Question

If $(1+i)(2+i)(3+i)\cdots \cdots (100+i)=a+ib,$ then the value of $2\cdot 5\cdot 10\cdots \cdots 10001$ is

• A. $\sqrt{a^2+b^2}$
• B. $a^2+b^2$
• C. $\sqrt{a^2-b^2}$
• D. $a^2-b^2$

Answer 1

The correct option is B $a^2+b^2$

Given: $(1+i)(2+i)(3+i)\cdots \cdots (100+i)=a+ib$

Applying modulus on both the sides, we get
$|(1+i)(2+i)(3+i)\cdots \cdots (100+i)|=|a+ib|\Rightarrow |(1+i)||(2+i)||(3+i)|\cdots \cdots |(100+i)|=|a+ib|\Rightarrow \sqrt{1^2+1^2}\sqrt{2^2+1^2}\cdots \cdots \sqrt{{100}^2+1^2}=\sqrt{a^2+b^2}\Rightarrow 2\cdot 5\cdot 10\cdots \cdots 10001=a^2+b^2$