I have this question in my homework where I have the following phasor of the electric field (we assume that all waves have $\omega$ frequency): $$\overline{\mathbf{E}}_{1}\left(x,y,z\right)=E_{0}\left[\hat{x}+\left(2+1.5j\right)\hat{y}+\left(2+3j\right)\hat{z}\right]e^{j\left(4x-4y+2z\right)}$$ and we are asked to find another plane wave $\overline{\mathbf{E}}_{2}\left(x,y,z\right)$ s.t their sum will carry power only in the $\hat{y}$ direction. I suggested the following plane wave: $$\overline{\mathbf{E}}_{2}\left(x,y,z\right)=E_{0}\left[-\hat{x}+\left(2+1.5j\right)\hat{y}-\left(2+3j\right)\hat{z}\right]e^{-j\left(4x+4y+2z\right)}$$ and I tried to calculate the average power by calculating: $$\vec{\boldsymbol{S}}_{tot} =\frac{1}{2}\Re\{{\overline{\mathbf{E}}_{tot}\times\overline{\mathbf{H}}_{tot}^{*}}\}$$and show that it has only a $\hat{y}$ non-zero component. I know this must be true but I probably have an arithmetic mistake (or something sillier). I don't know if it's legit to post my calculation here in hope that someone can catch my mistake but after $5$ hours of trying to figure out where's my mistake I am simply desperate. I'm sorry if it's too messy or too elaborate but I wrote it in LyX and converted it to Latex and paste it here. so here it is: \begin{align} \overline{\mathbf{E}}_{tot}\left(x,y,z\right) & =\overline{\mathbf{E}}_{1}\left(x,y,z\right)+\overline{\mathbf{E}}_{2}\left(x,y,z\right)= \\ & =E_{0}\left[\hat{x}+\left(2+1.5j\right)\hat{y}+\left(2+3j\right)\hat{z}\right]e^{j\left(4x-4y+2z\right)} \\ & \quad +E_{0}\left[-\hat{x}+\left(2+1.5j\right)\hat{y}-\left(2+3j\right)\hat{z}\right]e^{-j\left(4x+4y+2z\right)}=\\ & =E_{0}\left[\hat{x}\left(e^{j\left(4x+2z\right)}-e^{-j\left(4x+2z\right)}\right)+\left(2+1.5j\right)\hat{y}\left[e^{j\left(4x+2z\right)}+e^{-j\left(4x+2z\right)}\right] \right.\\ & \qquad \left. +\left(2+3j\right)\hat{z}\left(e^{j\left(4x+2z\right)}-e^{-j\left(4x+2z\right)}\right)\right]e^{-j4y}=\\ & =E_{0}\left[2j\sin\left(4x+2z\right)\hat{x}+\left(4+3j\right)\cos\left(4x+2z\right)\hat{y}+\left(-6+4j\right)\sin\left(4x+2z\right)\hat{z}\right]e^{-j4y} \end{align} The fields themselves: \begin{align} \overline{E}_{tot,x} & =2jE_{0}\sin\left(4x+2z\right)e^{-j4y} \\ \overline{E}_{tot,y} &=\left(4+3j\right)E_{0}\cos\left(4x+2z\right)e^{-j4y} \\ \overline{E}_{tot,z} &=\left(-6+4j\right)E_{0}\sin\left(4x+2z\right)e^{-j4y} \end{align}
Their derivatives:
\begin{align} \partial_{y}\overline{E}_{tot,z}&=\partial_{y}\left(\left(-6+4j\right)E_{0}\sin\left(4x+2z\right)e^{-j4y}\right)=\left(16+24j\right)E_{0}\sin\left(4x+2z\right)e^{-j4y} \\ \partial_{z}\overline{E}_{tot,y}&=\partial_{z}\left(\left(4+3j\right)E_{0}\cos\left(4x+2z\right)e^{-j4y}\right)=-\left(8+6j\right)E_{0}\sin\left(4x+2z\right)e^{-j4y} \\ \partial_{z}\overline{E}_{tot,x}&=\partial_{z}\left(2jE_{0}\sin\left(4x+2z\right)e^{-j4y}\right)=4jE_{0}\cos\left(4x+2z\right)e^{-j4y} \\ \partial_{x}\overline{E}_{tot,z}&=\partial_{x}\left(\left(-6+4j\right)E_{0}\sin\left(4x+2z\right)e^{-j4y}\right)=-\left(24-16j\right)E_{0}\cos\left(4x+2z\right)e^{-j4y} \\ \partial_{x}\overline{E}_{tot,y}&=\partial_{x}\left(\left(4+3j\right)E_{0}\cos\left(4x+2z\right)e^{-j4y}\right)=-\left(16+12j\right)E_{0}\sin\left(4x+2z\right)e^{-j4y} \\ \partial_{y}\overline{E}_{tot,x}&=\partial_{y}\left(2jE_{0}\sin\left(4x+2z\right)e^{-j4y}\right)=8E_{0}\sin\left(4x+2z\right)e^{-j4y} \end{align}
