Express the following complex number in the form a + ib : `[(1/5 + i7/5) + (4 + i1/3)] - (-4/3 + i)`

Explanation

Given , ` (1 + x)^(n) = C_(0) + C_(1) x + C_(2) x^(2) + …+ C_(n) x^(n)`
Integrating both sides w.r.t.x, within limits 0 to x , then we get
`int_(0)^(x) (1 + x)^(n) dx int_(0)^(x)(C_(0) + C_(1)x + C_(2)x^(2) + ...+ C_(n) x^(n))dx `
`((1 + x)^(n+1)-1)/((1+n)) = C_(0) x + (C_(1)x^(2))/(2) + (C_(2) x^(3))/(3) + ...+ (C_(n) x^(n+1))/(n+1)` ...(i)
and `(x +1)^(n) = C_(0) x^(n) + C_(1) x^(n-1) + C_(2) x^(n-2) + ...+ C_(n) ` ...(iii)
Multiplying Eqs (i) and (ii) , we get
` (1)/((n+1)) {(1 + x)^(2n+1) - (1 +x)^(n)}`
` = ( C_(0) x + (C_(1) x^(2))/(2) + (C_(2) x^(3))/(3) + ...+ (C_(n) x^(n+1))/(n+1))`
` xx(C_(0) x^(n) + C_(1) x^(n-1) + C_(2) x^(n-2) + ...+ C_(n))` ...(iii)
Now , coefficient of ` x^(n-1)` in RHS of Eq .(iii)
`= C_(0)^(2) + (C_(1)^(2))/(2) + (C_(2)^(2))/(3) + ...+ (C_(n)^(2))/(n+1) `
and coefficient of ` x^(n+1)` in RHS of Eq.(iii)
` = (1)/((n+1)) { ""^(2n+1)C_(n+1) -0}`
` = (1)/((n+1)) * ((2n +1)!)/((m+1)!n!) `
`= ((2n+1)!)/((n+1)!(n+1)!) = ((2n+1)!)/({(n+1)!}^(2))`
But Eq .(iii) is an identity , therefore coefficient of ` x^(n+1)` in
RHS of Eq.(iii) = coefficient of ` x^(n+1)` in LHS of Eq.(iii) .
` rArr C_(0)^(2) + (C_(1)^(2))/(2) + (C_(2)^(2))/(3) + ...+ (C_(n)^(n))/(n+1) = ((2n+1)!)/({(n+1)!}^(2))` .