If (3+4i)^{2025} = 5^{2023}(x+iy),then sqrt{x | vedclass

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(B) Given the equation $(3+4i)^{2025} = 5^{2023}(x+iy)$.Taking the modulus on both sides,we get $|(3+4i)^{2025}| = |5^{2023}(x+iy)|$.Since $|z^n| = |z

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$25$

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(B) Given the equation $(3+4i)^{2025} = 5^{2023}(x+iy)$.Taking the modulus on both sides,we get $|(3+4i)^{2025}| = |5^{2023}(x+iy)|$.Since $|z^n| = |z|^n$,we have $|3+4i|^{2025} = 5^{2023} \cdot |x+iy|$.We know that $|3+4i| = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5$.Substituting this into the equation,we get $5^{2025} = 5^{2023} \cdot \sqrt{x^2+y^2}$.Dividing both sides by $5^{2023}$,we get $\sqrt{x^2+y^2} = \frac{5^{2025}}{5^{2023}} = 5^{2025-2023} = 5^2 = 25$.

If $(3+4i)^{2025} = 5^{2023}(x+iy)$,then $\sqrt{x^2+y^2} = $

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