help-27

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[tall stirrup]

petition to stop writing cot(x) as ctg(x)

[devout snow]

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[plucky nexus]

sorry it`s in all of maths books in my country 😭

[plucky nexus]

.close

[devout snow]

Channel closed

Closed by @plucky nexus

Use .reopen if this was a mistake.

[olive snow]

Petition to write cot as 1/tan

[solar goblet]

pétition

[olive snow]

Frickin auto french corrector

The worst is à instead of a

[devout snow]

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[forest niche]

hi guys, i have a question regarding this exercice, is about complex numbers and i need help understanding how to do it (NOT THE SOLUTION), just how to start at least and clarify some concepts, because i don't understand them

[forest niche]

Let the complex number be: z = a + 3 + bi where z is a complex number, a is the first digit on the right of your UOC Campus IDP and b is any real number.
Determine all the complex numbers, z, that make the number c = iz /
z−2 a pure imaginary number.```

this is the problem, and what i don't understand, is that in my college book, it says this

To multiply a complex number by a real number, we multiply the real and imaginary parts of the complex number by the real number.
Example of a product of complex numbers
If we have z = 1 + 2i, to multiply it by a, a real number, we do:
a z = a (1 + 2i) = a 1 + a 2i = a + 2ai```

i understand how to multiply z1 * z2, but the thing is, in this case, is that i don't know how to start, because if i understood correctly, the "real" part of the z number, should be the result of a+3 right?

or am i bad? i would like to get this confirmed before continuing.

so for example the equation should be:

if a = 5:

(5 +3) + b * i

[devout snow]

@forest niche Has your question been resolved?

[forest niche]

no, nobody replied

[tall stirrup]

yes

you're correct that real part of z is a+3

c = iz/(z-2)

[forest niche]

okay, but its a + 3 or its the result of doing a + 3? that's one of the things i don't understand

[tall stirrup]

Do you know what the a is?

if yes, then its result of a+3

if no, then its a+3

[forest niche]

the a is a number forming the real part with the 3, and the a is a number that idk how to find it because its something about my id i guess

but its supposed that a = some number idk how to find (related to my student id, so for example a could be = to 5)

[tall stirrup]

yes, then z = 8 + bi

[forest niche]

okay, so in this case, 5 + 3 = 8, 8 is the real part and bi is the imaginary one right?

[tall stirrup]

yes, (btw, b is the imaginary part we say, not bi)

[forest niche]

okay, b is imaginary

i understand this

now, i don't understand the iz part, like what i'm supposed to do?

[tall stirrup]

it means that i is multiplied by z

so, in our case z = (a+3) + bi

[forest niche]

so i * 8 + bi?

[tall stirrup]

no i^2, its multiplied with imaginary part as well

so, 8i - b

[forest niche]

wait

i didn't understood

[tall stirrup]

ok

[forest niche]

iz = i * ((a+3) + bi)

right?

[tall stirrup]

wait

[woven radish]

Pro_Hecker

Pro_Hecker

[forest niche]

no but the i is on the left not the right

[tall stirrup]

do you get it now?

[forest niche]

it shouldn't be i((a+3) + bi)?

[tall stirrup]

commutative property of multiplication 3 * 5 = 5 * 3

[forest niche]

okay but can you use it like "normal" because im kinda new and it really confuses me

i thought it was something else

(not ur fault)

okay when bi² i guess its because b * i * i right?

[tall stirrup]

yes

[forest niche]

okay

[tall stirrup]

i^2 = -1

[forest niche]

now i understand that (a+3 = 8) so iz = i * (8 +b * i * i) right?

then i should get

8 + -b

[tall stirrup]

no, wrong parenthesis

i * (8) +b * i * i

[forest niche]

i * (8) + (b * i * i)

[tall stirrup]

yes

[forest niche]

alright

[tall stirrup]

8i - b

[forest niche]

si this is basically iz

wait

8i why

its because i * 8?

like we write 8i because its i * 8 ?

[tall stirrup]

yes

both are same

5 * 3 = 15
3 * 5 = 15

[forest niche]

yes i understand what do you mean

but if you tell me that bi is a imaginary number, then 8i should be it too because it has the i on the right

anyways i understand what you meant

[tall stirrup]

yes, 8 is the imaginary part of iz

[forest niche]

...

[tall stirrup]

but remember, the actual imaginary part is a+3

[forest niche]

can you elaborate please

no wait

[tall stirrup]

iz = (a+3)i - b

[forest niche]

why now a+3 is imaginary?

