyle \frac{\exp (2z)-2+\exp (-2z)}{4}}=1\end{array}$$