The modulus of left( frac{3 + 2i}{3 - 2i} rig | vedclass

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(A) Let $z = \frac{3 + 2i}{3 - 2i}$.Using the property of modulus,$|\frac{z_1}{z_2}| = \frac{|z_1|}{|z_2|}$.$|z| = \left| \frac{3 + 2i}{3 - 2i} \right

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(A) Let $z = \frac{3 + 2i}{3 - 2i}$.Using the property of modulus,$|\frac{z_1}{z_2}| = \frac{|z_1|}{|z_2|}$.$|z| = \left| \frac{3 + 2i}{3 - 2i} \right| = \frac{|3 + 2i|}{|3 - 2i|}$.Since $|a + bi| = \sqrt{a^2 + b^2}$,we have $|3 + 2i| = \sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}$.Similarly,$|3 - 2i| = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}$.Therefore,$|z| = \frac{\sqrt{13}}{\sqrt{13}} = 1$.

The modulus of $\left( \frac{3 + 2i}{3 - 2i} \right)$ is

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