You are correct in saying that you need a two tailed test. Now you want to test that the true proportion p0=0.5. Let X be the number of voters who voted for the candidate then X~N(np0,np0q0) where q0=1-p0 will approximate the binomial distribution that describes X. The test statistic is
Z = X-(np0) / ( sqrt(np0q0) )
but since the normal distribution is continous and the binomial is discrete you need to apply a continuity correction which involves adding or subtracting 0.5 from X. thus the test statistic is
Z = (X-0.5)-(np0) / ( sqrt(np0q0) )
=(1008-0.5)-(2011*0.5) / ( sqrt(2011*0.5*0.5) )
Where 1008 is obtained by finding 50.1% of 2011 and rounding up.
Did the question mean the test statistic is 0.009 and not 0.09?
Since your test is 2-tailed and say you want 95% confidence then look up the 97.5th percentile of the standard normal dis which is 1.96 and reject h0o=0.5 if the mod(Z)>1.96. Since our test statistic is 0.009 then we cannot reject h0.
the p-value for z=0.009 is approximately 1.
The conditions for the normal approximation to work are np0 and nq0 are both >5 which they are.
I hope this helps and that I have not mis-understood your question.