Solve Quadratic Equation: Exact Cartesian Form

[nk735]

Hello,
My question comes in two parts, I don't know if the first part is relevant to the second so i'll put it in anyway.

a. Express 1 + root(3)i in polar form

I can solve this to get:

2cis(pi/3)

My problem is with part b.

b. Solve the quadratic equation z^2 + 2z - root(3)i = 0, expressing your answers in exact cartesian form

I used the quadratic formula (I don't like completing the square) to get:

z = (-2 + root(4 + 4root(3)i))/2 and z = (-2 -root(4 + 4root(3)i))/2

However, I'm lost with the 'exact cartesian form' part.

Any help would be appreciated, thanks.

 

[Dick]

I think cartesian form just means express the answer in the form a+bi. You now have to express the square roots of the complex quantities in that form. It might actually have been easier to complete the square.

 

[Awsom Guy]

Ok I think You should get the real part and the imaginary part:
I think this is:
-2/2 and -2/2 are the reals.
Sqrt(4+4Sqrt3i)/2 and -Sqrt(4+4Sqrt3i)/2
Put these together and you should get a real part and an imaginary part.
I hope that helps. Don't take this as the real answer I might be wrong.
Check with your teacher or your tutor or whoever.
I'm pretty sure that's right, I'll keep thinking...hmmmm...

 

[Awsom Guy]

wait I see an error.
(-2 + root(4 + 4root(3)i))/2
The 4+4root... is wrong it should be 4-4root...
Remeber the equation is -b+/- root(b^2-4ac)/2a
Try again it might work.