If `I_(n)=int z^(n)e^(1//z)dz`, then show that `(n+1)! I_(n)=I_(0)+e^(1//z)(1!z^(2)+2!z^(3)+...+n!z^(n+1))`.

If the complex numbers z_1, z_2,.......z_n lie on te unit circle |z| = 1 then show that |z_1 + z_2 + ..... + z_n| = |z_1^(-1)1 + z_2^(-1) + ....... + z_n^(-1)|.

Define addition and multiplication of two complex numbers z_(1) and z_(2) . Hence show that: (i) R_(e)(z_(1)+z_(2))=R_(e)(z_(1))+R_(e)(z_(2)) (ii) I_(m)(z_(1)+z_(2))=I_(m)(z_(1))+I_(m)(z_(2)) (iii) R_(e)(z_(1)z_(2))=R_(e)(z_(1))R_(e)(z_(2))-I_(m)(z_(1))I_(m)(z_(2)) (iv) I_(m)(z_(1)z_(2))=R_(e)(z_(1))I_(m)(z_(2))+I_(m)(z_(1))R_(e)(z_(2)) .

if z_(1),z_(2),z_(3),…..z_(n) are complex numbers such that |z_(1)|=|z_(2)| =….=|z_(n)| = |1/z_(1) +1/z_(2) + 1/z_(3) +….+1/z_(n)| =1 Then show that |z_(1) +z_(2) +z_(3) +……+z_(n)|=1

If n|q and A={z in C:z^(n)=1},B{z:z^(p)=1} then

Let z=a+ib=re^(i theta) where a, b, theta in R and i=sqrt(-1) Then r=sqrt((a^(2)+b^(2)))=|Z| and theta=tan^(-1)((b)/(a))=arg(z) Now |z|^(2)=a^(2)+b^(2)=(a+ib)(a-ib)=zbar(z) rArr(1)/(2)=(bar(z))/(|z|^(2)) and |z_(1)z_(2)z_(3)......z_(n)|=|z_(1)||z_(2)||z_(3)|...|z_(n)| If |f(z)|=1 ,then f(z) is called unimodular. In this case f(z) can always be expressed as f(z)=e^(i alpha), alpha in R Also e^(i alpha)+e^(i beta)=e^(i((alpha+beta)/(2)))*2cos((alpha-beta)/(2)) and e^(i alpha)-e^(i beta)=e^(i((alpha+beta)/(2)))*2i sin((alpha-beta)/(2)) where alpha, beta in R Q:If Z_(1),Z_(2),Z_(3) are complex number such that |Z_(1)|=|Z_(2)|=|Z_(3)|=|Z_(1)+Z_(2)+Z_(3)|=1 , then |(1)/(Z_(1))+(1)/(Z_(2))+(1)/(Z_(3))| is

If n is a natural number gt 2 , such that z^(n) = (z+1)^(n) , then

If z_(1)=cos theta+i sin theta and 1,z_(1),(z_(1))^(2),(z_(1))^(3),......,(z_(1))^(n-1) are vertices of a regular polygon such that (Im(z_(1))^(2))/(ReZ_(1))=(sqrt(5)-1)/(2), then the value n is