This is a Doppler shift problem. Just to make sure you have the idea right, we'll derive
the equation here.
Question: At what frequency does the spaceship see the observer receiving information?
The distance between crests in the ship frame is simply the wavelength lambda. However, in the time it takes for the light to travel one wavelength, the earth will have moved. The time between crests is therefore:
\[ t = \frac{\lambda + vt}{c} \implies ct = \lambda + vt \implies t = \frac{\lambda}{c-v} \]
Since we are talking about electromagnetic radiation, lambda = c / frequency, so
\[ t = \frac{c}{c-v} \frac{1}{f} = \frac{1}{1-v/c} \frac{1}{f} \]
Now what about the observer?
The proper time for the reception of information is calculated in the observer's frame, because the receiver is stationary there. Therefore, the observer measures this time to be
\[t' = t/\gamma = \frac{\sqrt{1-v^2/c^2}}{1 - v/c} \frac{1}{f} = \sqrt{\frac{1+v/c}{1-v/c} } \frac{1}{f} \]
Therefore, the frequency measured by the observer will be
\[ f_o = \sqrt{\frac{1-\beta}{1+\beta} } f_s\]
where \[f_0, f_s, \beta \] are the observer frequency, the source frequency, and v/c respectively.
Plugging numbers in, that gives me ~82 Hz, a redshift, as expected.