Let be a complex number.
Consider the equation on the complex plane. How many zeroes does it have?
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Factoring: ( z − 1 ) ( z 2 + z + 1 ) / ( z − 1 ) ( z + 2 )
0/0 at z = 1. That is not a valid zero. The other two complex zeroes are valid, so it has two zeroes and one removable singularity.