2023 HSC Mathematics Extension II Question 16 c)

This is my solution for 2023 HSC Mathematics Extension II Question 16.c).
I don't know what happened to the audio input on my computer when I recorded this.

[smerriman]

Just saw this in the news and solved it myself without really needing to plot things on a circle at all. If z/w is cis (T) where T is from pi/2 to pi, then (xz + yw)/z = x + y(w/z) = x + y cis(-T). So basically, from the origin we move x right, and then y in the direction of -T, which is pointing towards the 3rd quadrant (southwesterly). To end up above and to the left of the origin, x and y must thus both be negative, and the limit on y is based on a right angled triangle with base -x, hypotenuse -y, and angle pi/2 - T.

[jamess]

My brain hurt but you explained it with surprising clarity.

[WillGray-pz]

Will the TGC collection be restocked? I've had a pair of black oxfords for 2 years from them and they're fantastic and I was seeing if I could get another pair.

[odysseymachine]

As a qualified Dog food tester i can confirm this solution is accurate.

[matmagix]

I'm curious, given this is a three mark question, how would a marker devote partial marks for an attempt even if the final solution is not reached? Would a mark be allocated for knowing that the angle between z and w should be obtuse?

[drtonyburns]

It will take some thought but it makes more sense than the one in SMH.

[stevendouglas]

Do other scenarios need the be considered? For example z in the first quadrant and w in the 4th quadrant?

[alexandros]

Did maths at Uni, and still don't know what he's talking about. Bit much to spring on students at an HSC exam!

[Fred-yqfs]

Granted, rote knowledge won't save you here, but it should not be that hard for advanced y12.
It's a very good test of complex numbers mastery, inequalities manipulation, and unit circle angles.
The ability to understand the problem is already a test.

Given |z|=|w|=1, we have
u = Arg(z/w) = Arg(z) - Arg(w) in the open interval (pi/2, pi)
so Cosu < 0 and Sinu > 0
z/w = Exp(iu)
w/z = exp(-iu)
Let's write r = (xz+yw)/z = x+yw/z
r = x+yCosu - iySinu
r = RExp(it)
r = RCost + iRSint
t is in the open interval (pi/2, pi)
Cost < 0 and Sint > 0

Sint = -ySinu
Given Sinu > 0, Sint > 0 => y < 0

Cost = x+yCosu < 0
yCosu < -x
Dividing by Cosu < 0 gives:
y > -1/Cosu * x
K=-1/Cosu > 0
so y > Kx

So we'll draw the line y = Kx and the region we are after is the region between the x axis and the kx line in the bottom plane.

[remipetit]

Great video!
But bring back the latex video 😂

[inny]

correction, W and Z are not complex numbers, they are part of the alphabet 😎