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+\def\arg{\mathop{arg}\nolimits}
+\let\rule\hrule
+\def\hrule{\goodbreak\medskip\rule\medskip\goodbreak}
+
+%page 293
+\noindent{\bf 2.}
+
+The graph of the contour image encircles the origin three times, giving
+$\Delta_C \arg f(z) = 2\pi\cdot 3 = 6\pi$ and, because it is analytic
+inside and on $C,$ $P = 0 \to 3 = Z - P = Z,$ or there are three zeroes,
+counting multiplicities, on $f(z)$ within $C.$
+
+\noindent{\bf 7.}
+
+\noindent{\it (a)}
+
+On $|z|=2,$
+$$36 = |9z^2| \geq 35 = 16 + 16 + 2 + 1 \geq |z^4 - 2z^3 + z - 1|,$$
+meaning the circle contains the same number of zeroes on the original
+and $9z^2,$ or 2, by Rouch\'e's theorem.
+
+\noindent{\it (b)}
+
+Similarly,
+$$32 = |z^5| \geq 29 = 24 + 4 + 1 \geq |3z^3+z^2+1|,$$
+giving 5 zeroes by the same method.
+
+\hrule
+%page 301
+\noindent{\bf 2.}
+
+$$w = iz + i = i(x+yi) + i = -y + (x+1)i = u+vi \to v = x+1,$$
+so if $x>0,$ $v>1.$
+
+\noindent{\bf 5.}
+
+$$w = (1-i)z = (1-i)(x+yi) = x - ix + yi + y = (x+y) + (y-x)i = u+vi,$$
+so by manipulating $u=x+y$ and $v=y-x,$
+$$y = (u+v)/2 > 1 \to u+v > 2 \leftrightarrow v > 2-u.$$
+
+\hrule
+%page 305
+\noindent{\bf 3.}
+
+$$y = {-v\over u^2+v^2} > c_2 \to c_2(u^2+v^2) + v < 0,$$
+which if $c_2>0$ becomes
+$$c_2u^2 + c_2(v+1/2c_2)^2 < 1/c_2^2,$$
+which is the interior of a circle.
+
+If $c_2 = 0,$ this is $v<0,$ or a half-plane, and if $c_2<0,$ the
+relation is inverted, and it is transformed to the exterior of a circle.
+
+\noindent{\bf 7.}
+
+This is first a leftward translation of $1$ and then the $w = 1/z$
+transform, or an inversion across $|z-1| = 1,$ then a reflection across
+the x-axis.
+
+\noindent{\bf 13.}
+
+With $z = r_0e^{i\theta},$
+
+$$w = z + {1\over z} = r_0e^{i\theta} + {1\over r_0}e^{-i\theta} =
+r_0(\cos\theta + i\sin\theta) + {1\over r_0}(\cos\theta - i\sin\theta) =
+u + vi$$
+
+$$u = (r_0+1/r_0)\cos\theta,\quad v = (r_0-1/r_0)\sin\theta.$$
+
+\hrule
+%page 311
+
+\noindent{\bf 3.}
+
+Since $0\mapsto\infty,$ $c\neq0$ and $d=0,$ giving
+$$w = {az+b\over cz}.$$
+$\infty\mapsto0,$ so $a=0,$ and $i\mapsto i$ gives
+$$i = {b\over ci} \to -c = b,$$
+thus
+$$w = {-c\over cz} = -{1\over z}$$
+
+\noindent{\bf 9.}
+
+If $f(0) = 0,$ and with $AB \neq 0,$
+$${w(w_2-w_3)\over (w-w_3)w_2} = {z(z_2-z_3)\over(z-z_3)z_2} \to
+Aw(z-z_3) = Bzw - Bzw_3 $$$$ w(A(z-z_3)-Bz) = - Bzw_3 \to w =
+{z\over (B-A)z/(Bw_3) - z_3/w_3},$$
+QED.
+
+\hrule
+%page 325
+
+\noindent{\bf 4.}
+
+$$w = e^z = e^{x+iy} = e^xe^{iy} = \rho e^{i\phi} \to \rho = e^x,\quad
+\phi = y.$$
+
+From the original constraints,
+$$0\leq\phi\leq\pi, \quad \rho\geq1.$$
+
+\def\li{\leavevmode\llap{\hbox to \parindent{\hfil$\bullet$\hfil}}}
+\noindent This gives boundaries:
+
+\li the inner semicircle of radius untiy, corresponding to the $x=0$
+boundary,
+
+\li the positive x-axis, corresponding to the $y=0$ boundary, and
+
+\li the negative x-axis, corresponding to the $y=\pi$ boundary.
+
+\noindent{\bf 9.}
+
+Using the following parametric representations,
+$$u = \sin x\cosh y,\quad v = \cos x\sinh y,$$
+the bottom of the rectangular region $y = a$ and $-\pi\leq x\leq\pi$
+becomes $u = \cosh a\sin x,$ $v = \sinh a\cos x,$ which is the
+parametrization of a complete ellipse with minor and major axes $\sinh
+a$ and $\cosh a,$ respectively.
+Similary, the upper bound parametrizes an ellipse with minor and major
+axes $\sinh b$ and $\cosh b.$
+This implies that the region between them continuously fills the space
+between them.
+
+The cut is between $x=\pi$ and $x=-\pi$ because $\sin\pi = \sin(-\pi)$
+and $\cos\pi=\cos(-\pi),$ meaning these boundaries overlap in the image.
+
+\bye