An algebra problem by Nick Okita

Algebra Level 4

Consider the complex polynomial p ( z ) = z 4 + a z 3 + 5 z 2 − i z − 6 p(z) = z^{4} + az^{3} + 5z^{2} - iz - 6 , where a is a complex constant. It's known that 2i is a root of p ( z ) = 0 p(z) = 0 , the other three roots are:

This question was taken from ITA (Instituto Tecnológico da Aeronáutica) Brazil 2013-2014 entrance exam.

-3i, -1, 1 -2i, -i, i -2i, -1, 1 -i, i, 1

Nick Okita by Nick Okita , 23, Brazil

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2 solutions

Since product of all roots is -6 and one of them is 2i. So product of remaining roots should be 3i. Only first option i.e. -3i, -1, 1 satisfy this condition. I know this is not a proper solution but you have to find some shortcut methods to solve a question of an objective exam to save some time.

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Nikhil Rajawat
Aug 30, 2014

Given that 2i is the root of p(z)=0.....this implies that a=-i. now substituting value of a in p(z) and dividing the p(z) by (z-2i)...we get , say g(z)=z^3+3iz^2-z-3i. now we can see that 1 is the root of g(z)=0. dividing g(z) with (z-1) we get say f(z)=z^2+(1+3i)z+3i and the roots of f(z)=0 can easily be calculated as -1 and -3i. hence 1 , -1 , -3i are the other three roots.

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good solution

will jain - 6 years, 7 months ago

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