it has a root in the form z=ia where a is a real number
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Substitute z=ia in the quartic:

a⁴+2ia³-5a²-6ia+6=0=(a²-2)(a²-3)+2ia(a²-3)

So, since a²-3 is a common factor both real and imaginary parts become zero, so a²=3 is a solution. Therefore a=±√3 and two solutions are z=±i√3.

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