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0

This is what I have typed so far in latex:

documentclass[10pt,letter]{article}





usepackage{amsmath}

usepackage{amsthm}

usepackage{mathtools}





usepackage{graphicx}





usepackage{setspace}

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





onehalfspacing







begin{document}







title{Homework Chapter 5}



author{}









maketitle 







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



paragraph{4.}

Let $alpha$ be a complex number. Show that if $(1+z)^{alpha}$ is taken as $e^{alpha operatorname{Log}(1+z)}$, then for $|z|< 1$\



$(1+z)^{alpha} = 1 + displaystylefrac{alpha}{1}z + frac{alpha(alpha -1)}{1cdot2}z^{2} + frac{alpha(alpha-1)(alpha-2)}{1cdot2cdot3}z^{3} + cdots$\



In general, 



begin{equation}



displaystylefrac{d^{j}}{dz^{j}}(1+z)^{alpha}=frac{alpha !(1+z)^{alpha - j}}{(alpha - j)!}



end{equation}

I get an error with the following code:

begin{equation}



displaystylefrac{d^{j}}{dz^{j}}(1+z)^{alpha}=frac{alpha !(1+z)^{alpha - j}}{(alpha - j)!}



end{equation}

I was wondering if someone could clarify why this happens?

equations

share|improve this question

asked 13 mins ago

K.MK.M

1305

New contributor

K.M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  You can't have a blank line inside an equation environment.
  
  – Phelype Oleinik
  10 mins ago
  

  
  
  
  

add a comment | 

0

This is what I have typed so far in latex:

documentclass[10pt,letter]{article}





usepackage{amsmath}

usepackage{amsthm}

usepackage{mathtools}





usepackage{graphicx}





usepackage{setspace}

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





onehalfspacing







begin{document}







title{Homework Chapter 5}



author{}









maketitle 







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



paragraph{4.}

Let $alpha$ be a complex number. Show that if $(1+z)^{alpha}$ is taken as $e^{alpha operatorname{Log}(1+z)}$, then for $|z|< 1$\



$(1+z)^{alpha} = 1 + displaystylefrac{alpha}{1}z + frac{alpha(alpha -1)}{1cdot2}z^{2} + frac{alpha(alpha-1)(alpha-2)}{1cdot2cdot3}z^{3} + cdots$\



In general, 



begin{equation}



displaystylefrac{d^{j}}{dz^{j}}(1+z)^{alpha}=frac{alpha !(1+z)^{alpha - j}}{(alpha - j)!}



end{equation}

I get an error with the following code:

begin{equation}



displaystylefrac{d^{j}}{dz^{j}}(1+z)^{alpha}=frac{alpha !(1+z)^{alpha - j}}{(alpha - j)!}



end{equation}

I was wondering if someone could clarify why this happens?

equations

share|improve this question

asked 13 mins ago

K.MK.M

1305

New contributor

K.M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  You can't have a blank line inside an equation environment.
  
  – Phelype Oleinik
  10 mins ago
  

  
  
  
  

add a comment | 

This is what I have typed so far in latex:

documentclass[10pt,letter]{article}





usepackage{amsmath}

usepackage{amsthm}

usepackage{mathtools}





usepackage{graphicx}





usepackage{setspace}

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





onehalfspacing







begin{document}







title{Homework Chapter 5}



author{}









maketitle 







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



paragraph{4.}

Let $alpha$ be a complex number. Show that if $(1+z)^{alpha}$ is taken as $e^{alpha operatorname{Log}(1+z)}$, then for $|z|< 1$\



$(1+z)^{alpha} = 1 + displaystylefrac{alpha}{1}z + frac{alpha(alpha -1)}{1cdot2}z^{2} + frac{alpha(alpha-1)(alpha-2)}{1cdot2cdot3}z^{3} + cdots$\



In general, 



begin{equation}



displaystylefrac{d^{j}}{dz^{j}}(1+z)^{alpha}=frac{alpha !(1+z)^{alpha - j}}{(alpha - j)!}



end{equation}

I get an error with the following code:

begin{equation}



displaystylefrac{d^{j}}{dz^{j}}(1+z)^{alpha}=frac{alpha !(1+z)^{alpha - j}}{(alpha - j)!}



end{equation}

I was wondering if someone could clarify why this happens?

equations

share|improve this question

asked 13 mins ago

K.MK.M

1305

New contributor

K.M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

documentclass[10pt,letter]{article}





usepackage{amsmath}

usepackage{amsthm}

usepackage{mathtools}





usepackage{graphicx}





usepackage{setspace}

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





onehalfspacing







begin{document}







title{Homework Chapter 5}



author{}









maketitle 







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



paragraph{4.}

Let $alpha$ be a complex number. Show that if $(1+z)^{alpha}$ is taken as $e^{alpha operatorname{Log}(1+z)}$, then for $|z|< 1$\



$(1+z)^{alpha} = 1 + displaystylefrac{alpha}{1}z + frac{alpha(alpha -1)}{1cdot2}z^{2} + frac{alpha(alpha-1)(alpha-2)}{1cdot2cdot3}z^{3} + cdots$\



In general, 



begin{equation}



displaystylefrac{d^{j}}{dz^{j}}(1+z)^{alpha}=frac{alpha !(1+z)^{alpha - j}}{(alpha - j)!}



end{equation}

begin{equation}



displaystylefrac{d^{j}}{dz^{j}}(1+z)^{alpha}=frac{alpha !(1+z)^{alpha - j}}{(alpha - j)!}



end{equation}

share|improve this question

asked 13 mins ago

K.MK.M

1305

New contributor

K.M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

asked 13 mins ago

K.MK.M

1305

New contributor

K.M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 13 mins ago

K.MK.M

1305

New contributor

K.M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 13 mins ago

K.MK.M

1305