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If $m_1, m_2, m_3$ and $m_4$ respectively denote the moduli of the complex numbers $1+4 i, 3+i, 1-i$ and $2-3 i$, then the correct one, among the following is
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Verified Answer
The correct answer is:
$m_3 < m_2 < m_4 < m_1$
Let $z_1=1+4 i, \quad z_2=3+i, \quad z_3=1-i$ and $z_4=2-3 i$
$$
\begin{array}{ll}
\therefore & m_1=\left|z_1\right|, m_2=\left|z_2\right|, m_3=\left|z_3\right| \\
\text { and } & m_4=\left|z_4\right| \\
\Rightarrow & m_1=\sqrt{1+4^2}, m_2=\sqrt{3^2+1^2}, \\
& m_3=\sqrt{1^2+1^2} \text { and } m_4=\sqrt{2^2+3^2} \\
\Rightarrow & m_1=\sqrt{17}, m_2=\sqrt{10}, m_3=\sqrt{2} \\
\text { and } & m_4=\sqrt{13} \\
\Rightarrow & m_3 < m_2 < m_4 < m_1
\end{array}
$$
$$
\begin{array}{ll}
\therefore & m_1=\left|z_1\right|, m_2=\left|z_2\right|, m_3=\left|z_3\right| \\
\text { and } & m_4=\left|z_4\right| \\
\Rightarrow & m_1=\sqrt{1+4^2}, m_2=\sqrt{3^2+1^2}, \\
& m_3=\sqrt{1^2+1^2} \text { and } m_4=\sqrt{2^2+3^2} \\
\Rightarrow & m_1=\sqrt{17}, m_2=\sqrt{10}, m_3=\sqrt{2} \\
\text { and } & m_4=\sqrt{13} \\
\Rightarrow & m_3 < m_2 < m_4 < m_1
\end{array}
$$
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