# Why do I get so many “underfull box” and “runaway argument” errors? [closed]

The code below was working for a while, but now, when I compile it, I get many "underfull box" warnings and "runaway argument" errors. Why?

\documentclass{article}

\usepackage{tikz}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{fancyhdr}
\usepackage[T1]{fontenc}
\usepackage{fourier}

% document size parameters

\setlength{\oddsidemargin}{0.0in}
\setlength{\evensidemargin}{0.0in}
\setlength{\textheight}{8.4in}
\setlength{\textwidth}{6.5in}
\setlength{\voffset}{-0.40in}
\setlength{\parindent}{0pt}
\setlength{\parskip}{6pt}

\usepackage{environ}

\pagestyle{fancyplain}
\lhead{\large{{\bf MATH 122: Complex Analysis} }}

% user defined macros

\renewcommand\Re{\operatorname{Re}}
\renewcommand\Im{\operatorname{Im}}
\newcommand\Arg{\operatorname{Arg}}

\newcommand{\secskip}{\vspace{6pt}}
\newcommand{\vect}[1]{\vec{#1}}

\begin{document}

\begin{Large}
\large
\flushright\textbf{{Kalvin Ogbuefi}}\\

\flushleft

\begin{center}
\textbf{Homework Assignment \# 4}\\
\textbf{Due: Wednesday, February 25, 2014}
\end{center}
\end{Large}

\begin{large}

When you are asked to prove or disprove a statement you \textbf{must} use appropriate proof structure with complete sentences.

\begin{enumerate}

\item Use the $\epsilon-\delta$ definition of limits to show the following limits:
\begin{enumerate}
\item $\lim_{z \to z_0} \Re(z) = \Re(z_0)$\\

Let $\epsilon > 0$, therefore we choose $\delta = \epsilon$\\
Then, if $0<|z-z_0|<\delta=\epsilon$, we have:\\
\begin{align*}
|Re(z)-Re(z_0)|<\epsilon \\
&=|Re(z-z_0)|<\epsilon\\
&=|Re(z-z_0)|\leq|z-z_0|<\delta=\epsilon\\
\text{as desired.}
\end{align*}

\item $\lim_{z \to 1} \frac{1}{(z-1)^3} = \infty$\\

Let $\epsilon > 0$, we choose $\delta = (\epsilon)^{1/3}$\\
Then, if $0<|z-1|<\delta$ we have:\\
$$|z-1|^{3}<\delta^{3}=\epsilon$$\\
and so:\\
$$|\frac{1}{z-1}|^{3}<\frac{1}{\epsilon}$$\\
\begin{align*}
\text{as desired.}
\end{align*}
\item $\lim_{z \to 0} \frac{\overline{z}^2}{z} = 0$\\

Let $\epsilon>0$, we choose $\delta=\epsilon$\\
Then for:\\
$$0<|z-0|=|z|<\delta=\epsilon$$
We have:
\begin{align*}
&|\frac{\overline{z}^{2}}{z}-0|\\
&=|\frac{\overline{z}^{2}}{z}|\\
&=\frac{|\overline{z}|^{2}}{|z|}\\
&=\frac{|{z}|^{2}}{|z|}\\
&=|z|<\epsilon\\
\text{as desired.}
\end{align*}

\end{enumerate}

\item Find the following limits:

\begin{enumerate}
\item $\lim_{z \to 2 + i} (z^2 - 4z + 2 + 5i)$\\

First we begin by expanding the complex numbers then apply limit:\\
\begin{align*}
&=\lim_{z \to 2 + i} (x^2-y^2+2ixy - 4x-4iy + 2 + 5i)\\
&=\lim_{z \to 2 + i} (x^2-y^2-4x+2+i(2xy-4y+5))\\
&=-3+i5\\
\text{as desired.}
\end{align*}

