\def\Re{\mathop{\rm Re}\nolimits}
\def\Im{\mathop{\rm Im}\nolimits}

% page 54
\noindent{\bf 7.}

By the triangle theorem,
$$|f(z)-w_0|+|w_0| \geq |f(z)| \to ||f(z)|-|w_0|| \leq
|f(z)-w_0|<\delta,$$
where $|z-z_0|<\epsilon$ by definition of the first limit.
% not sure how to prove the abs transformation

This shows that, as $z\to z_0,$ $|f(z)| \to |w_0|,$ because for all
$\epsilon,$ there exists $\delta$ such that $||f(z)|-|w_0||\leq\delta.$

\noindent{\bf 11.}

\noindent{\it (a)}

$\lim_{z\to\infty} T(z) = \infty$ is true iff $$\lim_{z\to 0}
T(z^{-1})^{-1} = 0 = \lim_{z\to 0} {d\over az^{-1}+b} = \lim_{z\to 0}
{d\over z^{-1}(a+bz)} = \lim_{z\to 0} {zd\over a+bz} =
(\lim_{z\to 0} zd)/(\lim_{z\to 0} a+bz).$$

$ad+bc \neq 0 \to ad \neq 0 \to a\neq0, d\neq0,$ so this becomes $0/a =
0,$ and the theorem is proved.

\noindent{\it (b)}

When $c\neq 0$, $\lim_{z\to\infty} T(z) = a/c$ if
$$\lim_{z\to 0} T(z^{-1}) = a/c = {az^{-1}+b\over cz^{-1}+d}
= {z^{-1}(a+bz)\over z^{-1}(c+dz)} = {a+bz\over c+dz} = {\lim_{z\to 0}
a+bz\over \lim_{z\to 0} c+dz} = a/c.$$

$\lim_{z\to -d/c} T(z) = \infty$ if $$\lim_{z\to -d/c} {cz+d\over az+b}
= 0 = {\lim_{z\to -d/c} cz+d\over\lim_{z\to -d/c} az+b} = {0\over k} = 0,$$

and $k\neq 0$ because $ad-bc\neq0 \to -ad/c + b \neq 0.$

% page 61
\noindent{\bf 2.}

\noindent{\it (a)}

$f'(z) = (3z^2)' + (-2z)' + 4' = 6z - 2,$ by the addition rule of
derivatives, $(af(x))' = af'(x),$ and the power rule.

\noindent{\it (b)}

By the chain rule, $g(w) = w^5,$ and $w = f(z) = 2z^2 + i,$ $$F'(z) =
g'(w)f'(z) = 5w^4 \cdot 4z = 20(2z^2+i)^4z,$$
with the power rule.

\noindent{\it (c)}

${z-1\over 2z+1} = (z-1)(2z+1)^{-1},$ so this can be solved with the
product and chain rules:
$${d\over dz} {z-1\over 2z+1} = (2z+1)^{-1} - {2(z-1)\over (2z+1)^2}.$$

\noindent{\it (d)}

With the product rule, and $f(z) = (1+z^2)^4$ and $g(z) = z^{-2},$
$$F'(z) = f'(z)g(z) + f(z)g'(z)
= {8z(1+z^2)^3\over z^2} - 2{(1+z^2)^4\over z^3}$$
because $f'(z) = 8z(1+z^2),$ by the chain rule with $h(z) = 1+z^2 \to
h'(z) = 2z$ and
$k(z) = w^4 \to k'(z) = 4w^3.$

\noindent{\bf 8.}

\noindent{\it (a)}

When $f(z) = \Re(z),$ $\delta w = \Re(z+\delta z) - \Re(z) = \Re(\delta
z).$ $f'(z) = \delta w/\delta z = \Re(\delta z)/\delta z$ is, if $z =
\epsilon,$ (and $\epsilon$ is real) 1, but if $z = i\epsilon,$ $f'(z) =
0/(i\epsilon) = 0.$ Therefore, the derivative DNE.

\noindent{\it (b)}

Similarly, $f(z) = \Im(z)$ gives $f'(z) = \Im(\delta z)/\delta z,$
which, for $z = i\epsilon,$ is 1, but for $z = \epsilon,$ $f'(z) = 0.$
Therefore, the derivative also DNE.

\bye