Solve for all complex numbers z such that z4+4z2+6=z.
Solution
Solution:
Rewrite the given equation as (z2+2)2+2=z. Observe that a solution to z2+2=z is a solution of the quartic by substitution of the left hand side into itself. This gives z=21±i7. But now, we know that z2−z+2 divides into (z2+2)2−z+2=z4+4z2−z+6. Factoring it out, we obtain (z2−z+2)(z2+z+3)=z4+4z2−z+6. Finally, the second term leads to the solutions z=2−1±i11.
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Source: MathNet,
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