Can Substituting -1 for i in Complex Cubic Equations Yield Accurate Solutions?

[Noir]

I know this isn't in the right format, but I figured I'd get a better answer here than anywhere else. In my last exam, there was a question asking to prove (a + bi - except there were values for a and b, but i forgot them) was a solution to a polynomial of the 3rd degree.

Said polynomial was complex. I sub'd it in and got 0, so it worked. Here's the catch. To find the other two solutions, I subbed -1 in for i so the equation wasn't complex. I have no idea why I did it, i just remember it working. Lo and behold I got it right, the cubic equation I got had the same solutions as the quadratic you would get if you found used the long division method. I wasn't thinking and didn't want to deal with the i's.

I want to know, does the above methord work? Subbing in -1 for i and then solving? If it does I just went from a B to an A and I'm happy. I tried looking on the internet, but couldn't find anything.

Cheers,

James.

 

[HallsofIvy]

Unfortunately, I have no idea what you are saying. Do you mean that the coefficients of the polynomial themselves were complex numbers? It would help a lot if you were to show what polynomial you are talking about. In general replacing "i" with "-1" will not give you anything worhwhile ("i" and "-1" are NOT equal and one CANNOT replace the other) so it sounds like you were just lucky in this particular example.

 

[Noir]

Sorry for any confusion. Much like this equation right here;
z^4 + 3iz^3 - (4 + i)z^2 - 3iz + 3 + i = 0
I think your right; I was just lucky my dodgy maths worked! I don't see how replacing i with -1 would, or even could work now...