No that's wrong, too.
The windings are symmetrical so the two poles each produce the same waveform - the normal waveform for this pole geometry at half the voltage.
With one pole offset A degrees the output waveform will become the sum of one half-waveform with the other shifted by (A*(poles/2)) degrees.
This is still symmetrical, but it's hard to generalize about it in the time domain. But it IS easy to genaralize about in the frequency domain, treating the waveform as a sum of a fundamental-frequency sine wave plus some harmonics. (This is especially useful for motors because the output waveform is a strong fundamental plus a rapidly decaying series of odd harmonics. The even harmonics canceled out due to the mirror symmetry of the waveform.)
The sum of two equal sine waves of voltage V offset by phase angle A is 2*v*cos((A*(poles/2))/2). So a small offset will slightly reduce the voltage of the fundamental. (For instance, a 5 degree offset in a four-pole motor will attenuate the fundamental by cos(5 degrees) = .9962, about half a percent.
The harmonics, however, get reduced by more, because the phase angle is larger for their higher frequency. It is multiplied by their harmonic number: The third harmonic voltage is 2*v*cos(3*A*(poles/2)/2), the fifth by 2*v*sin(5*A*(poles/2)/2), etc. (For four poles and a 5 degree offset the third harmonic is attenuated by about 3.5%, the fifth by 9.5%, the seventh by 18%, etc.) Eventually the sin function comes around back to 1 - and the associated harmonic may actually be attenuated less than the fundamental. But by that point you're FAR out into the harmonic tail. (For a 5 degree offset that's about the 26th harmonic, - which doesn't exists - but a slight angle error might give the 25th or 27th a half-percent boost over the fundamental.)
Net result is that the waveform will be symmetric, attenuated very slightly, and somewhat closer to a pure sine wave than the un-decogged rotor would have produced.