[tall stirrup]

[forest niche]

isn't supposed that a+3i = a + 3 * i?

[tall stirrup]

no, (a+3)i

don't forget the parenthesis (or brackets as they call it)

[forest niche]

okay but lets step back

please

an imaginary number for example is 7i

right?

[tall stirrup]

yes

Let's rewind and come back from the basics

[forest niche]

z = (real part is a+3 or 8) and imaginary one is bi

[tall stirrup]

so, $i = \sqrt{-1}$

[forest niche]

if i * (a+3) why is nos a+3 imaginary?

[woven radish]

Pro_Hecker

[tall stirrup]

now, a number cannot be squared to obtain a negative integer, so mathematicians invent a new number and call it i

[forest niche]

yes

i understand that i * i or i² = -1

no doubts there

[tall stirrup]

so, numbers like $i, 2i, \frac{3i}{4}$ are called imaginary numbers

[woven radish]

Pro_Hecker

[tall stirrup]

basically, any number that contains only i and nothing is added and subtracted from it is called imaginary number

now, we already know what real number is

so, complex number is made by adding( or subtracting) real and imaginary numbers

[forest niche]

okay

but the thing is

[woven radish]

Pro_Hecker

[forest niche]

i understood

like

imaginary is bi

and real is a

but the part that i don't understand is the following one

when you do iz = i * 8 + i * i * b, why now is 8i?

why 8 turned to be an imaginary number now?

[tall stirrup]

so, part which is not multiplied with i is called real part and part which is multiplied with i is called imaginary part

[tall stirrup]

8 is not a imaginary number per se, 8i is imaginary

it turned imaginary because it was multiplied by i

[forest niche]

yes but why it turned to be an imaginary number if in firs place wasn't?

oh

are you saying that anything * i = new imaginary number?

[tall stirrup]

real number * i = imaginary number

yes

[forest niche]

ok

and i guess that imaginary number * imaginary number = real number right

[tall stirrup]

yes

because i * i = -1

which is a real number

[forest niche]

ok

another question

we have bi as imaginary number

and when doing iz we are multiplying i * bi

so i * i * b

or i * b * i

right?

[tall stirrup]

both are the same

[forest niche]

okay perfect

so in this case

bi²

[tall stirrup]

yes

[forest niche]

means i * b * i

right

because we are multplying 2 times the i

[tall stirrup]

yes, bii, ibi iib, anyway you like

[forest niche]

yeah but about the bi²

its because bi² = i * b * i... wathever, it means that we are multiplying two times i

[tall stirrup]

yes, i^2 = -1

[forest niche]

ok

[tall stirrup]

so bi^2 = -b

[forest niche]

ok perfect

so now i think that the solution to iz could be something like

z = ((5+3) + bi)

so we do

iz = (i * (a+3)) + (i * b * i)

iz now turns to

iz = 8 * i = 8i

and i * b * i = -1 * b

so b * -1 = -b

8i + -b

right?

[tall stirrup]

yes

we generally write real part first

so, iz = -b + (a+3)i

[forest niche]

-b + 8i?

[tall stirrup]

yes

[forest niche]

its correct what i wrote or should i write it like you?

[forest niche]

i mean this

[tall stirrup]

Since you don't know a

you should write like that

[forest niche]

a is supposed to be something about my student id

(like i need to find this because i dont know how to find it in my campus)

that's why i said that we can imagine a = 5

[tall stirrup]

we can leave a as it is, then when you get the final answer just substitute whatever a equals to

[forest niche]

ok

now

z - 2

it sounds like

a + 3 * bi - 2

so

((a+3) + (b * i)) - 2

[tall stirrup]

open the brakets

[forest niche]

brackets?

{}?

[tall stirrup]

yes

[forest niche]

(a+3) + (b * i) - 2

[tall stirrup]

open those one too

[forest niche]

a + 3 + b * i - 2

[tall stirrup]

a + 3 + bi -2

now what is 3-2?

[forest niche]

1

[tall stirrup]

yes, so z-2 = (a+1) + bi

[forest niche]

okay but i didn't understood something

i understood that we substracted the 3 in a + 3

but if a = 5, then it should be 6

8 - 2 = 6

right?

