(z-2i) (z – 3) Given f (z) =- %3D z² (z +5)3 (z–2i)4 Which of the following is correct ? z = 0 is a pole of order 3 Oz= 2i is a pole of order 4 oz=5 is a pole of order 3 oz = 2i is a pole of order 3

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A: I have used the cauchy integral formula.

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A: ∫ydyy+2 Substitute, u+5=y+2 Then , du=12y+2dy ⇒2y+2 du =dy ⇒2u+5 du =dy and y=u+52-2=u2+10u+23

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