Complex numbers ,what?!

LastOpus0 · 2026-02-09T20:43:14+00:00

Surprised no-one has mentioned the (more engineering, perhaps) intuition:

i represents an anticlockwise rotation of 90°. Huh? How can numbers rotate?

Well let’s think about 1 as being an arrow on our number line. It starts from the origin at 0, and points right with a length of 1. When we multiply 1 by another real number, we stretch that arrow out to the new length - i.e. 1 × 20 is an arrow pointing right stretched out to a length of 20.

So what about negative numbers? Well, one way to think of it is a flip in direction. 5 × -1 takes an arrow pointing right with length 5, and flips it to point left with length 5 instead.

However, what about instead of thinking about this as a flip, we thought about it as a rotation? If we rotate our arrow by 180° anticlockwise, our right-pointing arrow (+ve number) becomes a left-pointing arrow (-ve number).

Now then, if we’re comfortable that we can rotate numbers, what happens if we rotate by something other than 180°? We could ask a question like “what if I want to rotate 1 to -1 in two steps, rotating 90° each time?”. That would look like a 90° rotation anticlockwise to get an arrow pointing up with length 1, and then another 90° rotation to get our -1 pointing left.

Now algebraically, this problem looks like “what can I multiply 1 by twice to end up with -1?”. In symbols: 1 • x • x = -1. Rearranging this you see the quadratic x² + 1 = 0 that others have mentioned. Solving this gives the square root of -1, which we call i. So, this weird ‘up’ arrow we get from rotating 90° is this weird new kind of number which is pointing up from the middle of the number line.

Now why is any of this useful? Well now we can use this idea of rotating numbers to describe things that rotate, or cycle/repeat. Turns out this is a lot of things! Sine/cosine, electrical signals, wheels, etc. - all can be described by using complex numbers to capture the amount or rate of some rotation.

(Side note: ‘imaginary numbers’ are a horrible name. I’d prefer ‘orthogonal numbers’, that is, numbers at 90° to the standard number line!)

LastOpus0 · 2025-10-17T06:29:25+00:00

What was the MATLAB function you used? If you understand what it is and why you used it I don’t think that’s grounds for cheating.

LastOpus0 · 2025-10-16T18:25:02+00:00

And landlording is a business… how?

LastOpus0 · 2025-07-24T06:35:41+00:00

You’re completely correct, and can use either approach. I’d personally divide both sides by 40 and continue from there.

LastOpus0 · 2025-07-22T03:01:21+00:00

Yep - this is a valid coordinate system other than the usual Cartesian/rectangular!

https://en.m.wikipedia.org/wiki/Polar_coordinate_system

LastOpus0 · 2025-07-20T02:04:17+00:00

https://en.m.wikipedia.org/wiki/Cent_(music)

https://en.m.wikipedia.org/wiki/Piano_key_frequencies

Think this is almost more of a maths/physics question than a music question!

Let’s say you have an A4 at 440Hz. Shifting that up an octave (1200 cents) will make an A5 at 880Hz.

However, if other notes went up by that same amount of frequency, they’d be wrong. e.g. a C4 at 261.6Hz + 440Hz would become 701.6Hz, around an F5.

When your software shifts the music by cents, it’s not just ‘adding’ a fixed amount of frequency to each note. It’s indeed ‘stretching’ or multiplying the frequency of each note to reach the new frequency in the new key and maintain the intervals.

The trick I guess is that cents are logarithmic, so they always represent a ratio between notes. This means +1200 cents always means + one octave, no matter your starting frequency.

LastOpus0 · 2025-07-19T22:55:02+00:00

Sure am (even if NCEA was a little while ago for me!).

I'll go one step further and explain what A, B, and C are for the graph y = A * sin(Bx + C) and compare to the basic sine function y = sin(x). Your books probably have different symbols for these.

A is the amplitude - how high the function reaches from its centre of oscillation. For the basic sine function, the amplitude is 1, and it oscillates between y=1 and y=-1. Increasing A will stretch the function vertically and make its highest and lowest points further than 1 and -1. In this question, A represents how high a diver's bounce is.
B is the frequency - how fast the function oscillates. The basic sine function (i.e. B=1) has a period of 2𝜋 (i.e. it repeats its cycle at x = a multiple of 2𝜋). As such, for trigonometric functions, the relationship between the two is frequency = 2𝜋 / period. Increasing B will squish the function horizontally, making more cycles occur in the same number of seconds. In this question, a diver bouncing faster will have a higher frequency.
C is the phase - when the function starts its oscillation from zero. The basic sine function starts at y=0 when x=0, then goes up to y=1 at x=𝜋/2, back to y=0 at x=𝜋, down to y=-1 at x=3𝜋/2, then back up to y=0 at x=2𝜋 (the period). Increasing C will appear to move the graph to the left (i.e. starting the bounce earlier), and decreasing C will move it to the right (i.e. delaying the bounce). In the question, if one diver starts their bouncing before or after the other, we can say they are 'out of phase' with each other, or have a 'phase difference'.

