For any complex number w=a+bi, where a,b∈R and i=−1. If w=cos⁡40∘+isin⁡40∘, then w+2w2+3w3+…+9w9−1 equals

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19sin⁡40∘sec⁡20∘

29sin⁡20∘

19cos⁡40∘

92cosec⁡20∘

answer is A.

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Detailed Solution

w=ei2π9 Let S=w+2w2+3w3+…+9w9→1⇒ws=w2+………..9w10→21-2⇒(1−w)s=w+w2+…..+w9−9w10 =0−9w10⇒s=−9w101−w⇒|s|=92sin⁡20∘⇒|s|−1=29sin⁡20∘

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