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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?



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\lhead{\large{{\bf MATH 122: Complex Analysis} }}
\rhead{\large{{\bf Spring Semester 2014}}}

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\flushright\textbf{{Kalvin Ogbuefi}}\\ 


\textbf{Homework Assignment \# 4}\\
\textbf{Due: Wednesday, February 25, 2014}


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


\item Use the $\epsilon-\delta$ definition of limits to show the following limits:
\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:\\ 
|Re(z)-Re(z_0)|<\epsilon \\
\text{as desired.}

\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:\\
and so:\\
\text{as desired.}
\item $\lim_{z \to 0} \frac{\overline{z}^2}{z} = 0$\\

Let $\epsilon>0$, we choose $\delta=\epsilon$\\ 
Then for:\\
We have:
\text{as desired.}


\item Find the following limits:

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

First we begin by expanding the complex numbers then apply limit:\\
&=\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))\\
\text{as desired.}

\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:\\
\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:\\

\item Let $f(z) = \frac{z^2}{|z|^2}$
\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\\
&=\lim_{x \to 0} \frac{x^2-x^2+(2x x))}{x^2+x^2} \\
&=\lim_{x \to 0} \frac{i2x^2}{2x^2}\\
\item Find the limit as $z \to 0$ along the parabola $y = x^2$.\\
Along $y=x^2$ we find:\\
&=\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}\\

\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.

\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:\\

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:\\
$lim_{z \to z_0} [(F(z)f(z)-f(z)W_0-F(z)w_0+2w_0W_0]=w_0W_0$\\
$=lim_{z \to z_0} [F(z)f(z)]-lim_{z \to z_0} W_0[lim_{z \to z_0}f(z)]-w_0[lim_{z \to z_0}F(z)]+lim_{z \to z_0} 2w_0W_0=w_0W_0$\\
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 $n=k+1$\\

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

\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:\\


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


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

\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:

\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. 



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closed as off-topic by Werner, Mico, Svend Tveskæg, karlkoeller, Guido Feb 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

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