# What is the wave form when the left hand circular polarized wave passes through 1/8 wave plate?

Consider an optical beam with left-handed circular polarization (LHCP), passing through an 1/8 wave plate.

I know that LHCP is written as $$\tag{1}\left( \begin{array}{c} 1 \\ i \end{array} \right)$$ in term of a Jones vector. So, I can calculate this problem by using the matrix form of the 1/8 wave plate. $$\tag{2} \left( \begin{array}{cc} 1 & 0 \\ 0 & \exp(i\pi/4) \end{array} \right) \left( \begin{array}{c} 1 \\ i \end{array} \right) = \left( \begin{array}{c} 1 \\ i\exp(i\pi/4) \end{array} \right) = \left( \begin{array}{c} \exp(i 0) \\ \exp(i3\pi/4) \end{array} \right) .$$

But, I don't know what is this result of wave form. Is this circular polarization or elliptical polarization?

Please let me know. Thank you.