[tall stirrup]

umm

yes

[forest niche]

so it could be something like (6) + (b * i)

is this correct? (we are assuming that we are using the result of a+3, but i understood what you did there)

[tall stirrup]

yes

(a+1) + bi

since a =5

[glass summit]

hello

[tall stirrup]

6+ bi

[devout snow]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[forest niche]

and i guess we did 3 - 2 because the only number to be substracted is 2

[glass summit]

what

[tall stirrup]

yes

[forest niche]

alright thanks

so the complete number should look like

(let me start paint)

[tall stirrup]

$\frac{iz}{z-2} = \frac{-b + (a+3)i}{(a+1) + bi}$

[forest niche]

oh damn

i don't know the commands let me try

[woven radish]

Pro_Hecker

[forest niche]

yes but in theory the first part (without optimization) would be like i writed no?

then after getting my result

you'll do the calculate

and do

[tall stirrup]

umm, why did you keep ibi

write it as -b

[forest niche]

because i didn't do the calculation yet

i wrote this because its how its writed before doing the calc

after doing the calc should be

(wait for my reply pls)

$iz = (a + 3)i + -b

[tall stirrup]

yes

[forest niche]

ok

[tall stirrup]

-b + (a+3)i

[forest niche]

now to "check if its 0" what the hells means that? this is not programming, how i'm supposed to check this?

[tall stirrup]

we write real part first, but that's not necessary at all

[forest niche]

thanks

because for my understanding i prefer this

[tall stirrup]

alright until now, we have this

there is a imaginary number in the denominator, we don't like it, how should we remove it?

[forest niche]

i don't know

[tall stirrup]

do you know how to rationalise a number?

[forest niche]

no

[tall stirrup]

what about complex conjugate?

[forest niche]

• above z?

ouch

(slash) above z?

[tall stirrup]

I'll be back in 5 mins

[forest niche]

alright dw

[tall stirrup]

alr i'm back

[forest niche]

yay

[tall stirrup]

so where were we?

ah yes the complex conjugate

[forest niche]

in rationalizing numbers

i searched and it seems its this thing

[tall stirrup]

yes

[forest niche]

oh $-z

[tall stirrup]

No, leave that

[woven radish]

Pro_Hecker

[forest niche]

yes i know this

like if the complex number is 6

the conjugate is -6

and vice versa

[tall stirrup]

no

[forest niche]

no?

ooh wait i missread

its only the imaginary part

this is from uni docs

[tall stirrup]

let z be complex number and $\bar{z}$ be its conjugate
$$z= a+bi \n \bar{z} = a-bi$$

[forest niche]

no wait

z = (a+3) + bi

sooo

i think its

conj z = (a+3) - bi?

[tall stirrup]

yes

[forest niche]

alr

but we already have -b

[tall stirrup]

what is the conjugate of z-2

[forest niche]

ooh no wait

z + 2

right?

[tall stirrup]

no

[forest niche]

aaa

wait

dont say nothing

[tall stirrup]

z -2 = (a+1) + bi

[forest niche]

wait

conj of z - 2 = (a+1) - bi

[tall stirrup]

yes

now we multiply the conjugate with numerator and denominator

[forest niche]

we multiply -bi or (a+1) - bi?

[tall stirrup]

$\frac{(-b + (a+3)i)( (a+1) - bi)}{((a+1) + bi)( (a+1) - bi)}$

[woven radish]

Pro_Hecker

[tall stirrup]

a+1 - bi

[forest niche]

question

why we conjugate?

[tall stirrup]

so that we can remove the i from denominator

we don't like it there

if we remove it from there, then we can separate our number into two parts

real and imaginary

[forest niche]

but denominator has only the i in bi

right?

[tall stirrup]

yes

but we have to multiply with conjugate, otherwise the i will remain there

we're simplifying the complex number

[forest niche]

and what's the problem about having the i?

oh ok

but its needed?

[tall stirrup]

we can't identify the real and imaginary parts

of iz/(z-2)

[forest niche]

alright

its needing there i guess

[tall stirrup]

I suggest you read some texts about it

[forest niche]

i have the text of my uni

the problem is that there are a lot of stuff "for convenience" but not everyone understand that "convenience"

about the final part, knowing if they are 0

how do you do that?

[tall stirrup]

yes,

when you multiply that will conjugate

you'll get a complex number in form (p+qi)

you try to get q = 0

[forest niche]

no wait

first

[forest niche]

how do you get the i out?

a * b = c?

[tall stirrup]

you can multiply complex numbers right?

