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

Jul 28, 2018

Using:

• ${\left\mid z \right\mid}_{1} \cdot {\left\mid z \right\mid}_{2} \cdot \ldots \cdot {\left\mid z \right\mid}_{n} = \left\mid {z}_{1} \cdot {z}_{2} \cdot \ldots \cdot {z}_{n} \right\mid$

• $\implies {\left\mid z \right\mid}_{1}^{2} \cdot {\left\mid z \right\mid}_{2}^{2} \cdot \ldots \cdot {\left\mid z \right\mid}_{n}^{2} = {\left\mid {z}_{1} \cdot {z}_{2} \cdot \ldots \cdot {z}_{n} \right\mid}^{2}$

And

• ${\left\mid z \right\mid}^{2} = z \cdot \overline{z}$

For the LHS:

${\left\mid \left(1 + i\right) \left(2 + i\right) \left(3 + i\right) \ldots . \left(n + i\right) \right\mid}^{2}$

$= {\left\mid 1 + i \right\mid}^{2} {\left\mid 2 + i \right\mid}^{2} {\left\mid 3 + i \right\mid}^{2.} \ldots {\left\mid n + i \right\mid}^{2}$

$= \left(1 + i\right) \overline{\left(1 + i\right)} \left(2 + i\right) \overline{\left(2 + i\right)} \left(3 + i\right) \overline{\left(3 + i\right)} \ldots \left(n + i\right) \overline{\left(n + i\right)}$

$= \left(1 + 1\right) \left(4 + 1\right) \left(9 + 1\right) \ldots \left({n}^{2} + 1\right) = 2.5 .10 \ldots . \left({n}^{2} + 1\right)$

For the RHS:

${\left\mid a + i b \right\mid}^{2} = {a}^{2} + {b}^{2}$

$\therefore$ LHS = RHS