If the point z=(1+i)(1+2i)(1+3i) ldots (1+10i | vedclass

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(C) The equation of a circle with center at the origin and radius $r$ is given by $|z| = r$.Given $z = (1+i)(1+2i)(1+3i) \ldots (1+10i)$.Taking the mo

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$2 	imes 5 	imes 10 	imes \ldots 	imes 101$

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(C) The equation of a circle with center at the origin and radius $r$ is given by $|z| = r$.Given $z = (1+i)(1+2i)(1+3i) \ldots (1+10i)$.Taking the modulus on both sides:$|z| = |1+i| \cdot |1+2i| \cdot |1+3i| \ldots |1+10i| = r$.Using the property $|a+bi| = \sqrt{a^2+b^2}$,we get:$|z| = \sqrt{1^2+1^2} \cdot \sqrt{1^2+2^2} \cdot \sqrt{1^2+3^2} \ldots \sqrt{1^2+10^2} = r$.$|z| = \sqrt{2} \cdot \sqrt{5} \cdot \sqrt{10} \ldots \sqrt{101} = r$.Squaring both sides,we obtain:$r^2 = 2 	imes 5 	imes 10 	imes \ldots 	imes 101$.

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

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