further maths argand diagram

[placid tulip]

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[cerulean shuttle]

the modulus of z2 is 2 i think

and |z1| is 1

[vocal briar]

Do you need to know what the modulus is tho?

Think of complex numbers as Ae^it

Like where A is the length or magnitude of those lines

[cerulean shuttle]

ye they use modulus argument form in the ms

but it makes no sense

[vocal briar]

Then what do you think the A for z_1 , Z_2 are related by

[cerulean shuttle]

since the argument isnt pi/4

[vocal briar]

But the angle between them is right?

[cerulean shuttle]

ye

[vocal briar]

Okay so let me draw a picture for you if it helps to make u understand better

Gimme a moment

[cerulean shuttle]

ok

[vocal briar]

Now what's Z_1/Z_2

[cerulean shuttle]

whats e

[vocal briar]

$\frac{z_1}{z_2} = \frac{|z_1| e^{i\theta_1}}{|z_2| e^{i\theta_2}}$

[ornate idol]

[redacted]

[vocal briar]

So are you familiar with polar form

For complex numbers

[cerulean shuttle]

no

[vocal briar]

Oh

I see

[cerulean shuttle]

we have not learnt this

[vocal briar]

My bad

So tell me what all things you know about

[cerulean shuttle]

like

[vocal briar]

You're familiar with trig identities?

[cerulean shuttle]

ye

[vocal briar]

Great

Let's compute (cos(k)+isin(k))(cos(l)+isin(l))

[cerulean shuttle]

this was the ms

[vocal briar]

Tell me what this turns out to be

[cerulean shuttle]

uh

i dtont think i know that one

or do u just want me to expand the brackets

[vocal briar]

Yes

[cerulean shuttle]

cos(k)cos(l) + i^2sin(k)sin(l) + cos(l)isin(k) + cos(k)isin(l)

oops

pressed enter

wait

[vocal briar]

Yeah take your time

[cerulean shuttle]

that?

[vocal briar]

Yes group the terms

[cerulean shuttle]

ohhh

[vocal briar]

Like the imaginary ones

And the real one

Yeah

And use trig identities

[cerulean shuttle]

cos(k)cos(l) - sin(k)sin(l) + i(cos(l)sin(k) + cos(k)sin(l))

so

[vocal briar]

Where's the i

cos(k)cos(l) - sin(k)sin(l) + i(cos(l)isin(k) + cos(k)isin(l))

Yes

[cerulean shuttle]

so its

cos(k+l) + isin(l + k)

?

[vocal briar]

Yep

Good job

Now we should try to compute what's 1/cos(k)+isin(k)

Now we should try to compute what's $\frac{1}{cos(k)+isin(k)}$

[ornate idol]

[redacted]

[cerulean shuttle]

uh

[vocal briar]

It's the same with simplifying a+ib

[cerulean shuttle]

ohh

[vocal briar]

1/a+ib

[cerulean shuttle]

so

[vocal briar]

Yep go ahead

[cerulean shuttle]

how do i write in like that

[vocal briar]

So here's a hint

[cerulean shuttle]

no i mean how do i write so it prints an image

[vocal briar]

Oh

In latex

Use dollars

[cerulean shuttle]

ill just do it normally

[vocal briar]

Go ahead

[cerulean shuttle]

(cosk-isink)/(cos^2k+sin^2k)

wait

no

its plus

[vocal briar]

Yeah

So the denominator is 1

[cerulean shuttle]

rightt

[vocal briar]

Hence you get cosk-isink

Great

[cerulean shuttle]

so its just cosk-isink

[vocal briar]

Now back to your problem

Write OP and OQ in those coordinate forms

Let me know if you didn't understand this

[cerulean shuttle]

yeah idk

[vocal briar]

Okay so back to the picture

Gimme a moment

[cerulean shuttle]

so k is pi/4?

[vocal briar]

No

But can you see why Z_1 is OP(cos(theta_1)+isin(theta_1))

And Z_2 is OQ(cos(theta_2)+isin(theta_2))

[cerulean shuttle]

ye thats the modulus argument form

[vocal briar]

Great so now compute z_1/z_2

Or z_2/z_1

[cerulean shuttle]

but u dont know the 2 thetas

[vocal briar]

You don't need to

Just simplify the expressions and let me know

The upshot is knowing theta_2-theta_1 is enough

Which is pi/4

If you're getting stuck in simplifying $\frac{OQ(cos(\theta_2)+isin(\theta_2))}{OP(cos(\theta_1)+isin(\theta_1))}$ then use the previously obtained relations we computed

[ornate idol]

[redacted]

[cerulean shuttle]

2(cosxcosy - sinxsiny + i(-cosxsiny + sinxcosy)/(cos^2y+sin^2y)

[vocal briar]

The bottom one becomes a + remember

Cause of the i

[cerulean shuttle]

oml

ye

[vocal briar]

Also note that OQ/OP=2

As per the problem

[cerulean shuttle]

so

[vocal briar]

Simplify

You should have something of the form A(cos(b-c)+isin(b-c))

[cerulean shuttle]

2(cos(x-y) + isin(x-y))

[vocal briar]

That means you messed up a sign somewhere

[cerulean shuttle]

oml

its the i^2

ye

ur rihgt

this topic is so stupid

[vocal briar]

It's not

You just need to find why

I showed you the two identities

That we computed

It's just a 2 liner after that

[cerulean shuttle]

and

x-y = pi/4

because

x is the arguemnt for z2

and y is the argument for z1

and then

its just

2(cospi/4 + isinpi/4)

[vocal briar]

Yep

[cerulean shuttle]

righttt

ok

[vocal briar]

So a comment i would like to make is

[vocal briar]

You did not need to do this

Instead

Use what we obtained

[cerulean shuttle]

so od i just memorise that

modulus(cos(a-b) +isin(a-b)) would be the 2

divided

ive seen questions like this before

[vocal briar]

$$\frac{OQ(cos(\theta_2)+isin(\theta_2))}{OP(cos(\theta_1)+isin(\theta_1))}$$

$$=(OQ(cos(\theta_2)+isin(\theta_2)))*\frac{1}{OP(cos(\theta_1)+isin(\theta_1))}$$

[ornate idol]

[redacted]

[vocal briar]

Now what was 1/cosk+i sink

[cerulean shuttle]

cos^2k + sin^2k

[vocal briar]

So that 1/thing becomes cos(-theta_1)+isin(theta_1)

No

[cerulean shuttle]

o

[vocal briar]

Here

[cerulean shuttle]

ohy e

so its like

cos2k-isink

[vocal briar]

Now what was (cos(a)+isin(a)) (cos(b)+isin(b))

[vocal briar]

Cosk-isink

[cerulean shuttle]

ye

i just worte it wrong

[vocal briar]

Here

Hence in just a single line you simplify the fraction (thanks to our previous computations) as 2(cos(pi/4)+isin(pi/4))

The upshot is these identities that we obtained

Can be used in many places

You don't always have to fall back to trig identities

That's it, just a comment

[cerulean shuttle]

ye

ok thank u bro

how do i +rep u

[vocal briar]

Np:)

Idk

Isc

Idc

You can just uh

Get back to work

[cerulean shuttle]

yo are u able to help with a mechanics q?

[vocal briar]

Post in help channels I'm Busy rn

Maybe someone else will

[sturdy kestrel]

run +close and it'll give you an option to rep

[cerulean shuttle]

+close

[woeful brook]

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[woeful brook]

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