\item $\lim_{z \to i} \frac{z^4 -1}{z-i}$\\

We begin by rewriting the limit according to the roots of unity:\\
$$\lim_{z \to i} \frac{(z-i)(z+i)(z+1)(z-1)}{(z-i)}$$
Simplifying yields:\\
$$\lim_{z \to i} (z+i)(z-1)(z+1)$$
Applying the limit will yield:\\
$$=-4i$$
\item $\lim_{z \to i} \frac{z^2 + 4z + 2}{z + 1}$\\
We begin by rewriting the limit:\\
$$\lim_{z \to i} \frac{x^2-y^2+4x+2+i(2xy+4y)}{x+1+iy}$$\\
Now we take the complex conjugate\\
$$\lim_{z \to i} \frac{x^2-y^2+4x+2+i(2xy+4y)}{x+1+iy} \frac{(x+1)-iy}{(x+1)-iy}$$
Simplifying yields:\\
$$\lim_{z \to i} \frac{(x^2-y^2+4x+2)+y(2xy+4y)+i((2xy+4y)(x+1)+y(x^2-y^2+4x+2))}{(x+1)^2+y^2}$$
Applying the limit will yield:\\
$$=\frac{5+i5}{2}$$
\end{enumerate}

\item Let $f(z) = \frac{z^2}{|z|^2}$
\begin{enumerate}
\item Find the limit as $z \to 0$ along the line $y = x$.

Let $f(z)=\frac{x^2-y^2+i(2xy)}{x^2+y^2}$\\

Along y=x, we have\\
\begin{align*}
&=\lim_{x \to 0} \frac{x^2-x^2+(2x x))}{x^2+x^2} \\
&=\lim_{x \to 0} \frac{i2x^2}{2x^2}\\
&=-i\\
\end{align*}
\item Find the limit as $z \to 0$ along the parabola $y = x^2$.\\
Along $y=x^2$ we find:\\
\begin{align*}
&=\lim_{x \to 0} \frac{x^2-x^4+(2x x^2))}{x^2+x^4} \\
&=\lim_{x \to 0} \frac{1-x^2+i2x}{1-x^2}\\
&=1\\
\end{align*}

\item What can you conclude about the limit of $f(z)$ as $z \to 0$?\\

Since the limits don't match the limit does not exist at that point.
\end{enumerate}

\item If $f(z)$ and $F(z)$ are two complex valued functions and
$$\lim_{z \to z_0} f(z) = w_0 \text{ and } \lim_{z \to z_0} F(z) = W_0$$
then show that
$$\lim_{z \to z_0} f(z) F(z) = w_0 W_0.$$\\

First of all, we have to use the $\epsilon-\delta$ definition of limits to use:\\
$$\lim_{z \to z_0} F(z) = W_0$$\\
There exists a $\delta$ s.t if $0<|z-z_0|<\delta$ then $|F(z)-W_0|<\sqrt{{\epsilon}}$\\
Likewise for:\\
$$\lim_{z \to z_0} f(z) = w_0$$\\
There exists a $\delta$ s.t if $0<|z-z_0|<\delta$ then $|f(z)-w_0|<\sqrt{{\epsilon}}$\\

So! If $\delta=min(\delta_1,\delta_2)$ , then:\\

\begin{align*}
&|f(z)-w_0||F(z)-W_0|\\
&=|(f(z)-w_0)(F(z)-W_0)|\\
&=|(f(z)-w_0)(F(z)-W_0)-0|\\
&<\sqrt{\epsilon}\sqrt{\epsilon}=\epsilon\\
\end{align*}
and so,\\
$$\lim_{z \to z_0} (f(z)-w_0)(F(z)-W_0)=0$$\\

Now to evaluate the limit using that answer we manipulate both sides to make the equation look like:\\
$$\lim_{z \to z_0} [(f(z)-w_0)(F(z)-W_0)+w_0W_0]=w_0W_0$$

