The question I have is regarding a solution to a later question (Q2). So in order for the question I have to make sense, unfortunately, I must typeset the previous questions.
(Q1)
We may represent a general electromagnetic plane wave by (real part of the complex exponentials): $$\vec E = \vec E_0\exp(i\vec k \cdot \vec r - i \omega t) \quad\text{&}\quad\vec B = \vec B_0\exp(i\vec k \cdot \vec r - i \omega t)$$ Show that Faraday's law $$\partial_t \vec B=-\nabla \times \vec E\tag{1}$$ becomes $$-i\omega \vec B_0=-i\vec k \times \vec E_0\tag{2}$$
I know I can write $$\partial_t \vec B=-i \omega\exp(i\vec k \cdot \vec r - i \omega t)\vec B_0$$ for the LHS of $(1)$
Letting $f=\exp(i\vec k \cdot \vec r - i \omega t)$ then $\partial_t \vec B=-i \omega f \vec B_0$ and $\vec E = f \vec E_0$ and the RHS of $(1)$ is $$\begin{align} -\nabla \times \vec E &=-\left(\nabla\times(f \vec E_0)\right)\\&=-\left(\nabla f \times \vec E_0 + f (\nabla \times \vec E_0)\right)\\&=-\left(i f\vec k \times \vec E_0+f(\nabla \times \vec E_0)\right)\end{align}$$
The only way I can reach $(2)$ is iff $$\nabla \times \vec E_0=\vec 0$$ which would be the case under the assumption that $\vec E_0$ and $\vec B_0$ are constant vectors. The question did not stipulate this is the case.
(Q2)
Circular Polarisation
i) Calculate the real part of the complex wave solution to Maxwell's equations in a vacuum: $$\vec E=E_0\left(\hat x + i \hat y\right)\exp[ik(z-ct)]$$ where $\hat x$ and $\hat y$ are the unit vectors in the x and y directions, respectively.
$$\mathcal R (\vec E)=\mathcal R\left\{E_0(\hat x+i\hat y) \left[\cos(k(z-ct))+i\sin(k(z-ct))\right]\right \}=E_0[\cos(k(z-ct))\hat x-\sin(k(z-ct))\hat y]$$
(this is all the background needed, I am now in a position to ask my question)
ii) Calculate the magnetic field of the wave
I'm going to assume that $\partial_r$ is a typo and should read $\partial_t$
In the solution the author writes
We can use the result from Question 1. In this case $\vec k=k \hat z$ and $\vec {E_0} =E_0(\hat x+i\hat y)$. Or you could use the usual form of the curl and note that since there are no $x$ or $y$ dependencies so $\nabla \times \vec E=-\partial_z E_y \hat x + \partial_z E_x \hat y$ $$\partial_t \vec B=-\nabla\times\vec E=-ikE_0\left[\hat z \times (\hat x+i\hat y)\exp[ik(z-ct)]\right]\tag{3}$$ and taking the real part
$$\bbox[5px,border:2px solid red]{\vec B =\frac{E_0}{c}\left[\hat x \times \left(\cos(k(z-ct))\hat x-\sin(k(z-ct))\hat y\right)\right]}$$
I basically don't really understand how the author arrived at that solution. So this is the way I would try to calculate the magnetic field:
I will make use of eqn $(2)$ (the hint given by the author), to find the amplitude of the magnetic field, $\vec B_0$, $$-i\omega \vec B_0=-i\vec k \times \vec E_0=-ik\hat z \times E_0(\hat x +i\hat y)$$ $$\implies \vec B_0 =E_0 \frac{k}{\omega}\hat z \times (\hat x +i\hat y)$$
Now, since $$\vec B=\vec B_0 \exp[ik(z-ct)]=E_0 \frac{k}{\omega}\hat z \times (\hat x +i\hat y)\exp[ik(z-ct)]$$
$$\implies \bbox[border:2px solid green]{\mathcal R (\vec B)=\frac{E_0}{c}\biggl(\hat z \times [\cos(k(z-ct))\hat x-\sin(k(z-ct))\hat y]\biggl)}$$
Where I have also used that $c=\omega/k$.
First of all, is my solution (in the green box) correct?
If both solutions are correct ($\hat x \to \hat z$), then, since the author is using Faraday's law the LHS of $(3)$ is the time derivative, so why is $(3)$ not being integrated wrt time (and then take real part)?
My other question is, it was my understanding that the magnetic field is perpendicular to the electric field, so why does the solution not just have a z-component?