[forest niche]

yes

z 1 and z2 for example

but a * b??

i mean

[forest niche]

you say i should multiply the first line

right?

so it could be z1 = left and z2 = right

then i do a complex number mult

[tall stirrup]

yes

[forest niche]

ok wait a few minutes

im doing the mult

wait.. there is no real part

aa wiat

wait

real part is the right one

a + 1

right?

[tall stirrup]

yes

[forest niche]

so we do

((a + 1) * - bi )

[tall stirrup]

hmm

[forest niche]

To multiply a complex number by a real number, we multiply the real and imaginary parts of the complex number by the real number.

Example of a product of complex numbers
If we have z = 1 + 2i, to multiply it by a, a real number, we do:
a · z = a · (1 + 2i) = a · 1 + a · 2i = a + 2ai

[tall stirrup]

we're multiplying a complex number by a complex number

[forest niche]

then i dont know how to do it

[tall stirrup]

hmm

[forest niche]

because the mult of complex numbers i got is this what my docs says

[forest niche]

hmm distributive property

[tall stirrup]

you said you can do z1 * z2

[forest niche]

yeah but i was thinking about this

anyways i got you

i understand what you mean

distributive property

like in this case..

(-b * a+1 )+ (a + 3 * - b)i

right?

[tall stirrup]

I don't think so

use parenthesis, I'm getting confused

[forest niche]

alright

(-b * (a+1)) + (-b * -bi) + ((a+3) * (a+1)) + ((a+3) * -bi))

is this right?

no wait

[tall stirrup]

no

[forest niche]

but i used the distributive property

[tall stirrup]

ah yes

I get it now

you didn't simplify it

Then it is right

[forest niche]

alright

once you have this what's next?

[tall stirrup]

when we simplify it, we'll get

wait for it

[forest niche]

oki

[tall stirrup]

$\frac{2b+(a^2+4a+b^2+3)i}{(a+1)^2+b^2}$

[woven radish]

Pro_Hecker

[tall stirrup]

now what is the imaginary part?

[forest niche]

hmmmmmmm

the right

where the i its placed

2b is the real

[tall stirrup]

yes imaginary part is a^2 + 4a + b^2 + 3

and what does it equal?

[forest niche]

you got me there

i don't know what to answer

[tall stirrup]

read the question

[forest niche]

yours?

[tall stirrup]

Let the complex number be: z = a + 3 + bi where z is a complex number, a is the first digit on the right of your UOC Campus IDP and b is any real number.
Determine all the complex numbers, z, that make the number c = iz /
z−2 a pure imaginary number.```

[forest niche]

or mine?

[tall stirrup]

OHHHH, my bad

this is a pure imaginary number

so, real part = 0

[forest niche]

yes

but i guess the calcs are the same aight?

[tall stirrup]

yes

[forest niche]

oki

[tall stirrup]

so, 2b =0

[forest niche]

no wait

how 2b = 0

2 * 0 = 0

[tall stirrup]

this is a pure imaginary number

[forest niche]

but b is not 0

[tall stirrup]

so, real part is zero

[forest niche]

i mean

a = 5
b = nothing```

how its b = 0??

[tall stirrup]

now, b is any real number

[forest niche]

i think its the same as

b = 1,2,3....999999999

[tall stirrup]

but now we now that c = this thing above

[tall stirrup]

no, b is anything from -infty to +infty

[forest niche]

ooh

so like

b = -99999 ... 1,2,3...999999

[tall stirrup]

now the question says c is purely imaginary

[forest niche]

no, it says that i need to "find the numbers that are purely imaginary"

or smth like this

[tall stirrup]

also includes, 0 , 0.3, 0.32343233333...., 1,141,,,,, pi, etc

[tall stirrup]

yes, so c is purely imaginary

we have to find it

[forest niche]

alright

but before continuing i need to ask something

[tall stirrup]

now for purely imaginary numbers real part is zero

[forest niche]

b = 0 because you choosed 0? you could choose 139139?

[tall stirrup]

no hold on with me

[forest niche]

e?

ah

ok

[tall stirrup]

do you get it?

[forest niche]

yes

for example

a + bi

a = 0

[tall stirrup]

ok, now real part what is it?

[forest niche]

bi = 5i

0 + 5i

real part of a +bi?

a is the real part

[tall stirrup]

real part of this

[forest niche]

2b

[tall stirrup]

yes and what should it equal

[forest niche]

0

[tall stirrup]

so that number is purely imaginary?

yes 2b = 0

so, b =0

[forest niche]

2 * 0 = 0 i guess?

