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If the point $z=(1+i)(1+2 i)(1+3 i) \ldots$ $(1+10 i)$ lies on a circle with centre at origin and radius $r$, then $r^2$ is equal to
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Verified Answer
The correct answer is:
$2 \times 5 \times 10 \times \ldots \times 101$
Any circle with centre at origin and radius $r$ has the equation
$$
\begin{array}{rlrl}
& & |z| & =r \\
\Rightarrow & & |1+i||1+2 i||1+3 i| \ldots|1+10 i| & =r \\
\Rightarrow & \sqrt{1^2+1^2} \cdot \sqrt{1^2+2^2} \cdot \sqrt{1^2+3^2} \ldots \sqrt{1^2+10^2} & =r \\
\Rightarrow & & \sqrt{2} \cdot \sqrt{5} \cdot \sqrt{10} \ldots \sqrt{101} & =r \\
& \therefore & r^2=2 \times 5 \times 10 \times \ldots \times 101
\end{array}
$$
$$
\begin{array}{rlrl}
& & |z| & =r \\
\Rightarrow & & |1+i||1+2 i||1+3 i| \ldots|1+10 i| & =r \\
\Rightarrow & \sqrt{1^2+1^2} \cdot \sqrt{1^2+2^2} \cdot \sqrt{1^2+3^2} \ldots \sqrt{1^2+10^2} & =r \\
\Rightarrow & & \sqrt{2} \cdot \sqrt{5} \cdot \sqrt{10} \ldots \sqrt{101} & =r \\
& \therefore & r^2=2 \times 5 \times 10 \times \ldots \times 101
\end{array}
$$
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