If x+iy = frac{(3+2i)(4-7i)(12+13i)}{(13-12i) | vedclass

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(C) Given,$z = x+iy = \frac{(3+2i)(4-7i)(12+13i)}{(13-12i)(2-3i)(11+3i)}$.Taking the modulus on both sides,we have $|z| = |x+iy| = \sqrt{x^2+y^2}$.Usi

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$\frac{1}{2}$

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(C) Given,$z = x+iy = \frac{(3+2i)(4-7i)(12+13i)}{(13-12i)(2-3i)(11+3i)}$.Taking the modulus on both sides,we have $|z| = |x+iy| = \sqrt{x^2+y^2}$.Using the property $|\frac{z_1 z_2 z_3}{z_4 z_5 z_6}| = \frac{|z_1| |z_2| |z_3|}{|z_4| |z_5| |z_6|}$,we get:$|z| = \frac{|3+2i| \cdot |4-7i| \cdot |12+13i|}{|13-12i| \cdot |2-3i| \cdot |11+3i|}$.Note that $|3+2i| = |2-3i| = \sqrt{3^2+2^2} = \sqrt{13}$.Also,$|12+13i| = |13-12i| = \sqrt{12^2+13^2} = \sqrt{313}$.Substituting these values,we get:$|z| = \frac{\sqrt{13} \cdot |4-7i| \cdot \sqrt{313}}{\sqrt{313} \cdot \sqrt{13} \cdot |11+3i|} = \frac{|4-7i|}{|11+3i|}$.$|z| = \frac{\sqrt{4^2+(-7)^2}}{\sqrt{11^2+3^2}} = \frac{\sqrt{16+49}}{\sqrt{121+9}} = \frac{\sqrt{65}}{\sqrt{130}} = \frac{\sqrt{65}}{\sqrt{2 	imes 65}} = \frac{1}{\sqrt{2}}$.Since $|z| = \sqrt{x^2+y^2}$,we have $\sqrt{x^2+y^2} = \frac{1}{\sqrt{2}}$.Squaring both sides,$x^2+y^2 = \frac{1}{2}$.

If $x+iy = \frac{(3+2i)(4-7i)(12+13i)}{(13-12i)(2-3i)(11+3i)}$,then $x^2+y^2=$

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