Question

If the roots of z2+ax+b=0z2+ax+b=0 are purely imaginary, then

Answer 1

Answer:

Let r be the purely imaginary root, so that r¯¯=−r. We have these conjugate equations:

ar2+br1+c=0anda¯¯¯r2−b¯¯r1+c¯¯=0

Here's a neat trick: Move the exponents on r to subscripts ...

ar2+br1+c=0anda¯¯¯r2−b¯¯r1+c¯¯=0

... and solve the linear system for them ...

r1=ac¯¯−a¯¯¯cab¯¯+a¯¯¯br2=−bc¯¯+b¯¯cab¯¯+a¯¯¯b

(Dealing with the case of ab¯¯+a¯¯¯b=0 is left as an exercise to the reader.) Now, return the subscripts to their positions as exponents ...

r1=ac¯¯−a¯¯¯cab¯¯+a¯¯¯br2=−bc¯¯+b¯¯cab¯¯+a¯¯¯b

... and simply observe the deep algebraic truth, r2=(r1)2 ...

(ac¯¯−a¯¯¯cab¯¯+a¯¯¯b)2=−bc¯¯+b¯¯cab¯¯+a¯¯¯b

Squaring and clearing fractions, we have

(ac¯¯−a¯¯¯c)2=−(ab¯¯+a¯¯¯b)(bc¯¯+b¯¯c)