Show that the product `[1+(1+i)/2][1+((1+i)/2)^2][1+((1+i)/2)^(2^2)]...[1+((1+i)/2)^(2^n)]` is equal to `(1-1/2^(2^n))(1+i)`

What is the conjugate of (2-i)/((1-2i)^(2))?

If S_(n)=1 + (1)/(2) + (1)/(2) + …..+ (1)/(2^(n-1)), (n in N) then …….

Conver the following in the polar form: (i) (1+7i)/(2-i)^(2) , (ii) (1+3i)/(1-2i)

If z=(1+7i)/((2-i)^(2) then......

Prove the following by using the principle of mathematical induction for all n in N (1-(1)/(2^2)) (1-(1)/(3^2))(1-(1)/(4^2))…….(1-(1)/(n^2)) = (n+1)/(2n) " " n >= 2

(1)/(1*2)+(1)/(2*3)+(1)/(3*4)+"…."+(1)/(n(n+1)) equals

Prove by mathematical induction that (1)/(1+x)+(2)/(1+x^2)+(4)/(1+x^4)+.....+(2^n)/(1+x^(2^n))=(1)/(x-1)+(2^(n+1))/(1-x^(2^(n+1))) where , |x|ne 1 and n is non - negative integer.

Given that, z=(1+2i)/(1-i)=((1+2i)(1+i))/((1-i)(1+i))

Using mathematical induction , show that (1-(1)/(2^2))(1-(2)/(3^2))(1-(1)/(4^2)).....(1-(1)/((n+1)^2))=(n+2)/(2(n+1)), forall n in N .

By using the principle of mathematical induction , prove the follwing : 1 + (1)/(1+2) + (1)/(1+2+3) + …..+ (1)/(1+2+…..+n) = (2n)/(n+1) , n in N