Complex number questions (1 Viewer)

[MatrixM]

I've got a test tomorrow and going through the past paper questions I was a lkittle unsure how to do these.
First-


And secondly:
The points z[1] = 4-i, z[2] = 2i and z[3] make an equilateral triangle. Find the position of z[3]

 

[deterministic]

(i) a+b=[sum of roots of z^7-1] -1 = 0-1 =-1

(ii) Easy, do it yourself. Just expand and simplify, remembering p^7=1 and 1+p+p^2+...+p^6=0

(iii) note sum of roots = - (a+b)=1
product of roots =ab from (ii)
So equation is x^2+x+ your answer from (ii)

(iv) Solve quadratic in (iii) using quadratic equation. equate real parts of p+p^2+p^4 with the real part of the roots you found in quad equation.

 

[rawrence]

And secondly:
The points z[1] = 4-i, z[2] = 2i and z[3] make an equilateral triangle. Find the position of z[3]

Equilateral triangle, angles of 60 degrees. Use rotation by 60 degrees i.e. multiple by cis pi/3

 

[deterministic]

or cis(-pi/3). There should be 2 possible solutions

 

[rawrence]

or cis(-pi/3). There should be 2 possible solutions

I stand corrected

 

[hscishard]

If you need a bit more info for iv)
The sum of roots is -1.
2cos2pi/7 + 2cos4pi/7 + 2cos8pi/7 = -1
Then divide by 2 on both sides to answer the question

 

[khorne]

product of roots is easier than you think. Use the fact that p^8 = p, p^9 = p^2 and p^10 = p^3, and you should find it. Additionally, keep in mind p^7 = 1