Final answer:
The complex function has singularities at points where sin(2iπz) equals zero, resulting in poles of order three. The numerator does not contribute new singularities since its zeros do not coincide with the sine function's zeros.
Step-by-step explanation:
The question asks to determine the nature of singularities of the given complex function (z-2i)³(z²+1)²/[sin(2iπz)]³. To analyze the singularities, we need to look for points where the denominator becomes zero, as those are the points that may cause the function to become undefined. The zeros of the sine function occur at z = n/i, where n is an integer. Hence, these points are the singularities of the given function.
To determine the nature of these singularities, we can classify them as either poles or essential singularities. A pole of order m occurs if near the singularity, the function can be written in the form (z-z_0)^-m times a holomorphic function. To find the order of the pole, we scrutinize the power of z in the denominator around the singularity. For each singularity z = n/i, as n varies over the integers, the function has a third-order pole because of the cubic power in the denominator. There could be additional singularities due to the numerator, which need to be analyzed separately to determine their nature.