If z_1 = 2 + 5i,z_2 = -1 + 4i,and z_3 = i,the | vedclass

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(A) Given: $z_1 = 2 + 5i$,$z_2 = -1 + 4i$,$z_3 = i$. First,calculate the numerator: $z_1 - z_3 = (2 + 5i) - i = 2 + 4i$. Next,calculate the denominato

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$\sqrt{2}$

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(A) Given: $z_1 = 2 + 5i$,$z_2 = -1 + 4i$,$z_3 = i$. First,calculate the numerator: $z_1 - z_3 = (2 + 5i) - i = 2 + 4i$. Next,calculate the denominator: $z_3 - z_2 = i - (-1 + 4i) = i + 1 - 4i = 1 - 3i$. Now,consider the fraction: $\frac{z_1 - z_3}{z_3 - z_2} = \frac{2 + 4i}{1 - 3i}$. Multiply the numerator and denominator by the conjugate of the denominator $(1 + 3i)$: $\frac{(2 + 4i)(1 + 3i)}{(1 - 3i)(1 + 3i)} = \frac{2 + 6i + 4i + 12i^2}{1^2 + 3^2} = \frac{2 + 10i - 12}{1 + 9} = \frac{-10 + 10i}{10} = -1 + i$. Finally,find the modulus: $\left| -1 + i \right| = \sqrt{(-1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2}$.

If $z_1 = 2 + 5i$,$z_2 = -1 + 4i$,and $z_3 = i$,then $\left| \frac{z_1 - z_3}{z_3 - z_2} \right| = $

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