homeworks and attendance quizzes for kang

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+\def\Log{\mathop{\rm Log}\nolimits}
+\let\rule\hrule
+\def\hrule{\medskip\rule\medskip}
+
+%page 195
+\noindent{\bf 1.}
+
+First,
+$$\cosh z = \sum_{n=0}^\infty {z^{2n}\over (2n)!}$$
+
+Substituting $z^2$ for $z$ and then multiplying each term by $z$ will
+give $z\cosh(z^2).$
+
+$$z{(z^2)^{2n}\over (2n)!} = {zz^{4n} \over (2n)!} = {z^{4n+1}\over
+(2n)!}.$$
+
+This gives
+$$z\cosh(z^2) = \sum_{n=0}^\infty {z^{4n+1}\over (2n)!}$$
+This applies over the entire plane because this function is entire.
+
+\noindent{\bf 3.}
+
+The Maclaurin expansion of
+$$f(z) = {z\over z^4+4} = {z\over 4}\cdot {1\over 1+(z^4/4)}$$
+may be found by substituting $-z^4/4$ in the place of $z$ in the
+standard $1/(1-z)$ Maclaurin expansion, then multiplying by $z/4.$
+
+This gives
+$$f(z) = \sum_{n=0}^\infty {z^{4n+1}\over 4^{n-1}}$$
+
+\noindent{\bf 7.}
+
+$$f(z) = \sin z = {e^{iz} - e^{-iz}\over 2}.$$
+$$f^{(n)}(z) = i^n{e^{iz} - (-1)^ne^{-iz}\over 2}.$$
+Where $n = 2k+1,$ where $k$ is an integer, $f^{(n)}(0)$ simplifies to
+$0$ because $e^{i0} - (-1)^{2k+1}e^{-i0} = 1 - 1 = 0.$
+Similarly, where $n = 2k,$ $f^{(n)}(0)$ simplifies to $i^{2k} = (-1)^k$
+because $e^{i0} - (-1)^{2k}e^{-i0} = 1 + 1 = 2,$ which substituted into
+the original equation becomes $i^{2k}{2\over 2}.$
+
+This provides an alternate justification of the Maclaurin expansion
+given in Section 64.
+Using the standard Maclaurin sum,
+$$f(z) = \sum_{n=0}^\infty {f^{(n)}(0)z^n\over n!} = \sum_{k=0}^\infty
+{f^{(2k)}(0)z^{2k}\over (2k)!} + \sum_{k=0}^\infty
+{f^{(2k+1)}(0)z^{2k+1}\over (2k+1)!} = \sum_{k=0}^\infty
+{(-1)^kz^{2k}\over (2k)!}$$
+
+\hrule
+%page 205
+\noindent{\bf 2.}
+
+$$f(z) = {1\over 1+z} = {1\over z}\cdot {1\over 1+(1/z)}$$
+
+By substitution into the standard $1/(1-z)$ expansion (with $-1/z$ being
+substituted for $z$),
+$${1\over 1+(1/z)} = \sum_{n=0}^\infty (-1)^n(1/z)^n =
+\sum_{n=0}^\infty {(-1)^n\over z^n}.$$
+$${1\over z} \cdot {1\over 1+(1/z)} = \sum_{n=0}^\infty {(-1)^n\over
+z^{n+1}} = \sum_{n=1}^\infty {(-1)^{n-1}\over z^n}$$
+
+The standard $1/(1-(-1/z))$ applies because
+$1<|z|<\infty \to |-1/z| < 1.$
+
+\noindent{\bf 3.}
+
+$${1\over z(1+z^2)} = {1\over z^3(1+(1/z^2))}$$
+
+With $1<|z|<\infty \to |-1/z^2|<1,$
+$${1\over (1-(-1/z^2))} = \sum_{n=0}^\infty (-1)^n z^{-2n}.$$
+$${1\over z^3(1-(-1/z^2))} = \sum_{n=0}^\infty (-1)^n z^{-2n-3}.$$
+
+\noindent{\bf 5.}
+
+% I know how to do this one. Just formulate 1/(z-1) and repeat for
+% 1/(z-2)
+
+$$f(z) = {-1\over (z-1)(z-2)} = {1\over z-1} - {1\over z-2}.$$
+
+Within $|z|<1,$
+$${1\over z-1} = -{1\over 1-z} = \sum_{n=0}^\infty (-1)z^n.$$
+On $|z|>1,$
+$${1\over z-1} = {1\over z}\cdot{1\over 1 - (1/z)} =
+\sum_{n=0}^\infty z^{-(n+1)}.$$
+
+Within $|z|<2,$
+$${1\over z-2} = -{1\over 2}\cdot{1\over 1 - z/2} =