[tall stirrup]

$\frac{2(0)+(a^2+4a+(0)^2+3)i}{(a+1)^2+(0)^2}$

[woven radish]

Pro_Hecker

[tall stirrup]

now we put zero instead of b in complex number c

[forest niche]

so....

in this case

if i understood correctly

we would put a = 5, so 4a is 4 | 5

(4 * 5)

[tall stirrup]

yes

[forest niche]

then should be 2 * (0) + (5² + 4 | 5 + (0²) + 3)i

/ 5 + 1² + (0²)

[tall stirrup]

yes

[forest niche]

ok

[tall stirrup]

use brackets pls

[forest niche]

this?

{}

[tall stirrup]

(5+1)^2

not 5 + 1^2

[forest niche]

i got you

[tall stirrup]

$\frac{(a^2+4a+3)i}{(a+1)^2}$

[woven radish]

Pro_Hecker

[forest niche]

i dont got you now

[tall stirrup]

we simplified this

[forest niche]

hm alright

i don't think i like simplifications

haha

[tall stirrup]

I have one question

Let the complex number be: z = a + 3 + bi where z is a complex number, a is the first digit on the right of your UOC Campus IDP and b is any real number.
Determine all the complex numbers, z, that make the number c = iz /
z−2 a pure imaginary number.``` How did you colour this?

[forest niche]

i used c

i think

[tall stirrup]

this is your final answer, just find a and substitute it in

[forest niche]

substitute?

i dont understand what you mean

[tall stirrup]

replace

[forest niche]

ah

the a you mean?

[tall stirrup]

si

[forest niche]

alr

i have another question

when seeing something like this

[tall stirrup]

si, bien

[forest niche]

u speak spanish or joking

[tall stirrup]

se un poco espanol

[forest niche]

ghahahahaha

alr

osea cuando ves esto

como sabes que tienes que hacer todo eso que me enseñaste arriba?

[tall stirrup]

practice

experiencia

[forest niche]

yeah but someone has to determine that in order to resolve this kind of problems you need to start from somewhere

like linear equations

[tall stirrup]

most of the problems have a pattern, and we have things we know

we try to connect them and solve

[forest niche]

alright another question

when we removed the i

[forest niche]

.

[tall stirrup]

yes

[forest niche]

how did we remove exactly?

i mean we multiplied the two complex numbers

but in which point does it remove?

[tall stirrup]

the denominator was $((a+1) + bi)((a+1)-bi)$

[woven radish]

Pro_Hecker

[tall stirrup]

this was after we multiplied by complex conjugate

[forest niche]

sorry i dont understand

[tall stirrup]

so, using distributive property $(a+1)^2+ (a+1)bi - (a+1)bi-bi^2)$

[woven radish]

Pro_Hecker

[tall stirrup]

do you remember this step?

we multiplied top and bottom by (a+1 -bi)

[forest niche]

yes

[tall stirrup]

we get this if we use distributive property to simplify

[forest niche]

with top and bottom you mean this right?

[tall stirrup]

yes

[tall stirrup]

no, numerador y denominador

[forest niche]

but i did this

actually

and we were multiplying top

the two comples numbers

[tall stirrup]

You were asking how we removed i in the denominator

[forest niche]

yes

leave it i dont understand and im tired

thank you very much for your help

[tall stirrup]

si, hasta luego

!done

[devout snow]

If you are done with this channel, please mark your problem as solved by typing .close

[tall stirrup]

.close

[devout snow]

Channel closed

Closed by @tall stirrup

Use .reopen if this was a mistake.

[forest niche]

tomorrow i will check this again if i have problems

now i want to rest a bit

[devout snow]

Available help channel!

Send your question here to claim the channel.

Remember:
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[forest niche]

alright i have another question sorry

[forest niche]

.reopen

@tall stirrup am sorry bro, but can you please explain again to me when and where did we removed the i?

[surreal linden]

I can explain

[forest niche]

alright thanks!

y?

[tall stirrup]

Okay he got timed out it seems

so lets continue

[forest niche]

Alr

[tall stirrup]

the problem asked us to find a complex number c such that its pure imaginary

so, c should have real part 0

[forest niche]

Yes

[tall stirrup]

but to identify real and imaginary parts, we had to simplify the complex number

so, we multiplied numerator and denominator by denominator's complex conjugate

[forest niche]

Okay wait

Hang there

[forest niche]

How did we get there in first place

I understand that

The first one is the result of i * z

Right?

