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

Question

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

anonymous · Answer

@myininaya

phi · Answer

You assumed z was pure imaginary, and found a consistent solution with b=12:

so one root is 12i
You can divide the original by z-12i  to get  z^2+2z+2 =0
you can factor this to find the remaining two roots