Question

If $(1-i)(1+2i)(1+3i)....(1+ni)=\alpha +i\beta$ then $2\cdot 5\cdot 10....(1+n^2)=$

• A. $\alpha -i\beta$
• B. ${\alpha }^2-{\beta }^2$
• C. ${\alpha }^2+{\beta }^2$
• D. None of these

Answer 1

Answer: (C)

${\alpha }^2+{\beta }^2$

• We know, $|x+iy|=\sqrt{x^2+y^2}$

Given $(1+i)(1+2i)(1+3i).............(1+ni)=\alpha +i\beta$

$\Rightarrow |(1+i)(1+2i)(1+3i).............(1+ni)|=|\alpha +i\beta |$

$\Rightarrow |1+i||1+2i|..........|1+ni|=\sqrt{{\alpha }^2+{\beta }^2}$

$\Rightarrow \sqrt{2}\sqrt{5}\sqrt{10}.......\sqrt{1+n^2}=\sqrt{{\alpha }^2+{\beta }^2}$

squaring on both the sides we get

$\mathrm{2.5.10........}(1+n^2)={\alpha }^2+{\beta }^2$