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+### Statement
+
+$12.1.3.$ [Insert the problem statement]
+
+### Solution
+
+For a plane sinusoidal wave, the general equation of the electromagnetic wave is given by:
+
+
+\begin{equation}
+E(z,t)=E_0\sin(At+Bz+\varphi_0)
+\end{equation}
+
+where $A$,$B$ and $\varphi_0$ are constants.
+
+
+
+From the condition of the problem we know that:
+
+\begin{equation}
+E(z,0)=E_0\sin(\frac{2\pi}{\lambda}z)
+\end{equation}
+
+
+
+Using eq(1) we get:
+
+
+\begin{equation}
+E_0\sin{(Bz+\varphi_0)}=E_0\sin(\frac{2\pi}{\lambda}z)
+\end{equation}
+
+
+By comparing the phases we obtain that:
+
+\begin{equation}
+B=\frac{2\pi}{\lambda};\varphi_0=0
+\end{equation}
+
+
+
+As the wave travels in the positive direction:
+
+\begin{equation}
+\frac{dz}{dt}=c
+\end{equation}
+
+
+Setting the phase($\varphi=At+Bz+\varphi_0$)to a constant value allows us to derive the phase velocity($c$):
+
+
+\begin{equation}
+A+B\frac{dz}{dt}=\frac{d\varphi}{dt}=0
+\end{equation}
+
+
+
+So we get the constant $A$:
+
+\begin{equation}
+A=-\frac{2\pi c}{\lambda}
+\end{equation}
+
+
+
+Plugging all the constants into the eq(1) gives us:
+
+\begin{equation}
+E(z,t)=E_0\sin(\frac{2\pi}{\lambda}(z-ct))
+\end{equation}
+
+
+
+#### Answer
+
+[Insert a concise answer or boxed result]