Quadratic equation with complex coefficients

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  • #2
verty
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I'm interested to know, why did you use polar coordinates? Would it not be easier to let z = a + b.i, then solve for a and b?
 
  • #3
I thought using polar coordinates would be easiest to eliminate the square root of the complex number.

I don't know if its right or not, thats why I wanted someone to check it.
 
  • #4
Mentallic
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You want to get rid of the square root of the determinant, so let [tex]\sqrt{-96+32i}=a+ib[/tex] on squaring both sides, we solve [tex]-96+32i=a^2-b^2+2abi[/tex]

Thus you have two equations to solve, [tex]-96=a^2-b^2[/tex] and [tex]32=2ab[/tex] since the real and imaginary coefficients must be equal.

But first you may want to check if you can simplify -96+32i. Notice 96=32*3
 

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