If the roots of the equation (z-4)^3=8 i are | vedclass

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(C) Given the equation $(z-4)^3=8 i$. Let $w = z-4$,then $w^3 = 8 i = 8 e^{i \pi/2}$.The roots are $w_k = 2 e^{i(\pi/2 + 2k\pi)/3}$ for $k=0, 1, 2$.Fo

Vedclass · Accepted Answer

$2 \sqrt{13}$

Vedclass · Answer

(C) Given the equation $(z-4)^3=8 i$. Let $w = z-4$,then $w^3 = 8 i = 8 e^{i \pi/2}$.The roots are $w_k = 2 e^{i(\pi/2 + 2k\pi)/3}$ for $k=0, 1, 2$.For $k=0$,$w_0 = 2 e^{i\pi/6} = 2(\cos(\pi/6) + i\sin(\pi/6)) = 2(\frac{\sqrt{3}}{2} + i\frac{1}{2}) = \sqrt{3} + i$.For $k=1$,$w_1 = 2 e^{i(5\pi/6)} = 2(\cos(5\pi/6) + i\sin(5\pi/6)) = 2(-\frac{\sqrt{3}}{2} + i\frac{1}{2}) = -\sqrt{3} + i$.For $k=2$,$w_2 = 2 e^{i(9\pi/6)} = 2 e^{i(3\pi/2)} = 2(0 - i) = -2 i$.Since $z = w+4$,the roots are $z_0 = 4+\sqrt{3}+i$,$z_1 = 4-\sqrt{3}+i$,and $z_2 = 4-2 i$.Comparing these with $a-2 i, b+i, c+i$,we identify $a=4, b=4+\sqrt{3}, c=4-\sqrt{3}$.Then $abc = 4(4+\sqrt{3})(4-\sqrt{3}) = 4(16-3) = 4(13) = 52$.Therefore,$\sqrt{abc} = \sqrt{52} = \sqrt{4 	imes 13} = 2 \sqrt{13}$.

If the roots of the equation $(z-4)^3=8 i$ are $a-2 i, b+i$,and $c+i$,then $\sqrt{a b c}=$

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