Roots of a equation with complex numbers

subopolois
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Homework Statement


how do i find the roots of this: x^2-(2-1)x+(3-i)=0


Homework Equations


-b+-sqrtb^2-4ac/2a (quardaric equation)


The Attempt at a Solution


 
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What have you done so far?

And why did you write it as "(2-1)x" and not "x"? Did you mean "(2-i)x" instead? Why can't you just apply the formula?
 
Defennder said:
What have you done so far?

And why did you write it as "(2-1)x" and not "x"? Did you mean "(2-i)x" instead? Why can't you just apply the formula?

and what formula would that be?
 
subopolois said:
and what formula would that be?

Use the 'quadratic equation'.
 
The quadratic formula that YOU gave in your first post! Be careful about taking the square root. That's the only "hard" part.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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