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Question

The value of z satisfying the equation logz+logz2+...+logzn=0

A
cos4mπ(n(n+1)isin4mπn(n+1),m=1,2,....
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B
cos4mπn(n+1)+isin4mπn(n+1),m=1,2,...
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C
0
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D
sin4mπn(n+1)+icos4mπn(n+1),m=1,2,....
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Solution

The correct option is B cos4mπn(n+1)+isin4mπn(n+1),m=1,2,...
We have,
logz+logz2+...+logzn=0

log(zz2zn)=0

logzn(n+1)2=0

zn(n+1)2=1

zn(n+1)2=e2mπ,m=0,1,2,

z=e4mπn(n+1)

z=cos4mπn(n+1)+isin4mπn(n+1),m=0,1,2,

Hence, option (A) is correct.

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