Crthuky

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

documentclassarticle
usepackage[utf8]inputenc
usepackage[english]babel

usepackageamsthm
usepackageamsmath
usepackage[left=1.5in, right=1.5in, top=0.5in]geometry


newtheoremdefinitionDefinition
newtheoremtheoremTheorem 
theoremstyleremark

begindocument
 titleExtra Credit
 maketitle

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

 beginequation
 f(z_0) + f'(z_0)(z-z_0) + fracf''(z_0)2!(z-z_0)^2 + cdots = sum_n=0^infty fracf^(n)(z_0)n!(z-z_0)^n
 endequation

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

 begintheorem
 If f is analytic inside and on the simple closed positively oriented contour $Gamma$ and if $z_0$ is any point inside $Gamma$, then
 beginequation
 f^(n)(z_0) = fracn!2pi i int_Gamma fracf(zeta)(zeta - z_0)^n+1dzeta hspace1cm (n=1,2,3, cdots )
 endequation
 endtheorem
 hrulefill



begintheorem
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. 
endtheorem

beginproof
Suppose that the function textitf is analytic in the disk $|z-z_0|<R'$. We can define a positively oriented contour $C$ as $$ C:=Bigz:.$$ Letting $zeta$ be an arbitrary point on $C$ and applying Theorem 1 to $(1)$, we get 

beginequation
sum_n=0^infty frac(z-z_0)^n2pi i int_C fracf(zeta)(zeta - z_0)^n+1dzeta.
endequation\ 

Or equivalently, we have that


beginalign*
 sum_n=0^infty frac(z-z_0)^n2pi i int_C fracf(zeta)(zeta - z_0)^n+1dzeta
 &= frac12pi i sum_n=0^infty int_C frac(z-z_0)^nf(zeta)(zeta - z_0)^n+1dzeta 
 \ &= frac12pi iint_Csum_n=0^infty frac1zeta - z_0fracf(zeta)frac(zeta - z_0)^n(z-z_0)^ndzeta 
 \ &= frac12pi i int_C fracf(zeta)zeta - z_0sum_n=0^inftyleft( fracz-z_0zeta - z_0right)^n dzeta
 
endalign*

The above image is created using

beginalign*
 sum_n=0^infty frac(z-z_0)^n2pi i int_C fracf(zeta)(zeta - z_0)^n+1dzeta
 &= frac12pi i sum_n=0^infty int_C frac(z-z_0)^nf(zeta)(zeta - z_0)^n+1dzeta 
 \ &= frac12pi iint_Csum_n=0^infty frac1zeta - z_0fracf(zeta)frac(zeta - z_0)^n(z-z_0)^ndzeta 
 \ &= frac12pi i int_C fracf(zeta)zeta - z_0sum_n=0^inftyleft( fracz-z_0zeta - z_0right)^n dzeta
 
endalign*

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

beginalign*
 sum_n=0^infty frac(z-z_0)^n2pi i int_C fracf(zeta)(zeta - z_0)^n+1dzeta
 &= frac12pi i sum_n=0^infty int_C frac(z-z_0)^nf(zeta)(zeta - z_0)^n+1dzeta 
 \ 
&= frac12pi iint_Csum_n=0^infty frac1zeta - z_0fracf(zeta)displaystylefrac(zeta - z_0)^n(z-z_0)^ndzeta 
 \ &= frac12pi i int_C fracf(zeta)zeta - z_0sum_n=0^inftyleft( fracz-z_0zeta - z_0right)^n dzeta
 
endalign*

or simply dfrac as you can use amsmath package.