z 2 + 2z − i sqroot(3) = 0

[ggxoxo]

Why won't the quad formula work for z 2 + 2z − i sqroot(3) = 0

[ggxoxo]

It's VCAA 2006 last qn btw

[Phy124]

It will work, but it won't give the answer in cartesian form.

Your first step in answering the question will be using the quadratic formula, to have an answer in the form
Your second step will be converting these answers to cartesian form.

HINT: you showed that in part a, use this and apply de Moivre's theorem, to derive your answers in cartesian form

edit: misspelled cartesian and couldn't help myself

[Somye]

because when you apply the formula, you're going to have root(4+4i root(3)), and this in itself becomes another equation you have to solve...

[ggxoxo]

Thank you so much rangaaaaa and Somye!!!!!

[b^3]

EDIT: Typed this out, so will post anyway... but SPOLIER ALERT THEN, try it first with the above guys info





It will give you a square root under a square root which we don't want, so we have to convert it to polar form and put it through that way.

Now converting the thing under the square root to polar form.

Then we have

[Special At Specialist]

b^3 was correct:

z^2 + 2z – i*sqrt(3) = 0
(z + 1)^2 – 1 – i*sqrt(3) = 0
(z + 1)^2 = 1 + i*sqrt(3)
(z + 1)^2 = 2cis(pi/3)
z + 1 = sqrt(2)cis(pi/6) or z + 1 = sqrt(2)cis(-5pi/6)
z = -1 + sqrt(2)(sqrt(3)/2 + ½ i) or z = -1 + sqrt(2)(-sqrt(3)/2 – ½ i)
z = -1 ± (sqrt(6)/2 + sqrt(2)/2 i)
So the two solutions are:
z = -1 + sqrt(6)/2 + (sqrt(2)/2)i or z = -1 - sqrt(6)/2 - (sqrt(2)/2)i

[b^3]

Remember to read the question

Why won't the quad formula work for z 2 + 2z − i sqroot(3) = 0

z^2 – 2z – i*sqrt(3) = 0

EDIT: 2100^th post.