Question

If the point $z=(1+i)(1+2i)(1+3i).....(1+10i)$ lies on a circle with centre at origin and radius $r$, then $r^2=$

• A. $10!$
• B. $2\times 3\times 4\times ......\times 10$
• C. $2\times 5\times \mathrm{10.....}\times 101$
• D. $11!$

Answer 1

The correct option is C $2\times 5\times \mathrm{10.....}\times 101$

$z$ lies on the circle centered at origin.

Therefore, distance of $z$ frorm origin is equal to the radius of the circle.

$⟹|z-0|=r$

$⟹|(1+i)(1+2i)(1+3i)...(1+10i)-0|=r$

$⟹|(1+i)||(1+2i)||(1+3i)|...|(1+10i)|=r$ ...($|ab|=|a||b|$)

$⟹\left(\sqrt{1^2+1^2}\right)\left(\sqrt{1^2+2^2}\right)\left(\sqrt{1^2+3^2}\right)...\left(\sqrt{1^2+{10}^2}\right)=r$

$⟹\left(\sqrt{2}\right)\left(\sqrt{5}\right)\left(\sqrt{10}\right)...\left(\sqrt{101}\right)=r$

Squaring above equation we get,

$⟹2\times 5\times \mathrm{10...}\times 101=r^2$

Therefore, answer is option (C)