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

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∘