# Thread: Complex Numbers - Finding Roots

1. ## Complex Numbers - Finding Roots

Hi! I'm stuck on this question!

(a) For β = 1 - i√3, write the product z - β and z - conjugateβ as a quadratic expression in z, with real coefficients.

I got z^2 - 2z + 4

(b) (i) Express β in mod-arg form. Ans: 2cis(-π/3)
(ii) Find β^2 and β^3. Ans: 4cis(-2π/3) and 8cis(π)
(iii) Hence show that β is a root of z^3 - z^2 + 2z + 4 = 0. Stuck on this part, "hence"?

(c) Let the three roots be a, b, c. Let a be the point in the 1st quadrant, B the point on the real axis. Let c be the other root.

(i) Find the lengths AB and CB.
(ii) Describe the triangle ABC

2. Originally Posted by classicstrings
Hi! I'm stuck on this question!

(a) For β = 1 - i√3, write the product z - β and z - conjugateβ as a quadratic expression in z, with real coefficients.

I got z^2 - 2z + 4

(b) (i) Express β in mod-arg form. Ans: 2cis(-π/3)
(ii) Find β^2 and β^3. Ans: 4cis(-2π/3) and 8cis(π)
(iii) Hence show that β is a root of z^3 - z^2 + 2z + 4 = 0. Stuck on this part, "hence"?
You have shown that (z - beta) (z - beta') = z^2 - 2z +4, so:

(z+1)(z-beta)(z-beta') = (z+1)(z^2 - 2z +4) = z^3 - z^2 + 2z + 4

Hence z=beta is a root of z^3 - z^2 + 2z + 4=0.

RonL

4. Originally Posted by classicstrings
Hi! I'm stuck on this question!

(a) For β = 1 - i√3, write the product z - β and z - conjugateβ as a quadratic expression in z, with real coefficients.

I got z^2 - 2z + 4

(b) (i) Express β in mod-arg form. Ans: 2cis(-π/3)
(ii) Find β^2 and β^3. Ans: 4cis(-2π/3) and 8cis(π)
(iii) Hence show that β is a root of z^3 - z^2 + 2z + 4 = 0. Stuck on this part, "hence"?

(c) Let the three roots be a, b, c. Let a be the point in the 1st quadrant, B the point on the real axis. Let c be the other root.

(i) Find the lengths AB and CB.
(ii) Describe the triangle ABC
Attached is a diagram of the position of the three points in the complex
plane (the roots of z^3 - z^2 + 2z + 4 = 0), from which you should be able
to answer (c) parts i and ii.

RonL