De Broglie wave length of electron Consider an electron with total energy $E>V_2$ in a potential well with
$$V(x)= \begin{cases} 
      \infty & x< 0 \\
      V_1 & 0< x< L \\
      V_2 & x>L 
   \end{cases}
 $$
where $V_2>V_1>0$.
We can determine that
$$\phi_E(x)= \begin{cases} 
      0 & x< 0 \\
      A\sin(kx)+B\cos(kx) & 0< x< L \\
      Ce^{qx}+De^{-qx} & x>L 
   \end{cases}
 $$
where $k^2=\frac{2m_e E}{\hbar}$ and $q=k\sqrt{V_1-E}$.
We can also apply the boundary condition at $x=0$ to determine that $$\phi_e(x)=A\sin(kx)$$
for $x\in[0,L]$.
We can also apply boundary conditions at $x=L$ to find that
$$A\sin(kL)=De^{-qL}$$
$$Ak\cos(kL)=-Dqe^{-qL}$$
(since $C=0$ due to the corresponding positive exponent), and
$$k\cot(kL)=-q$$
I need to calculate the de Broglie wavelength of the electron in the regions defined by the potential, if $V_1 = 10.0$ eV, $V_2 = 20.0$ eV, $E=30$ eV.
For the region where $0\le x\le L$, if I'm correct, the de Broglie $\lambda$ is determined by $\frac{h}{p}$, where $p=\hbar k=\sqrt{2m_e E}$, and so we can find $\lambda_{\text{de Broglie}}=8.962639\times 10^{-18}$ m.
For the region where $x<0$, we can't find this length because of the infinite potential.
For the region where $x>L$, if $p=\hbar q$, then it appears that $p$ is complex-valued, and then I'm in doubt whether I was correct to have dropped the coefficient $C$ above in my derivations of equations.
Would appreciate some help.
 A: As indicated in the answer to your previous question, since as $E(=30\:\mathrm{eV})$ is higher than $V_2(=20\:\mathrm{eV})$, that particle is not bound, it's not an eigenstate of the system's Schrödinger equation (it's a scattered state). Its wave function for $x \to +\infty$ would something like:
$$\psi=c_1e^{-ikx}+c_2e^{+ikx}$$
... where both complex parts represent the incoming and reflected waves respectively. That's essentially the wave function of a free particle.
Its de Broglie wave length where $x>L$, is calculated using:
$$p=\hbar k=\sqrt{2m_e (E-V_2)}$$
But for bound particles with $E<V_2$, the wave function $\psi$ (noted as $\phi$ by you) is Real over the entire domain $0\leq x \leq+\infty$.
For $0 \leq x \leq L$, then $p=\hbar k=\sqrt{2m_e (E-V_1)}$ (bound particle).

and then I'm in doubt whether I was correct to have dropped the coefficient $C$ above in my derivations of equations.

You're entirely correct to drop $C$ ($C=0$) for bound particles but note that not doing so wouldn't make $\psi(x)$ Complex in that area of the domain. Classically a particle with $E<V_2$ could not enter into that $x \leq L$ area but a quantum particle tunnels into the classically forbidden area.

1D SE solution for free particle of energy $E$ in constant potential field $V$:
$$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V\psi=E\psi$$
$$\frac{d^2\psi}{dx^2}+\frac{2m}{\hbar^2}(E-V)\psi=0$$
$$k^2=\frac{2m}{\hbar^2}(E-V)$$
$$\frac{d^2\psi}{dx^2}+k^2\psi=0$$
For:
$$k>0$$
$$\implies \psi(x)=c_1e^{-ikx}+c_2e^{+ikx}$$
