Determine the number of zeros and poles inside (counted with multiplicity) of the function z6 sin(TZ) (1 z) (2 z) (3 z7) f(z) inside the disc z Z 4 5 We have that the zeros of the divisor are given as z 0 (multiplicity 7), z 1 (multiplicity 1), z 2 (multiplicity 1), z 3, and z 4 Nevertheless, when z 1, we have that both the dividend and divisor are equal to zero and so those two points are neither zeros nor poles As such, I reckon the number of zeros and poles (counted with multiplicity) inside the disc z z 4 5 are 7 3 2 13 and 2 7 9 Is this a fair argument

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Determine the number of zeros and poles inside (counted with multiplicity) of the function z6 sin(TZ) (1 - z) (2 - z) (3 z7)

Determine the number of zeros and poles inside (counted with multiplicity) of the function z6 sin(TZ) (1 - z) (2 - z) (3 z7) f(z) inside the disc {z: |Z|

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