In the exam document class, how to use EnvUplevel for solution that spans multiple pages?

Question

In the exam document class, how to use EnvUplevel for solution that spans multiple pages? Here is my code. (It works perfectly if the solution is within a page) (I am using page size a5)

\documentclass[answers]{exam}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
% \Aboxed in the following
\usepackage{mathtools}
% page size
\usepackage{geometry}
\geometry{
    papersize={140mm,210mm},
    twoside=True,
    inner=15mm,
    outer=15mm,
    bindingoffset=7mm,
    top=20mm,
    bottom=22.5mm

}
\extrawidth{.5in}
\header{\oddeven{}{\thepage}}%
    {{\small 18MAT41 - Jan/Feb 2021}}%
    {\oddeven{\thepage}{}}
\footer{}{{\small Visit entuition.org to unlock all QRs}}{}
\begin{document}
\begin{questions}
% Question 1 
% \renewcommand{\thequestion}{5}
\question 
\begin{parts}
% Question 1a
\part
    Show that $w = \log z$, $z \neq 0$ is analytic and hence find $\displaystyle{dw \over dz}$.
% Solution 1a
\begin{EnvUplevel}
\begin{TheSolution}
        Given,
    [w = \log z \tag{1}] 
    We know, 
    [ z = r e^{i \theta} ]
    Substituting $z$ in (1),
    \begin{align}
        w &= \log ( r e^{i \theta}) \
        w &= \log r + \log e^{i \theta} \
        w &= \log r + i \theta \tag{2}
    \end{align}
    We know,
    [ w = u+ iv \tag{3}]
    Comparing (2) and (3),
    \begin{align}
            u &= \log r           &     v &= \theta
    \end{align}
    Partially differentiate with respect to $r$,
    \begin{align}
            u_r &= {1 \over r} &        v_r &= 0 \tag{4}
    \end{align}
    Partially differentiate with respect to $\theta$,
    \begin{align}
            u_{\theta} &= 0    &        v_{\theta} &= 1 \tag{5} 
    \end{align}
        We know CR equations are,
        \begin{align}
            r u_r &= v_{\theta} \
            r v_r &= -u_{\theta}
        \end{align}
        From (4) and (5),
        \begin{align}
            r u_r &= r \cdot {1 \over r} = 1 = v_\theta \
            r v_r &= r \cdot 0 = 0 = -u_\theta
        \end{align}
        The CR equations are satisfied. Hence $w = \log z$ is analytic.\
        We know, 
        [{dw \over dz} = e^{-i \theta}(u_r + iv_r)]
        Substituting $u_r$ and $v_r$ from (4),
        \begin{align}
            {dw \over dz} &= e^{-i \theta} \left( {1 \over r} + i , 0 \right) \
            {dw \over dz} &= {1 \over re^{i \theta}} \
            \Aboxed{{dw \over dz} &= {1 \over z}}
        \end{align}
\end{TheSolution}
\end{EnvUplevel}
\end{parts}
\end{questions}
\end{document}

Answer 1

EnvUpLevel puts the contents into a \vbox, which cannot be broken. This uses a \trivlist.

\documentclass[answers]{exam}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
% \Aboxed in the following
\usepackage{mathtools}
% page size
\usepackage{geometry}
\geometry{
    papersize={140mm,210mm},
    twoside=True,
    inner=15mm,
    outer=15mm,
    bindingoffset=7mm,
    top=20mm,
    bottom=22.5mm
}
\extrawidth{.5in}
\header{\oddeven{}{\thepage}}%
    {{\small 18MAT41 - Jan/Feb 2021}}%
    {\oddeven{\thepage}{}}
\footer{}{{\small Visit entuition.org to unlock all QRs}}{}
% compute \leftmargin for each level
\settowidth{\leftmargin}{10.\hskip\labelsep}%
\edef\exammargini{\the\leftmargin}%
\settowidth{\leftmargin}{(m)\hskip\labelsep}%
\edef\exammarginii{\the\leftmargin}%
\settowidth{\leftmargin}{vii.\hskip\labelsep}%
\edef\exammarginiii{\the\leftmargin}%
\settowidth{\leftmargin}{($\psi$)\hskip\labelsep}%
\edef\exammarginiv{\the\leftmargin}%
\makeatletter
\newenvironment{myuplevel}{\ifnum@listdepth>0
    \leftmargin=\csname exammargin@roman@listdepth\endcsname\relax
    \advance@totalleftmargin-\leftmargin
  \fi
  \trivlist
  \ifnum@listdepth>0
    \leftskip=-\csname exammargin@roman@listdepth\endcsname\relax
    \advance@listdepth-1
  \fi
  \item}{\endtrivlist}
\makeatother
\begin{document}
\begin{questions}
% Question 1 
% \renewcommand{\thequestion}{5}
\question 
\begin{parts}
% Question 1a
\part
    Show that $w = \log z$, $z \neq 0$ is analytic and hence find $\displaystyle{dw \over dz}$.
% Solution 1a
\begin{myuplevel}
\begin{TheSolution}
        Given,
    [w = \log z \tag{1}] 
    We know, 
    [ z = r e^{i \theta} ]
    Substituting $z$ in (1),
    \begin{align}
        w &= \log ( r e^{i \theta}) \
        w &= \log r + \log e^{i \theta} \
        w &= \log r + i \theta \tag{2}
    \end{align}
    We know,
    [ w = u+ iv \tag{3}]
    Comparing (2) and (3),
    \begin{align}
            u &= \log r           &     v &= \theta
    \end{align}
    Partially differentiate with respect to $r$,
    \begin{align}
            u_r &= {1 \over r} &        v_r &= 0 \tag{4}
    \end{align}
    Partially differentiate with respect to $\theta$,
    \begin{align}
            u_{\theta} &= 0    &        v_{\theta} &= 1 \tag{5} 
    \end{align}
        We know CR equations are,
        \begin{align}
            r u_r &= v_{\theta} \
            r v_r &= -u_{\theta}
        \end{align}
        From (4) and (5),
        \begin{align}
            r u_r &= r \cdot {1 \over r} = 1 = v_\theta \
            r v_r &= r \cdot 0 = 0 = -u_\theta
        \end{align}
        The CR equations are satisfied. Hence $w = \log z$ is analytic.\
        We know, 
        [{dw \over dz} = e^{-i \theta}(u_r + iv_r)]
        Substituting $u_r$ and $v_r$ from (4),
        \begin{align}
            {dw \over dz} &= e^{-i \theta} \left( {1 \over r} + i , 0 \right) \
            {dw \over dz} &= {1 \over re^{i \theta}} \
            \Aboxed{{dw \over dz} &= {1 \over z}}
        \end{align}
\end{TheSolution}
\end{myuplevel}
\end{parts}
\end{questions}
\end{document}