[Complex Numbers] Just need some confirmation. (1 Viewer)

[Pyrobooby]

Question from 2003 HSC paper:
Q2 (e)

Suppose that the complex number z lies on the unit circle, and 0≤arg (z)≤pi/2. Prove that 2arg(z+1)=arg(z).




In OABC:
OA // BC (since BC is a translation of OA 1 unit to the right)
AB // OC (since the segment between z and z+1 is parallel to the x-axis)
Hence OABC is a parallelogram.

In OAB and OBC:
OB is a shared side.
OA = OC (equal radii of unit circle)
AB = BC (opposite sides of a parallelogram are equal)

Hence OAB is congruent to OBC.

.'.< AOB = < BOC (corresponding angles in congruent triangles)
.'.< AOC = 2 < BOC (adjacent angles)
but < BOC = arg (z+1) amd < AOC = arg (z)

.'. arg (z) = 2 arg (z+1)

Is the above proof correct?

Thank you in advance.

 

[Drongoski]

The vectors corresponding to the 2 complex numbers: z and 1 are adjacent sides of //gram (OABC) with vector OB representing z+1 the included diagonal. (or: I could have simply stated: "OABC is a //gram since diagonal z+1 is vector sum of z and 1") Now |z| = |1| = 1 (i.e, OA = OC); That means the //gram OABC have equal adjacent sides. Therefore all 4 sides of OABC are equal. .: it is a rhombus. Diagonals of rhombus bisect their internal angles.

In this problem therefore diagonal OB bisects angle AOC. i.e. angle AOC = 2 x angle BOC.

i.e. arg(z) = 2 arg(z+1)



Edit

Your proof is not wrong.

All I'm suggesting is you can state straightaway OABC is //gram. (no need to prove it) following the parallellogram law of vector addition.. Also OA = OC means it is a rhombus. A rhombus has certain special properties like the diagonals bisect the angles of the //gram and they bisect each other at right angles (but you don't need this here).

 

[ohexploitable]

You could go straight from this:

In OABC:
OA // BC (since BC is a translation of OA 1 unit to the right)
AB // OC (since the segment between z and z+1 is parallel to the x-axis)
Hence OABC is a parallelogram.

To:

Diagonals of a parallelogram bisect the vertices,
.'. 2arg(z+1)=arg(z)

 

[Drongoski]

You could go straight from this:



To:

Diagonals of a parallelogram bisect the vertices,
.'. 2arg(z+1)=arg(z)

Not quite: bisects only if adjacent sides of //gram are equal; i.e. a rhombus.

 

[Pyrobooby]

Ahh right, thank you guys. I find geometry really iffy and I believe this is what I would have done in an exam situation.

Thanks again.

 

[Drongoski]

A lot of people nowadays are not very good at geometry because it is no longer a big part of school maths. But you have challenging geometry questions in MX2.

If you really want to be good at it, and provided you have what it takes, I can help.