• anonymous
$z ^{3}+2(1−6i)z ^{2}+2(1−12i)z−24i=0$ $−ib ^{3}−2(1−6i)b ^{2}+2(1−12i)bi−24i=0$ $−ib ^{3}−2b ^{2}+12ib ^{2}+2bi+24b−24i=0$ $\Re=−2b ^{2}+24b=0$ $\Re=−2b(b−12)=0$ $\Im=(−b ^{3}+12b ^{2}+2b−24)i=0$ $\Im=(−b ^{2}+2)(b−12)i=0$ Clean imaginary root $z=12i$ How do I continue from here, the root with both Re and Im part, how do I find it?
Mathematics

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