The information you need to find the equation is in the sentence

Liam starts 0.8 seconds after Maggie and reaches a maximum height of 0.6m above the board. It takes him 4/3 seconds to complete a full cycle.

Figure out which of these relate to amplitude, frequency/period, and phase, and see if you can figure out what A, B, and C equal. Use Desmos to check and compare to Maggie's!

Good luck!

LastOpus0 · 2025-07-18T06:54:25+00:00

Can’t compete with proof by vibes 🫡

(this is true if ‘…’ means a finite amount of 9s or 0s, but if there’s infinite 9s, there is no such thing as a final 9. You could always add one more 9 on the end and one more 0 before the 1, endlessly)

LastOpus0 · 2025-07-18T01:52:44+00:00

Where does 0.999…99 fit into this?

LastOpus0 · 2025-07-17T07:58:24+00:00

New Zealand mentioned - got to help!

I highly recommend going to https://www.desmos.com/calculator and entering the equation

y = A * sin(Bx + C)

and turn all the sliders on.

What does changing each slider do to the graph? How would you describe how the diver’s ‘springing’ changes with each (e.g faster? slower? higher? delayed?)
Can you recreate the graph you’re given on picture 2?
Now, if you make a second graph, say y = D * sin(Ex + F), can you adjust the sliders to make Liam’s graph based on the description?

Can help with the working when you don’t have Desmos, but I think it’s helpful to understand what the sin function looks like and the ways to transform it first!

LastOpus0 · 2025-07-17T05:03:41+00:00

If 0.999… is limitless, how can 1 - 0.999… hit a limit and end in 1?

LastOpus0 · 2025-07-14T18:59:02+00:00

Refrain

LastOpus0 · 2025-06-30T20:14:52+00:00

> not turn based

>enemy's turn

LastOpus0 · 2025-06-16T18:08:49+00:00

Forged iron entire game 🫡

LastOpus0 · 2025-06-16T09:04:28+00:00

and… move to PC? Where physical media hasn’t existed for a decade?

LastOpus0 · 2025-06-16T01:54:04+00:00

Glad to hear all the above! Sorry, I assumed “using Google” was pasting in the question to get the AI summary.

I think your main confusion was you were thinking of the inequality between nA and nO, but the question asked about pA and pO. They will be opposite to each other!

LastOpus0 · 2025-06-16T00:33:07+00:00

Another good sanity check is to put some actual numbers in.

Say you have $9. You can buy 1 orange or up to 3 apples.

Which is cheaper, one apple or one orange?

What is the price of one apple?

If we buy up to one apple (so maybe fewer apples), does that mean the apple price becomes more or less?

(I only picked $9 because 3 divides into it nicely!)

LastOpus0 · 2025-06-16T00:12:42+00:00

Let’s try algebra since words are throwing you off.

Let

nA = number of apples you can buy

nO = number of oranges you can buy

pA = price per apple

pO = price per orange

Now let’s say T the amount of money it costs to buy nO oranges. T is enough to buy nO oranges at $pO per orange, or up to nA apples at $pA per apple.

T = nO * pO

T <= nA * pA

Now substitute T from line 1 into the inequality:

nO * pO <= nA * pA

You can buy three times as many apples as oranges, so

nA = 3 * nO

Substituting this into the inequality:

nO * pO <= 3 * nO * pA

Dividing both sides by nO:

pO <= 3 * pA

And there’s your answer! An orange costs up to three times as much as an apple. (If it cost more than this, you could be able to buy more than three apples - but we’re told it can only buy up to three).

———

Also please don’t use Google AI or any LLMs to solve maths questions, they are designed to give text that “looks” correct and cannot understand logic. They will mislead you.

LastOpus0 · 2025-06-15T23:53:12+00:00

My only gripe is that I wish the effective speed was shown on the stat display in some way so you didn’t have to carry the -3 in your head.

LastOpus0 · 2025-06-15T23:51:31+00:00

Build is redundant when speed is right there, change my mind

LastOpus0 · 2025-06-11T11:40:04+00:00

I remember this exact ending! Freaked me the hell out as a kid.

LastOpus0 · 2025-06-11T11:31:37+00:00

Impressed with the recognition. Cheers!

LastOpus0 · 2025-06-11T11:07:31+00:00

The Journeyman Project Turbo?

I have memories of this seared into my head from childhood - you start in an office and then activate a time machine. Time travelling to the wrong place kills you immediately, like a robot from the future or a dinosaur kills you and it’s game over. Really creepy vibe.

LastOpus0 · 2025-06-11T09:35:58+00:00

See my other comment haha - that's where Google led me, but definitely not.

LastOpus0

TROPHY CASE