The magnetic field: \begin{align*} \Longrightarrow\,\,\,\overline{\mathbf{H}}_{tot}\left(x,y,z\right) & =-\frac{1}{j\omega\mu}\nabla\times\overline{\mathbf{E}}_{tot}=\frac{j}{\omega\mu}\begin{vmatrix}\hat{x} & \hat{y} & \hat{z}\\ \partial_{x} & \partial_{y} & \partial_{z}\\ \overline{E}_{tot,x} & \overline{E}_{tot,y} & \overline{E}_{tot,z} \end{vmatrix}\\ & =\frac{j}{\omega\mu}\left[ \hat{x}\left(\partial_{y}\overline{E}_{tot,z}-\partial_{z}\overline{E}_{tot,y}\right)+\hat{y}\left(\partial_{z}\overline{E}_{tot,x}-\partial_{x}\overline{E}_{tot,z}\right) \right.\\ & \qquad \qquad \left. +\hat{z}\left(\partial_{x}\overline{E}_{tot,y}-\partial_{y}\overline{E}_{tot,x}\right) \right] =\\ & =\frac{jE_{0}}{\omega\mu}\left[\left(16+24j+8+6j\right)\sin\left(4x+2z\right)\hat{x}+\left(4j+24-16j\right)\cos\left(4x+2z\right)\hat{y} \right.\\ & \qquad \qquad \left. -\left(16+12j+8\right)\sin\left(4x+2z\right)\hat{z}\right]e^{-j4y}=\\ & =\frac{jE_{0}}{\omega\mu}\left[\left(24+30j\right)\sin\left(4x+2z\right)\hat{x}+\left(24-12j\right)\cos\left(4x+2z\right)\hat{y} \right.\\ & \qquad \qquad \left. -\left(24+12j\right)\sin\left(4x+2z\right)\hat{z}\right]e^{-j4y} \end{align*}
and its components,
\begin{align} \overline{H}_{tot,x}&=-\frac{E_{0}}{\omega\mu}\left(30-24j\right)\sin\left(4x+2z\right)e^{-j4y} \\ \overline{H}_{tot,y}&=\frac{E_{0}}{\omega\mu}\left(12+24j\right)\cos\left(4x+2z\right)e^{-j4y} \\ \overline{H}_{tot,z}&=\frac{E_{0}}{\omega\mu}\left(12-24j\right)\sin\left(4x+2z\right)e^{-j4y} \\ \overline{H}_{tot,x}^{*}&=-\frac{E_{0}^{*}}{\omega\mu}\left(30+24j\right)\sin\left(4x+2z\right)e^{j4y} \\ \overline{H}_{tot,y}^{*}&=\frac{E_{0}^{*}}{\omega\mu}\left(12-24j\right)\cos\left(4x+2z\right)e^{j4y} \\ \overline{H}_{tot,z}^{*}&=\frac{E_{0}^{*}}{\omega\mu}\left(12+24j\right)\sin\left(4x+2z\right)e^{j4y} \end{align}
giving the Poynting vector
\begin{align*} \vec{\boldsymbol{S}}_{tot} & =\frac{1}{2}\overline{\mathbf{E}}_{tot}\times\overline{\mathbf{H}}_{tot}^{*}=\frac{1}{2}\begin{vmatrix}\hat{x} & \hat{y} & \hat{z}\\ \overline{E}_{tot,x} & \overline{E}_{tot,y} & \overline{E}_{tot,z}\\ \overline{H}_{tot,x}^{*} & \overline{H}_{tot,y}^{*} & \overline{H}_{tot,z}^{*} \end{vmatrix}=\\ & =\frac{1}{2}\left[\hat{x}\left(\overline{E}_{tot,y}\cdot\overline{H}_{tot,z}^{*}-\overline{E}_{tot,z}\cdot\overline{H}_{tot,y}^{*}\right)+\hat{y}\left(\overline{E}_{tot,z}\cdot\overline{H}_{tot,x}^{*}-\overline{E}_{tot,x}\cdot\overline{H}_{tot,z}^{*}\right) \right. \\ & \qquad \left. +\hat{z}\left(\overline{E}_{tot,x}\cdot\overline{H}_{tot,y}^{*}-\overline{E}_{tot,y}\cdot\overline{H}_{tot,x}^{*}\right)\right] \end{align*}
