First, note that that the transfer function you give has the pole (and zero too) in the right hand plane, i.e., the system it describes is unstable. I suspect you meant:
\$H(s) = \dfrac{s + 1055}{s+67}\$
However, even this is not in standard form. Putting this transfer function into standard form yields:
\$ H(s) = \dfrac{s + 1055}{s + 67} = \dfrac{1055}{67} \dfrac{\frac{s}{1055} + 1}{\frac{s}{67} + 1}\$
So, now you see where the undesired gain has come from. Knowing (only) the pole and zero, you should guess instead:
\$H(s) =\dfrac{\frac{s}{1055} + 1}{\frac{s}{67} + 1}\$
