\magnification=\magstep1
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\noindent
{\it Comments and Solutions}
\bigskip
\noindent
3J:  Since $h(z) = \pi^2\csc^2(\pi z) - \sum_{n=-\infty}^\infty (z - n)^{-2}$, you also 
need to show that the first term is periodic with period $\pi$.  This comes from the 
trig identity (addition formula) $\sin(z + w) = \sin(z)\cos(w) + \sin(w)\cos(z)$.  So 
$$\sin(\pi z + \pi) = \sin(\pi z)\cos(\pi) + \sin(\pi)\cos(\pi z) = -\sin(\pi z).$$
Hence
$$\pi \csc^2(\pi z + \pi) = {\pi\over \sin^2(\pi z + \pi)} = {\pi \over (-\sin(\pi z))^2} = 
\pi \csc^2(\pi z).$$

Then, you also need to show that $h(z/2) + h((z+1)/2) = 4h(z)$.  In fact, each of the two
parts in the definition of $h(z)$ satisfies the same identity.  The infinite series part
works like this:  substitute $z/2$ and $(z + 1)/2$, then rearrange as follows:
$$\eqalign{
\sum_{n=-\infty}^\infty \left({z\over 2} - n\right)^{-2} + \sum_{n=-\infty}^\infty \left({ z+1\over 2} - n\right)^{-2}
&= 4\sum_{n=-\infty}^\infty (z - 2n)^{-2} + 4\sum_{n=-\infty}^\infty (z - (2n-1))^{-2}\cr
&= 4\sum_{k=-\infty}^\infty (z - k)^{-2},\cr}$$
since the first sum gives all the terms for even $k$ and the second sum gives all the 
terms for odd $k$.  The series can be rearranged to get this since they are absolutely convergent 
on all compact subsets of ${\bf C}$. 

\bigskip
The other term satisfies the same identity because of trigonometric identities, including 
the addition formula for the $\sin$ function again.  We have
$$\eqalign{
\pi^2\csc^2(\pi z/2) + \pi^2\csc^2(\pi (z + 1)/2) &= 
{\pi^2\over \sin^2(\pi z/2)} + {\pi^2 \over \sin^2(\pi z/2 + \pi/2)}\cr
&= {\pi^2\over \sin^2(\pi z/2)} + {\pi^2 \over \cos^2(\pi z/2)}\cr
&= {\pi^2 (\cos^2(\pi z) + \sin^2(\pi z))\over (\sin(\pi z/2)\cos(\pi z/2))^2}\cr
&= {\pi^2 \over \sin^2(\pi z)/4}\cr
&= {4\pi^2 \over \sin^2(\pi z)}\cr
&= 4\pi^2 \csc^2(\pi z).\cr}$$

\noindent
3P:  You showed correctly that $\wp'(z)$ and 
$$f(z) = {\sigma(z - \omega_1/2) \sigma(z - \omega_2/2) \sigma(z - \omega_3/2)\over 
\sigma(\omega_1/2)  \sigma(\omega_2/2) \sigma(\omega_3/2) \sigma^3(z)}$$
have the same poles and zeroes with the same orders.  But then you need to think about
the Laurent series expansions of the two sides at $z = 0$ to see the correct constant $c$
so that $\wp'(z) = c f(z)$.  You can do that as follows.  Since $\sigma(z)$ has a
single zero at $0$ in the usual period parallelogram, and 
$$\sigma(z) = z \prod_{\omega \in \Omega} \left(1 - {z\over \omega}\right)
\exp\left({z\over \omega} + {z^2\over 2\omega^2}\right)$$
we can write $\sigma(z) = z h(z)$
where $h(z)$ is holomorphic for all $z$.  Therefore, $\sigma'(z) = zh'(z) + h(z)$ and we have
$\sigma'(0) = 0 + h(0) = 1$.  Hence
$$\sigma(z) = z + {\rm h.o.t.}$$
(higher order terms).  Similarly, expanding at $z = 0$, and using the fact that $\sigma$ is 
an odd function,
$$\sigma(z - \omega_i/2) = \sigma(-\omega_i/2) + {\rm h.o.t.} = -\sigma(\omega_i/2) + {\rm h.o.t.}.$$
Hence 
$$f(z) = {- 1 + {\rm h.o.t} \over z^3 (1 + {\rm h.o.t})^3} = {-1\over z^3} + {\rm h.o.t.}.$$
Comparing this with the Laurent expansion of $\wp'(z)$:
$$\wp'(z) = {-2\over z^3} + {\rm h.o.t.}$$
we see that $c = 2$.  

\bigskip
\noindent
3S:  For this you need to use 3.14.3 and 3.14.4.   The function $\zeta(z - b)$ has
a simple pole at $b$ with residue 1, but it's not elliptic (i.e. not $\Omega$-periodic).
However, 3.14.3 says that a sum $\sum_i c_i \zeta(z - b_i)$ is elliptic as long
as $\sum_i c_i = 0$.  Hence using that and the rest of 3.14.4, an appropriate function is 
$$f(z) = \zeta(z - b_1) - \zeta(z - b_2) + 2\wp(z - b).$$

\bigskip
\noindent
3U:  I think you misread the question here.  The first part:  ``Prove that 
$$\wp(u) - \wp(v) = {\sigma(v - u) \sigma(v + u) \over \sigma^2(u) \sigma^2(v)}.''$$
is a separate statement that needs to be done first.  (The other parts follow
from that; they don't prove this part.)  The way to show this is pretty similar to 3P
above.  The idea is that if you fix $v \ne 0$ and think of $u$ as varying, then 
the left side $\wp(u) - \wp(v)$ has a pole of order 2 at $u = 0$ and zeros at
$u = \pm v$ (recall that $\wp$ is even).  The right side has zeroes at $u = \pm v$ and 
a pole of order 2 at $u = 0$ as well.  Moreover the Laurent expansions of both sides
at $u = 0$ 
start ${1\over u^2} + {\rm h.o.t}$, so this time the constant is $1$.  
\bigskip
\noindent
Lisa,
\bigskip
I'm not sure what happened here, but this was not your best work.  You were missing
several big chunks of what were not very difficult problems here.   I'm guessing
you might have been ``crunched for time'' due to other things the past week.  In 
any case, your course grade for independent study will be a B+.  If you have any 
questions, please contact me.  
\bigskip
\noindent
John Little
\end 
