The radius of the circle represented by (1+i) | vedclass

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(B) Given the equation $(1+i)(1+3i)(1+7i) = x+iy$. First,multiply the first two complex numbers: $(1+i)(1+3i) = 1 + 3i + i + 3i^2 = 1 + 4i - 3 = -2 +

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$10\sqrt{10}$

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(B) Given the equation $(1+i)(1+3i)(1+7i) = x+iy$. First,multiply the first two complex numbers: $(1+i)(1+3i) = 1 + 3i + i + 3i^2 = 1 + 4i - 3 = -2 + 4i$. Now,multiply the result by the third complex number: $(-2+4i)(1+7i) = -2 - 14i + 4i + 28i^2 = -2 - 10i - 28 = -30 - 10i$. Thus,$x = -30$ and $y = -10$. The radius of the circle represented by the complex number $z = x+iy$ in the Argand plane is given by $|z| = \sqrt{x^2 + y^2}$. $|z| = \sqrt{(-30)^2 + (-10)^2} = \sqrt{900 + 100} = \sqrt{1000} = 10\sqrt{10}$.

The radius of the circle represented by $(1+i)(1+3i)(1+7i) = x+iy$ is $(i = \sqrt{-1})$.

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