Cfrgtkky

Th following is the full code that produces the image shown below:

documentclass{article}

usepackage[utf8]{inputenc}

usepackage[english]{babel}



usepackage{amsthm}

usepackage{amsmath}

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





newtheorem{definition}{Definition}

newtheorem{theorem}{Theorem} 

theoremstyle{remark}



begin{document}

    title{Extra Credit}

    maketitle



    begin{definition}

        If f is analytic at $z_0$, then the series



        begin{equation}

            f(z_0) + f'(z_0)(z-z_0) + frac{f''(z_0)}{2!}(z-z_0)^2 + cdots = sum_{n=0}^{infty} frac{f^{(n)}(z_0)}{n!}(z-z_0)^n

        end{equation}



        is called the Taylor series for f around $z_0$.

    end{definition}



    begin{theorem}

        If f is analytic inside and on the simple closed positively oriented contour $Gamma$ and if $z_0$ is any point inside $Gamma$, then

        begin{equation}

            f^{(n)}(z_0) = frac{n!}{2pi i} int_{Gamma} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta hspace{1cm} (n=1,2,3, cdots )

        end{equation}

    end{theorem}

    hrulefill







begin{theorem}

If f is analytic in the disk $|z-z_0|<R'$, then the Taylor series $(1)$ converges to $f(z)$ for all $z$ in this disk. 

end{theorem}



begin{proof}

Suppose that the function textit{f} is analytic in the disk $|z-z_0|<R'$. We can define a positively oriented contour $C$ as $$ C:=Big{z:|z-z_0|=frac{R + R'}{2}, 0<R< R'  Big}.$$ Letting $zeta$ be an arbitrary point on $C$ and applying Theorem 1 to $(1)$, we get 



begin{equation}

sum_{n=0}^{infty} frac{(z-z_0)^{n}}{2pi i} int_{C} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta.

end{equation}\ 



Or equivalently, we have that





begin{align*}

  sum_{n=0}^{infty} frac{(z-z_0)^{n}}{2pi i} int_{C} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta

  &= frac{1}{2pi i} sum_{n=0}^{infty} int_{C} frac{(z-z_0)^{n}f(zeta)}{(zeta - z_0)^{n+1}}dzeta 

  \ &= frac{1}{2pi i}int_{C}sum_{n=0}^{infty} frac{1}{zeta - z_0}frac{f(zeta)}{frac{(zeta - z_0)^n}{(z-z_0)^n}}dzeta                   

  \ &= frac{1}{2pi i} int_{C} frac{f(zeta)}{zeta - z_0}sum_{n=0}^{infty}left( frac{z-z_0}{zeta - z_0}right)^{n} dzeta

                   

end{align*}

The above image is created using

begin{align*}

  sum_{n=0}^{infty} frac{(z-z_0)^{n}}{2pi i} int_{C} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta

  &= frac{1}{2pi i} sum_{n=0}^{infty} int_{C} frac{(z-z_0)^{n}f(zeta)}{(zeta - z_0)^{n+1}}dzeta 

  \ &= frac{1}{2pi i}int_{C}sum_{n=0}^{infty} frac{1}{zeta - z_0}frac{f(zeta)}{frac{(zeta - z_0)^n}{(z-z_0)^n}}dzeta                   

  \ &= frac{1}{2pi i} int_{C} frac{f(zeta)}{zeta - z_0}sum_{n=0}^{infty}left( frac{z-z_0}{zeta - z_0}right)^{n} dzeta

                   

end{align*}

which is extracted from the full code above. How can I resize the second line of the image, so that all the characters are the same size?

You want to

begin{align*}

  sum_{n=0}^{infty} frac{(z-z_0)^{n}}{2pi i} int_{C} frac{f(zeta)}{(zeta - z_0)^{n+1}}dzeta

  &= frac{1}{2pi i} sum_{n=0}^{infty} int_{C} frac{(z-z_0)^{n}f(zeta)}{(zeta - z_0)^{n+1}}dzeta 

  \ 

&= frac{1}{2pi i}int_{C}sum_{n=0}^{infty} frac{1}{zeta - z_0}frac{f(zeta)}{displaystylefrac{(zeta - z_0)^n}{(z-z_0)^n}}dzeta                   

  \ &= frac{1}{2pi i} int_{C} frac{f(zeta)}{zeta - z_0}sum_{n=0}^{infty}left( frac{z-z_0}{zeta - z_0}right)^{n} dzeta                  

end{align*}

or simply dfrac as you can use amsmath package.

Please observe also that one backslash before end{align*} was removed.