Mrthfx

This is the following code that I have so far:

documentclass{article}

usepackage[left=1.5in, right=1.5in, top=0.5in]{geometry}



usepackage{amsmath}

usepackage{mathtools}

usepackage{amsthm}



usepackage{graphicx}

usepackage{setspace}



onehalfspacing



DeclareMathOperator{Log}{Log}



begin{document}



title{Homework Chapter 5}

author{}



maketitle 



section*{Section 5.2: 4, 5, and 7}



paragraph{4.}

Let $alpha$ be a complex number. Show that if $(1+z)^{alpha}$ is taken 

as $e^{alpha Log(1+z)}$, then for $|z|< 1$

begin{equation*}

(1+z)^{alpha} = 1 + frac{alpha}{1}z + 

frac{alpha(alpha -1)}{1cdot2}z^{2} +

frac{alpha(alpha-1)(alpha-2)}{1cdot2cdot3}z^{3} + dotsb

end{equation*}

In general, 

begin{equation}

frac{d^{j}}{dz^{j}}(1+z_0)^{alpha}=

frac{alpha! , (1+z_0)^{alpha - j}}{(alpha - j)!}

end{equation}

We can prove result (1) by induction. For our base case, we let $j=1$. Hence,

begin{equation}

frac{d}{dz}(1+z_0)^{alpha} = alpha (1+z_0)^{alpha - 1}

end{equation}

Now for our induction hypothesis, we let $j=k$ and assume that

begin{equation}

frac{d^{k}}{dz^{k}}(1+z_0)^{alpha}=

frac{alpha! , (1+z_0)^{alpha - k}}{(alpha - k)!}

end{equation}

And so,

begin{equation}

frac{d}{dz}Big[frac{d^{k}}{dz^{k}}(1+z_0)^{alpha}Big] = frac{d}{dz}Big[frac{alpha! , (1+z_0)^{alpha - k}}{(alpha - k)!}Big] = frac{alpha !, (1+z_0)^{alpha -k - 1}}{(alpha -k -1)!} = frac{alpha !, (1+z_0)^{alpha-(k+1)}}{[alpha - (k+1)]!}

end{equation}

Since (3) holds for $j = k + 1$, it follows that (1) holds for all positive integers $j$. QED

Expanding (1), we have 

begin{equation}

frac{alpha! , (1+z_0)^{alpha - j}}{(alpha - j)!} = frac{alpha (alpha-1) dots (alpha-j)! , (1+z_0)^{alpha - j}}{(a-j)!} = alpha  (alpha - 1) dots (alpha - j + 1) (1+z_0)^{alpha - j}.

end{equation}

The Taylor series for $e^{alpha Log(1+z)}$ about $z_0=0$ is thus

begin{equation}

1 + sum_{j=1}^{infty} frac{alpha , (alpha - 1) dots (alpha - j + 1)}{j!}z^{j}

end{equation}



end{document}

How do I create vertical space between
Since (3) holds for $j = k + 1$, it follows that (1) holds for all positive integers $j$. QED

and

Expanding (1), we have 

I don't think it's necessary to create extra vertical whitespace above "Expanding (1), we have". A paragraph break (with indentation of the string "Expanding (1), we have") should be enough for your readers to note a "break" in the action.

Do replace the two instances of dots with either dotsb or cdots, though.

documentclass{article}

usepackage[hmargin=1.5in, top=0.5in]{geometry}



%%usepackage{amsmath}

usepackage{mathtools} % loads "amsmath" automatically

DeclareMathOperator{Log}{Log}

usepackage{amsthm,graphicx} 

usepackage{setspace}

onehalfspacing



begin{document}



title{Homework Chapter 5}

author{}



maketitle



section*{Section 5.2: 4, 5, and 7}



paragraph{4.}

Let $alpha$ be a complex number. Show that if $(1+z)^{alpha}$ is taken

as $e^{alpha Log(1+z)}$, then for $|z|< 1$,

begin{equation*}

(1+z)^{alpha} = 1 + frac{alpha}{1}z +

frac{alpha(alpha -1)}{1cdot2}z^{2} +

frac{alpha(alpha-1)(alpha-2)}{1cdot2cdot3}z^{3} + dotsb

end{equation*}

In general,

begin{equation} label{eq:1}

frac{d^{j}}{dz^{j}}(1+z_0)^{alpha}=

frac{alpha! , (1+z_0)^{alpha - j}}{(alpha - j)!}

end{equation}

% A blank line creates a paragraph break:



We can prove result eqref{eq:1} by induction. For our base case, 

we let $j=1$. Hence,

begin{equation}

frac{d}{dz}(1+z_0)^{alpha} = alpha (1+z_0)^{alpha - 1}

end{equation}

Now for our induction hypothesis, we let $j=k$ and assume that

begin{equation}label{eq:3}

frac{d^{k}}{dz^{k}}(1+z_0)^{alpha}=

frac{alpha! , (1+z_0)^{alpha - k}}{(alpha - k)!},.

end{equation}

And so,

begin{equation} 

frac{d}{dz}Bigl[frac{d^{k}}{dz^{k}}(1+z_0)^{alpha}Bigr] 

= frac{d}{dz}Bigl[frac{alpha! , (1+z_0)^{alpha - k}}{(alpha - k)!}Bigr] 

= frac{alpha !, (1+z_0)^{alpha -k - 1}}{(alpha -k -1)!} 

= frac{alpha !, (1+z_0)^{alpha-(k+1)}}{[alpha - (k+1)]!}

end{equation}

Since eqref{eq:3} holds for $j = k + 1$, it follows that eqref{eq:1} 

holds for all positive integers~$j$. QED

% A blank line creates a paragraph break:



Expanding eqref{eq:1}, we have

begin{equation}

frac{alpha! , (1+z_0)^{alpha - j}}{(alpha - j)!} 

= frac{alpha (alpha-1) dotsb (alpha-j)! , (1+z_0)^{alpha - j}}{(a-j)!} 

= alpha  (alpha - 1) dotsb (alpha - j + 1) (1+z_0)^{alpha - j}.

end{equation}

The Taylor series for $e^{alpha Log(1+z)}$ about $z_0=0$ is thus

begin{equation}

1 + sum_{j=1}^{infty} frac{alpha , (alpha - 1) dotsb (alpha - j + 1)}{j!}z^{j}

end{equation}



end{document}