+\sum_{n=0}^\infty {(-1)z^n\over 2^{n+1}}.$$
+On $|z|>2,$
+$${1\over z-2} = {1\over z}\cdot{1\over 1-(2/z)} = \sum_{n=0}^\infty
+2^n/z^(n+1)$$
+
+Within $|z|<1$ and $|z|<2,$ ($D_1$) these become
+$$\sum_{n=0}^\infty z^n\left({1\over 2^{n+1}} - 1\right).$$
+
+Within $|z|>1$ and $|z|<2,$ ($D_2$) these become
+$$\sum_{n=0}^\infty {z^n\over 2^{n+1}} + \sum_{n=1}^\infty {1\over z^n}.$$
+
+Within $|z|>1$ and $|z|>2,$ ($D_3$) these become
+$$\sum_{n=1}^\infty {1 - 2^{n-1}\over z^n}$$
+
+\hrule
+%page 218
+\noindent{\bf 1.}
+
+The first derivative of $1/(1-z)$ is $1/(1-z)^2,$ which will have the
+Maclaurin series equal to the termwise differentiation of the $1/(1-z)$
+Maclaurina series. Except $n = 0,$ (for which $z^n = z^0$ has first
+derivative 0) the first derivative of $z^n$ is $nz^{n-1}.$
+This gives summation
+$${1\over (1-z)^2} = \sum_{n=1}^\infty nz^{n-1} = \sum_{n=0}^\infty
+(n+1)z^n.$$
+
+Similarly, the second derivative of $1/(1-z)$ is $2/(1-z)^3.$
+The second derivative of $z^n$ for $n\geq2$ is $n(n-1)z^{n-2},$ giving
+summation
+$${2\over (1-z)^3} = \sum_{n=2}^\infty n(n-1)z^{n-2} = \sum_{n=0}^\infty
+(n+2)(n+1)z^n.$$
+
+\noindent{\bf 5.}
+
+$${\cos z\over z^2 - (\pi/2)^2} = {\cos z\over (z-\pi/2)(z+\pi/2)} =
+{\cos z\over \pi(z-\pi/2)} - {\cos z\over \pi(z+\pi/2)} =
+{-\sin(z-\pi/2)\over \pi(z-\pi/2)} - {\sin(z+\pi/2)\over \pi(z+\pi/2)}.$$
+
+As shown in the example in Section 71, $\sin w/w$ is analytic, so this
+function is analytic.
+
+\noindent{\bf 6.}
+
+With $|z-1|<1,$
+$$\int_1^z {1\over w}dw = \sum_{n=0}^\infty \int_1^z (-1)^n (w-1)^n dw
+\to \Log z = \sum_{n=0}^\infty (-1)^n (z-1)^{n+1}/(n+1) =
+\sum_{n=1}^\infty {(-1)^{n+1}(z-1)^n\over n}.$$
+
+\hrule
+%page 224
+\noindent{\bf 1.}
+
+On $0<|z|<1,$
+$$e^z = 1 + z + z^2/2 + z^3/6 + \cdots$$
+
+$${1\over z(z^2+1)} = 1/z - z + \cdots$$
+
+Taking their product gives
+
+$${e^z\over z(z^2+1)} = 1/z + 1 + z/2 + z^2/6 - z - z^2 + \cdots =
+1/z + 1 - z/2 - 5z^2/6 + \cdots$$
+
+\hrule
+%page 237
+\noindent{\bf 1.}
+% I don't understand residues at all.
+
+\noindent{\it (a)}
+
+$${1\over z+z^2} = {(1+z)^{-1}\over z},$$
+giving a residue of $1+z,$ or, at $z=0,$ $1.$
+
+\noindent{\it (b)}
+
+$$z\cos\left({1\over z}\right) = \sum_{n=0}^\infty (-1)^n {z^{1-2n}\over
+(2n)!}.$$
+The value at $z^{-1}$ or $1-2n = -1 \to n = 1$ is $-1/2.$
+
+\noindent{\it (c)}
+
+$${z-\sin z\over z} = 1 - \sum_{n=0}^\infty (-1)^n {z^{2n}\over (2n+1)!}$$
+has residue of $0$ because this is strictly a Taylor series.
+
+\noindent{\it (d)}
+
+Noting that $\cot z = 1/z - z/3 - z^3/45 + \ldots,$
+$${\cot z\over z^4} = {1\over z^4}(1/z - z/3 - z^3/45 + \ldots) =
+{1\over z^5} - {1\over 3z^3} - {1\over 45z} + \ldots,$$
+giving a residue of $-1/45.$
+
+\noindent{\it (e)}
+
+$${\sinh z\over z^4(1-z^2)} =
+z^{-4}(z+{z^3\over 3!}+\cdots)(1+z^2+z^4+\cdots) =
+z^{-3} + z^{-1}/6 + z^{-1} + z/6 + z + z^3/6,$$
+giving a residue of $7/6.$
+
+\bye