Line-breaking two equations within equation array

Question

I have the code

\begin{eqnarray}
    &&\nonumber
    \left.\begin{array}{l}
        & \textbf{E}_{\textrm{tr}}  (x,y,z)= \left(\ts \textbf{s} + \tp \textbf{p}_+\right) \, \exp\lec{i   {\ko} \les\left({x\cos\psi+y\sin\psi}\right)\sin\theta_{\rm inc}+(z-N  P )   \cos \theta_{\rm inc}\ris }\ric
        \\[5pt]
        \textbf{H}_{\textrm{tr}} (x,y,z) = 
        \eta_0^{-1}\left(\ts \textbf{p}_+ - \tp \textbf{s}\right)
        \, \exp\lec{i   {\ko} \les\left({x\cos\psi+y\sin\psi}\right)\sin\theta_{\rm inc} +(z-N P)   \cos \theta_{\rm inc}\ris
        }\ric
    \end{array}\right\}\,,
    \\[5pt]
    &&
    \qquad\qquad\qquad z>N P \,.
\end{eqnarray} 

which produces the output

No matter what I try, I cannot get this to render more nicely. I would like for each equation to break after the plus sign, and for the equation number to appear to the right of the bracket.

Answer 1

In order to make your code compilable, I had to come up with emergency definitions of \ts, \tp, \ko, \ric, \ris, \lec, and \les.

I've also simplified your code by replacing the eqnarray environment with a far more flexible aligned environment and by getting rid of all interior \left and \right sizing directives. (Hint: they do nothing except mess up the horizontal spacing.) Next, I replaced the 4 instances of \textbf with \mathbf. Finally, I would state z>N P in ordinary, i.e., inline, math on a line following the displayed equations.

\documentclass{article}
\usepackage{amsmath} % for 'aligned' environment
%% check if the following definitions are appropriate
\providecommand\ts{t_s}
\providecommand\tp{t_p}
\providecommand\ko{k_0}
\providecommand\lec{\bigl\{}
\providecommand\ric{\bigr\}}
\providecommand\les{\bigl[}
\providecommand\ris{\bigr]}
\newcommand\minc{{\mathrm{inc}}}
\newcommand\mtr{{\mathrm{tr}}}

\begin{document}
\begin{equation}
\left.\begin{array}{@{}l}
\begin{aligned}
\textbf{E}_\mtr (x,y,z)
    &= (\ts \mathbf{s} + \tp \mathbf{p}_+)  
      \exp\lec i \ko \les({x\cos\psi+y\sin\psi})\sin\theta_\minc \\
    &\qquad+(z-N P ) \cos \theta_\minc\ris \ric
    \\[\jot]
\textbf{H}_\mtr (x,y,z) 
    &= \eta_0^{-1}(\ts \mathbf{p}_+ - \tp \mathbf{s})
      \exp\lec i \ko \les({x\cos\psi+y\sin\psi})\sin\theta_\minc \\
    &\qquad+(z-N P) \cos \theta_\minc\ris\ric
\end{aligned}
\end{array}\right\}
\end{equation} 
and $z>N P$.
\end{document}

Answer 2

I would write your equation as follows:

\documentclass{article}
\usepackage{bm}
\usepackage{mathtools}

\begin{document}
    \begin{gather}
\begin{rcases}
    \bm{E}_{\mathrm{tr}}(x,y,z) = \left(t_s\bm{s} + t_p \bm{p}_{+}\right) A(\psi,\theta)  &   \\
    \bm{E}_{\mathrm{tr}}(x,y,z) = \eta_0^{-1}\left(t_s\bm{p}_{+} - t_p \bm{s}\right) A(\psi,\theta)  
\end{rcases}   
\shortintertext{where}
A(\psi,\theta) = \exp\Bigl\{i k_o\bigl[(x\cos\psi + y\sin\psi)\sin\theta_{\mathrm{inc}} +
                       (z - NP)\cos\theta_{\mathrm{inc}}\bigr]\Bigr\}
    ,\quad    z > N P ,  \notag
    \end{gather} 
\end{document}

Answer 3

I would definitely use a shorthand for the long repeated part in the two equations. I'd also avoid the big brace, because the equation number midway will help readers in knowing that the number refers to both.

\documentclass{article}
\usepackage{amsmath}

\begin{document}

\begin{align}
  &\begin{alignedat}{2}
    &\mathbf{E}_{\mathrm{tr}}&&(x,y,z)
    = (t_s \mathbf{s} + t_p \mathbf{p}_+)f(x,y,z)
   \\
    &\mathbf{H}_{\mathrm{tr}}&&(x,y,z)
    = \eta_0^{-1}(t_s\mathbf{p}_+ - t_p\mathbf{s})f(x,y,z)
  \end{alignedat}
  \\
  & f(x,y,z)=
    \exp\{ik_0 [(x\cos\psi+y\sin\psi)\sin\theta_{\mathrm{inc}}+(z-NP)\cos\theta_{\mathrm{inc}}]\},
    \notag
  \\
  &z>NP. \notag
\end{align}

\end{document}

I prefer to leave a small gap between E_tr and the parenthesis rather than a hole in front of the E.

Note that the input has been simplified a lot, without useless braces and the obfuscation given by \tp, \ts, \ric and so on.

If you really want the big brace:

\documentclass{article}
\usepackage{amsmath}

\begin{document}

\begin{align}
  &
  \left.\kern-\nulldelimiterspace\!\! % alignment tricks
  \begin{alignedat}{2}
    &\mathbf{E}_{\mathrm{tr}}&&(x,y,z)
    = (t_s \mathbf{s} + t_p \mathbf{p}_+)f(x,y,z)
   \\
    &\mathbf{H}_{\mathrm{tr}}&&(x,y,z)
    = \eta_0^{-1}(t_s\mathbf{p}_+ - t_p\mathbf{s})f(x,y,z)
  \end{alignedat}
  \right\rbrace,
  \\
  & f(x,y,z)=
    \exp\{ik_0 [(x\cos\psi+y\sin\psi)\sin\theta_{\mathrm{inc}}+(z-NP)\cos\theta_{\mathrm{inc}}]\},
    \notag
  \\
  &z>NP. \notag
\end{align}

\end{document}

Beware that \textbf in equations is generally wrong and that \rm has been deprecated for 30 years.