Complex Numbers

[ninbam1k]

Hey guys, came across this problem and didn't know how to approach it. Thought maybe some VNers would be able to help me so here it is:

If Arg(z+a) = pi/6, Arg(z-a) = 2pi/3, find z where a is an element of R.



The points O, A, B represent complex numbers 0, a, and b respectively in the complex plane.

a. If a^2 + b^2 = 0, prove that OAB is an isosceles right angled triangle.

b. If OAB is an equilateral triangle, prove that a^2 + b^2 = ab

I have the answers but I don't know how to work it out :/

Thanks in advance

[brightsky]

For the first question, draw a graph, it makes it much easier.
tan(Arg(z+a)) = tan(pi/6) = y/(x+a) = 1/sqrt(3)
tan(Arg(z-a)) = tan(2pi/3) = y/(x-a) = -sqrt(3)
Rest should be trivial.

For question two:
We let a = x + yi, b = p + qi.
(x+yi)^2 + (p + qi)^2 = 0
x^2 + 2xyi - y^2 + p^2 + 2pqi - q^2 = 0
(x^2 - y^2 + p^2 - q^2) + (2xy + 2pq)i = 0
Hence we want x^2 - y^2 + p^2 - q^2 = 0
x^2 - y^2 = q^2 - p^2
We also want, 2xy = -2pq, meaning p = -y, q = x or p = y, q = -x.
Hence subbing in, we have b = -y + xi or b = y - xi, which are a turned 90 degrees clockwise or anticlockwise.

[evaever]

Q2  a^2=-b^2, so a=ib or a=-ib
Meaning a is the 90 deg anticlockwise rotation of b about o, or a is the 90 deg clockwise rotation of b about o
so OAB is isos right triangle

[evaever]

Do it in polar form
let b=rcis@, so a=rcis(@+pi/3)
de moivre.....
compound angle formula etc