From 61b48fd8baae321daf294ebf670ef4906240d260 Mon Sep 17 00:00:00 2001
From: Holden Rohrer <hr@hrhr.dev>
Date: Wed, 5 May 2021 19:03:09 -0400
Subject: homeworks and attendance quizzes for kang

---
 kang/hw3.tex | 84 ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 84 insertions(+)
 create mode 100644 kang/hw3.tex

(limited to 'kang/hw3.tex')

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+\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
-- 
cgit