Thread: Complex Number Question.

1. Complex Number Question.

The vertices O, A, B of a triangle in the Argand diagram are the points corresponding to the numbers 0, 1 , 1 + i respectively. Show that when the point z describes the perimeter of the triangle OAB, then locus of the points $\displaystyle z^2$ consists of part of the real axis part of a parabola and part of the imaginary axis. Sketch this locus.

[FP3 review exercise question 51]
I barely understand what this question is asking. excuse my ignorance if this is totally wrong, but isn't z just constant ? as it is the perimeter of a fixed triangle. I am a bit lost on this.

Bobak

2. Originally Posted by bobak
I barely understand what this question is asking. excuse my ignorance if this is totally wrong, but isn't z just constant ? as it is the perimeter of a fixed triangle. I am a bit lost on this.

Bobak
It's a poorly worded question alright. Here's my take on it:

Consider the values of z along the side OA:
z = a + 0i, 0 < a < 1,
=> z^2 = a^2, 0 < a < 1. These points obviously lie on part of the real axis. The part from 0 to 1.

Consider the values of z along the side OB:
z = a + ai, 0 < a < 1,
=> z^2 = 2ai, 0 < a < 1. These points obviously lie on part of the imaginary axis. The part from 0 to 2i.

Consider the values of z along the side AB:
z = 1 + ai, 0 < a < 1,
=> z^2 = (1 + ai)^2 = 1 - a^2 + 2ai, 0 < a < 1. So:

x = 1 - a^2 .... (1)

y = 2a => y/2 = a .... (2)

Treat (1) and (2) as parametric equations and eliminate a by substituting (2) into (1):

x = 1 - (y/2)^2, which is the equation of a sideways parabola. The domain is 0 < x < 1. The range is 0 < y < 2.

So as z TRACES the perimeter of the triangle, the locus of z^2 does indeed consist of part of the real axis, part of a parabola and part of the imaginary axis. The sketch of this locus should be straightforward.

3. Thanks a lot Mr F. the locus is a slice of cheese brilliant!

4. Originally Posted by bobak
Thanks a lot Mr F. the locus is a slice of cheese brilliant!
Or a piece of cake lol!