Let z be a function defined as zn=12iy[(−1)nn!(x−iy)n+1−(−1)nn!(x+iy)n+1], where i=√−1,n∈N and x,y∈R−{0}. If zn be the function of θ, where θ=tan−1yx, then which of the following is/are correct
A
z6(π2)=−4516 when y=2
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B
z6(π2)=4516 when y=2
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C
limθ→0yθz5z4=−6
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D
z4,z5,z6 are in G.P.
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Solution
The correct options are Az6(π2)=−4516 when y=2 Climθ→0yθz5z4=−6 zn=12iy[(−1)nn!(x−iy)n+1−(−1)nn!(x+iy)n+1]
Putting x=rcosθ,y=rsinθ Where r=√x2+y2,θ=tan−1(y/x)