Question

If (1+i)(2+i)(3+i)....(n+i)=a+ib then show that 2.5.10....(n^2+1)=a^2+b^2 ?

Answer 1

Using:

• `absz_1 * absz_2 * ... * absz_n = abs ( z_1 * z_2* ...* z_n)`

• `implies absz_1^2 * absz_2^2 * ... * absz_n^2 = abs ( z_1 * z_2* ...* z_n)^2`

And

• `abs(z)^2 = z * bar z`

For the LHS:

`abs((1+i)(2+i)(3+i)....(n+i))^2`

`= abs(1+i)^2abs(2+i)^2abs(3+i)^2....abs(n+i)^2`

`= (1+i)bar((1+i)) (2+i)bar((2+i)) (3+i)bar((3+i))... (n+i)bar((n+i))`

`= (1+1) (4+1) (9+1)... (n^2+1) = 2.5.10....(n^2+1)`

For the RHS:

`abs(a + i b)^2 = a^2 + b^2`

`:.` LHS = RHS