Let z be a complex number with |z| = 6. Let A be the point representing z. Let B be the point representing (1 + i)z.
a)
i. Find |(1 + i)z|. Done. The answer is .
ii. Find |(1 + i)z - z|. Done. The answer is 6.
iii. Prove that OAB is an isosceles right-angled triangle. Is this just a matter of a^{2} + b^{2} = c^{2}?
b) Let z_{1} and z_{2} be non-zero complex numbers satisfying z_{1}^{2 }- 2z_{1}z_{2} + 2z_{2}^{2} = 0.
If z_{1} = $\alpha$ z_{2}:
i) show that $\alpha$ = 1 + i or 1 - i. Done.
ii) for each of these values of $\alpha$ describe the geometrical nature of the triangle whose vertices are the origin and the points representing z_{1} and z_{2}. How would I go about doing this?
a) i) In complex numbers, each complex number works the same way as a vector. Segment OA is z. Segment OB is (1 + i)z. Using vector algebra, that means AB = OB - OA = (1 + i)z - z, thus they form three sides of a triangle. Because the three sides are in the ratio $\displaystyle \begin{align*} 6 : 6 : 6\sqrt{2} = 1 : 1 : \sqrt{2} \end{align*}$, the triangle is isosceles right angled.