Question

(1-i) (1-2i)..... (1-ni) =x-iy then 2.5.10.••••(1+n^2)

Answer 1

Given (1-i)(1-2i)(1-3i).............(1-ni)=x-iy
(1-i)(1-2i)(1-3i).............(1-ni)=x-iy
⇒|(1-i)(1-2i)(1-3i).............(1-ni)|=|x-iy|
⇒|(1-i)(1-2i)(1-3i).............(1-ni)|=|x-iy|
⇒|1-i||1-2i|..........|1-ni|=root(x^2+y^2)−−−−−−
⇒|1-i||1-2i|..........|1-ni|=root(x^2+y^2)
⇒√2–√5–√10−−.......(√1+n^2)−−−−−(√x2+y2)−−−−−−
squaring on both the sides we get
2.5.10........(1+n2)=x^2+y^2