Quadratic equation with complex coefficients

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 9K views
elimenohpee
Messages
64
Reaction score
0

Homework Statement


Solve the quadratic equation

z^2 + 4(1 + i(3^0.5))z - 16 = 0


Homework Equations





The Attempt at a Solution


I think I've done this correctly, I just wanted to verify.
I've only done the solution for k=0

http://i129.photobucket.com/albums/p201/elimenohpee182/Capture-1.png
 
Physics news on Phys.org
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?
 
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, that's why I wanted someone to check it.
 
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