If we foil the inside and use the properties of limits we'll find the solution to be:\\
$lim_{z \to z_0} [(F(z)f(z)-f(z)W_0-F(z)w_0+2w_0W_0]=w_0W_0$\\
$=lim_{z \to z_0} [F(z)f(z)]-lim_{z \to z_0} W_0[lim_{z \to z_0}f(z)]-w_0[lim_{z \to z_0}F(z)]+lim_{z \to z_0} 2w_0W_0=w_0W_0$\\
$$w_0W_0-2w_0W_0+2w_0W_0=$$\\
$$w_0W_0$$\\
So therefore through using the limit definition were we able to prove this limit.
\item Use mathematical induction, and the property you proved in the previous exercise to show:
$$\lim_{z \to z_0} z^n = z_0^n.$$\\

First we begin with the base case where $n=1$\\
$$lim_{z \to z_0} z^1=z_0^1$$\\

With the Inductive hypothesis we assume $n=k$ so $lim_{z \to z_0} z^k=z_0^k$.\\
Therefore we use this if $n=k+1$\\

$\lim_{z \to z_0} z^{k+1}=\lim_{z \to z_0} z^{k}z^1$\\
=$(\lim_{z \to z_0} z^k)(\lim_{z \to z_0} z^1)$\\
=$z_0^{k}z_0$\\
=$z_0^{k+1}$\\

\item  Let $f(z)$ be continuous for all values of $z$. Prove that $g(z) = f(\overline{z})$ is also continuous for all $z$.
Since this is true:\\
$$lim_{z \to z_0} f(z)=f(z_0)$$\\

We want to show that this is also true!:\\

$$lim_{z \to z_0} f(\overline{z})=f(\overline{z_0})$$\\

So, $|f(z)-f(z_0)|<\epsilon and |f(\overline{z})-f(\overline{z_0)|$\\

So, since:\\

\begin{align*}
&|z-z_0|\\
&=|\overline{z-z_0}|\\
&=|\overline{z}-\overline{z_0}|\\
\end{align*}

We have $0<|\overline{z}-\overline{z_0}|<\delta$ and so\\

$f(\overline{z})-f(\overline{z_0})<\epsilon$

\item Determine where the following functions are continuous:
\begin{enumerate}
\item $\frac{z+1}{z^2+1}$
\item $\frac{z^2 + 6z + 5}{z^2 + 3z + 2}$
\end{enumerate}

\item Are there any points where $f(z) = (\overline{z})^2$ is differentiable?  By definition the derivative (if it exists) is defined by the following limit:

$$\lim_{\Delta z \to 0} \frac{ f(z_0 + \Delta z) - f(z_0)}{ \Delta z}$$

Decide for yourself if the limit exists by considering the following:

\begin{enumerate}
\item Determine the limit as $z \to z_0$ along vertical lines:

$$\lim_{\Delta z \to 0} \frac{f(z_0 + \Delta z) - f(z_0)}{ \Delta z} = \lim_{\Delta y \to 0} \frac{ f\left( x_0 + i (y_0 + \Delta y)\right) - f(x_0,y_0)}{ i \Delta y}$$

\item Determine the limit as $z \to z_0$ along horizontal lines:

$$\lim_{\Delta z \to 0} \frac{f(z_0 + \Delta z) - f(z_0)}{ \Delta z} = \lim_{\Delta x \to 0} \frac{ f\left( (x_0 + \Delta x) + i y_0\right) - f(x_0,y_0)}{ \Delta x}$$

\item Decide for yourself where $f(z)$ is differentiable.
\end{enumerate}

\end{enumerate}

\end{large}

\end{document}

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## closed as off-topic by Werner, Mico, Svend Tveskæg, karlkoeller, GuidoFeb 26 '14 at 6:57

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See line 212 (should be \overline{z_0}). –  Werner Feb 26 '14 at 5:49
This question appears to be off-topic because it is very localized and is not likely to help a broader audience. –  Werner Feb 26 '14 at 5:50
Welcome to TeX.SX! All the underfull boxes are due to improper use of \\. Use blank lines (or \par) to start new lines and \bigskip when you need more spacing. –  karlkoeller Feb 26 '14 at 6:01