Question

If $(2+i)(2+2i)(2+3i).....(2+ni)=x+iy$ then the value of $\mathrm{5.8.13.....}(4+n^2)$.

Answer 1

Given,

$(2+i)(2+2i)(2+3i).....(2+ni)=x+iy$

Now taking modulus both sides we get,

$|(2+i)(2+2i)(2+3i).....(2+ni)|=|x+iy|$

or, $|(2+i)|.|(2+2i)|.|(2+3i)|.....|(2+ni)|=|x+iy|$ [ Since $|ab|=|a|.|b|$]

Now squaring both sides we get,

or, $|(2+i){|}^2.|(2+2i){|}^2.|(2+3i){|}^2.....|(2+ni){|}^2=|x+iy{|}^2$

or, $\mathrm{5.8.13......}(4+n^2)=x^2+y^2$.