with components
\begin{align*} S_{tot,x} & =\frac{1}{2}\left(\overline{E}_{tot,y}\cdot\overline{H}_{tot,z}^{*}-\overline{E}_{tot,z}\cdot\overline{H}_{tot,y}^{*}\right)=\\ & =\frac{1}{2}\left(\left(\left(4+3j\right)E_{0}\cos\left(4x+2z\right)e^{-j4y}\right)\cdot\left(\frac{E_{0}^{*}}{\omega\mu}\left(12+24j\right)\sin\left(4x+2z\right)e^{j4y}\right) \right. \\ & \qquad \left. -\left(\left(-6+4j\right)E_{0}\sin\left(4x+2z\right)e^{-j4y}\right)\cdot\frac{E_{0}^{*}}{\omega\mu}\left(12-24j\right)\cos\left(4x+2z\right)e^{j4y}\right)=\\ & =\frac{\left|E_{0}\right|^{2}}{2\omega\mu}\left(\left(4+3j\right)\cdot\left(12+24j\right)-\left(-6+4j\right)\cdot\left(12-24j\right)\right)\sin\left(4x+2z\right)\cos\left(4x+2z\right)=\\ & =\frac{\left|E_{0}\right|^{2}}{2\omega\mu}\left(-24+132j-24-192j\right)\sin\left(4x+2z\right)\cos\left(4x+2z\right)=\\ & =-\frac{\left|E_{0}\right|^{2}}{2\omega\mu}\left(24+60j\right)\sin\left(8x+4z\right) \end{align*}
and
\begin{align*} S_{tot,y} & =\frac{1}{2}\left(\overline{E}_{tot,z}\cdot\overline{H}_{tot,x}^{*}-\overline{E}_{tot,x}\cdot\overline{H}_{tot,z}^{*}\right)=\\ & =\frac{1}{2}\left(\left(\left(-6+4j\right)E_{0}\sin\left(4x+2z\right)e^{-j4y}\right)\cdot\left(-\frac{E_{0}^{*}}{\omega\mu}\left(30+24j\right)\sin\left(4x+2z\right)e^{j4y}\right) \right. \\ & \qquad \left. -\left(2jE_{0}\sin\left(4x+2z\right)e^{-j4y}\right)\cdot\left(\frac{E_{0}^{*}}{\omega\mu}\left(12+24j\right)\sin\left(4x+2z\right)e^{j4y}\right)\right)=\\ & =\frac{\left|E_{0}\right|^{2}}{2\omega\mu}\left(-\left(-6+4j\right)\cdot\left(30+24j\right)-2j\cdot\left(12+24j\right)\right)\sin^{2}\left(4x+2z\right)=\\ & =\frac{\left|E_{0}\right|^{2}}{2\omega\mu}\left(276+{24j}+48-{24j}\right)\sin^{2}\left(4x+2z\right)=\\ & =\frac{162\left|E_{0}\right|^{2}}{\omega\mu}\sin^{2}\left(4x+2z\right) \end{align*}
and
\begin{align*} S_{tot,z} & =\frac{1}{2}\left(\overline{E}_{tot,x}\cdot\overline{H}_{tot,y}^{*}-\overline{E}_{tot,y}\cdot\overline{H}_{tot,x}^{*}\right)=\\ & =\frac{1}{2}\left(\left(2jE_{0}\sin\left(4x+2z\right)e^{-j4y}\right)\cdot\left(\frac{E_{0}^{*}}{\omega\mu}\left(12-24j\right)\cos\left(4x+2z\right)e^{j4y}\right) \right. \\ & \qquad \left. -\left(\left(4+3j\right)E_{0}\cos\left(4x+2z\right)e^{-j4y}\right)\cdot\left(-\frac{E_{0}^{*}}{\omega\mu}\left(30+24j\right)\sin\left(4x+2z\right)e^{j4y}\right)\right)=\\ & =\frac{\left|E_{0}\right|^{2}}{2\omega\mu}\left(2j\cdot\left(12-24j\right)+\left(4+3j\right)\cdot\left(30+24j\right)\right)\sin\left(4x+2z\right)\cos\left(4x+2z\right)=\\ & =\frac{\left|E_{0}\right|^{2}}{2\omega\mu}\left(48+24j+48+186j\right)\sin\left(4x+2z\right)\cos\left(4x+2z\right)=\\ & =\frac{\left|E_{0}\right|^{2}}{2\omega\mu}\left(48+105j\right)\sin\left(8x+4z\right) \end{align*}
So, as you can see the real parts of the $\hat{x}$ and $\hat{z}$ components are not $0$.
Any help would be much appriciated.