MX2 Geometrical Properties question HELP! (1 Viewer)

[ledamn]

I need help with this question, if anyone wants to solve. It's a Complex Number question.
1. A(z) and B(w) form an equilateral triangle with the origin. With argw>argz
a) Express w in terms of z. Which is w=zcis60, which is pretty easy.
b) Prove that z^2 + w^2 = zw <- Can anyone work this solution?

2. The origin and the points representing the complex number z, 1/z and (z+1/z) are joined to form a quadrilateral. Write down the co-ordinates of z such that quadrilateral is:
a) A rhombus
b) A square

 

[Carrotsticks]

Question 1 (b): Cube both sides and notice that cis(180) is equal to -1. You will have the sum of two cubes, which you can then factorise.

Question 2 (a): I don't think there is a unique answer. The best we can do to impose that the quadrilateral formed is a rhombus is to say that z and 1/z have equal moduli (ie: both have moduli equal to 1). So Z can be any point on the unit circle (having Z lie on the axes results in a degenerate rhombus) and 1/Z will be its conjugate, so Z+1/Z will lie on the real axis.

Question 2 (b): We must have the conditions above (since all squares are rhombuses), as well as imposing that the vectors are 90 degrees apart. There are four possible locations of Z (one for each quadrant). They are all lying on the unit circle at an angle of 45 degrees from any set of axes, and length 1 from the origin. So the possible locations are

 

[ledamn]

Thank you!!