Find roots of cubic polynomial with complex coefficient

  • Thread starter BearY
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  • #1
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8

Homework Statement


Find roots of
$$
-\lambda ^3 +(2+2i)\lambda^2-3i\lambda-(1-i) = 0
$$

Homework Equations




The Attempt at a Solution


I tried my old trick
I tried to separating the 4 terms into 2 pairs and try to find a common factor in the form of ##\lambda + z## between them,
$$
-\lambda ^2 (\lambda -2-2i) - 3i\lambda -1+i
$$
It doesn't seem to work.
$$
-\lambda (\lambda^2 +3i) - (2+2i)\lambda -1+i
$$
Nope.
And I am out of tricks.
Should I keep trying to factorize it or there is something else I should do about polynomial with complex coefficient?
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement


Find roots of
$$
-\lambda ^3 +(2+2i)\lambda^2-3i\lambda-(1-i) = 0
$$

Homework Equations




The Attempt at a Solution


I tried my old trick
I tried to separating the 4 terms into 2 pairs and try to find a common factor in the form of ##\lambda + z## between them,
$$
-\lambda ^2 (\lambda -2-2i) - 3i\lambda -1+i
$$
It doesn't seem to work.
$$
-\lambda (\lambda^2 +3i) - (2+2i)\lambda -1+i
$$
Nope.
And I am out of tricks.
Should I keep trying to factorize it or there is something else I should do about polynomial with complex coefficient?
You can use Cardano's formula for the solution of a cubic. It works for any coefficients, not just for real ones.
 
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  • #3
LCKurtz
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The roots are ##1,~i,~1+i##.
 
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  • #4
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The shortest way is usually to guess one root, divide the polynomial by the corresponding linear factor and solve the quadratic rest polynomial. That ##1## is a root helps a lot.
 
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