We can write 2=e^(ln(2)) and we can write z=a+ib.
e^(aln(2))×e^(ibln(2)). Let r=e^(aln(2)).
But e^(ibln(2)) can be expressed as cosθ+isinθ where θ=bln(2).
So we have re^iθ to replace the original 2^z, where r is the modulus |2^z|.
So r=1=e^(aln(2)) making a=0. Since a is the real part of z, Re(z)=0, answer A.