Solution

For a plane sinusoidal wave, the general equation of the electromagnetic wave is given by:
$$E(z,t) = E_0 \sin(At + Bz + \varphi_0)$$
where $A$, $B$ and $\varphi_0$ are constants.

From the condition of the problem (based on the wave profile at $t=0$), we know that the initial spatial distribution is:
$$E(z,0) = E_0 \sin\left(\frac{2\pi}{\lambda} z\right)$$
where $\lambda$ is the wavelength.

By substituting $t=0$ into the general equation, we equate the two expressions:
$$E_0 \sin(Bz + \varphi_0) = E_0 \sin\left(\frac{2\pi}{\lambda} z\right)$$

By comparing the phases, we obtain the spatial constant and the initial phase:
$$B = \frac{2\pi}{\lambda}, \quad \varphi_0 = 0$$

Since the wave travels in the positive $z$-direction with a phase velocity $c$, the phase of the wave $\varphi = At + Bz + \varphi_0$ must remain constant for a fixed point on the wave profile. Taking the time derivative of the phase yields:
$$\frac{d\varphi}{dt} = A + B \frac{dz}{dt} = 0$$

Given that the wave propagation speed is $\frac{dz}{dt} = c$, we get:
$$A + Bc = 0 \implies A = -Bc = -\frac{2\pi c}{\lambda}$$

Finally, plugging the derived constants $A$, $B$, and $\varphi_0$ back into the general equation gives us:
$$E(z,t) = E_0 \sin\left(-\frac{2\pi c}{\lambda} t + \frac{2\pi}{\lambda} z\right) = E_0 \sin\left[\frac{2\pi}{\lambda}(z - ct)\right]$$