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frx

  • 3 years ago

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

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  1. frx
    • 3 years ago
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    @myininaya

  2. phi
    • 3 years ago
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    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

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