In a video (http://youtu.be/r_gBQ_qhg8U?t=9m58s) it's stated that a matrix element of an imaginary operator acting on a real wave function is zero, i.e. $$\langle\text{real}|\text{imaginary}|\text{real}\rangle ~=~ 0,$$ and I don't really understand why.
When we make an actual calculation, won't the $i$ from $l_z$ simply move in front of the integral and have no influence on its actual value?
I don't know the term "imaginary operator". I take this to be an antihermitian operator, which eigenvalues are purely imaginary. Then the statement is clearly not true. Take as a counter example any hermitian operator $\hat{A}$ and real wavefunction $\psi$ with $\langle \psi | \hat{A} |\psi\rangle = A_\psi \neq 0$. $A_\psi$ is of course real. Take now $\hat{B} = i \hat{A}$, which is antihermitian. It follows $$\langle \psi | \hat{B} |\psi\rangle =i \langle \psi | \hat{A} |\psi\rangle=i A_\psi\neq 0$$
In the video Prof. G. Rangarajan is considering the expectation value
$$\mathbb{R}~\ni~ \langle \psi |\underbrace{\hat{L}_z}_{\text{self-adj.}} |\psi \rangle ~=~ \int\! d^3r ~\underbrace{\overline{\psi({\bf r})}}_{\text{real}} \underbrace{ (-i\hbar) \frac{\partial}{\partial \varphi}}_{\text{imaginary}} \underbrace{\psi({\bf r})}_{\text{real}} ~\in~i\mathbb{R}. $$
In other words, it should be both real and imaginary. He concludes that it must be zero.
