Question

If $(1+i)(2i+1)(1+3i)....(1+ni)=x+iy,$ then $\mathrm{2.5.10....}(1+n^2)$ is equal to

• A. $1$
• B. $i$
• C. $x^2+y^2$
• D. $1+n^2$

Answer 1

Answer: (C)

Given, $(1+i)(2i+1)(1+3i)...(1+ni)+x+iy$ ..(i)
$\because$ $(a_1+i{b}_1)(a_2+i{b}_2)...(a_n+i{b}_n)=A+iB$
$\Rightarrow$ $r_1.r_2.....r_n[\cos ({\theta }_1+{\theta }_2+.....+{\theta }_n)$
$+i\sin ({\theta }_1+{\theta }_2+.....+{\theta }_n)]$
$=A+iB$
$\Rightarrow$ $r_1.r_2....r_n=\sqrt{A^2+B^2}$ and $\tan ({\theta }_1+{\theta }_2+.....+{\theta }_n)=\frac{B}{A}$ ..(iii)
Now, from Eq. (ii), we get
$\sqrt{1^2+1^2}.\sqrt{2^2+1^2}.\sqrt{3^2+1^2}.....\sqrt{1^2+n^2}=\sqrt{x^2+y^2}$
$\Rightarrow$ $\sqrt{2}.\sqrt{5}.\sqrt{10}....\sqrt{1+n^2}=\sqrt{x^2+y^2}$
$\Rightarrow$ $\mathrm{1.2.5.10....}(1+n^2)+x^2+y^2$