[tall stirrup]

yes

[forest niche]

Ok

[tall stirrup]

the numerator is the result of iz

[forest niche]

But the first denominator is z - 2

Without co jugate

Right?

[tall stirrup]

yes

[forest niche]

Alright

So now the part i dont understand

How we get the text on the image

We conjugate yes

But just because we wanted or what?

[tall stirrup]

we wanted the denominator to be real

[forest niche]

Ok

And to be real we have to remove the i

How we removed the i?

[tall stirrup]

When a complex number is multiplied by its conjugate

it results in a real number

let's see this with an example

let $z = a+bi$, so $\bar{z} = a-bi$

[woven radish]

Pro_Hecker

[forest niche]

Yes

[tall stirrup]

$z\bar{z}=(a+bi)(a-bi)$

[woven radish]

Pro_Hecker

[forest niche]

We did the same with the denominayor

[tall stirrup]

yes

[woven radish]

Pro_Hecker

[forest niche]

Okay so

In order to do this

We needed to conjugare the denominator

Right?

[tall stirrup]

yes

we need to multiply the denominator by its conjugate

[forest niche]

Okay but

Denominator is the down one

I mean

The conjugator

[tall stirrup]

yes

[forest niche]

Step 1: we get the nominatir and denominator, step 2: we conjugate the denominator, step3: we multiply the co jugate denominator for the nominator

Right?

[tall stirrup]

step1: we obtain c = iz/z-2 in terms of and b
step2: multiply the numerator and the denominator by the conjugate of denominator
step3:the real part of c so obtained = 0

[forest niche]

Okay when we multiply the nominator * denominator

We removed the i?

Can you do it plz?

[tall stirrup]

Sorry, I don't get what you mean

[forest niche]

I mean

[forest niche]

We did this

Conjuged denominator x nominator right?

[tall stirrup]

yes

[forest niche]

So when we removed the i? Yes we multiplied using the distributive property

[tall stirrup]

yes

[forest niche]

But when exaclty we removed the i?

[tall stirrup]

when we simplified this

[tall stirrup]

to this

[forest niche]

But there are two i

Like we just quite one by "simplifying"?

[tall stirrup]

3i + 2i = 5i

like this two i become one i

[forest niche]

Hmmmm

[tall stirrup]

we just added some stuff

[forest niche]

In multiplications is the same?

[tall stirrup]

yes

[forest niche]

Oooooooooh

So

When we multiply 3ix4i

We can simplify it to 3 x 4?

3x4i?

[tall stirrup]

3*4i^2

so -12

[forest niche]

Oooooo

I think i understand

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[heavy sail]

Am I going wrong anywhere

[livid geyser]

!help

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[livid geyser]

Anyways

ur value for a is wrong.

a + a + 2a + 2a = 6a

[heavy sail]

Ye I just realised lol

[heavy sail]

Wait I’m still confused what do I do after that

[livid geyser]

same thing you did but replace 4a with 6a

[heavy sail]

Ohk ye I got it

[livid geyser]

[heavy sail]

Thx

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[devout snow]

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[brisk heart]

Is this the correct way and How do I continue with it

[brisk heart]

Im not really sure if my derivatives are right either

I am kinda confused how to treat U1 and U2

the statement U1'e^-3x + U2'e^x = 0 is from the Variation of Parameters method

<@&286206848099549185>

oh and the homog. part it is yc = C1e^-3x + C2e^x

[devout snow]

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[brisk heart]

<@&286206848099549185>

[devout snow]

@brisk heart Has your question been resolved?

[brisk heart]

<@&286206848099549185>

[brisk heart]

<@&286206848099549185>

[devout snow]

@brisk heart Has your question been resolved?

[devout snow]

@brisk heart Has your question been resolved?

[tawny pewter]

From what I recall from my intro ode course, for the variation if parameters method you have to memorize a formula for u_1 and u_2 unless you want to redo the derivation of these formula for your example. For your problem, we were also shown a method to guess wisely that y_p has the form below.

g=(-3x-1)cos(3x)

[brisk heart]

thank you man

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[plush perch]

how would i solve this

[plush perch]

this is what i have so far

[uneven talon]

So just to make sure, this is $3\tan{(\frac12 \theta)} - 2 = 0$

[woven radish]

hiidostuff

[uneven talon]

Is this right?

[plush perch]

no idea

all i know is you add 2 then divide 3 to the other side

[uneven talon]

True

[plush perch]

so you get tan1/2theta=2/3

[uneven talon]

Well if your equation is right, then yes you're right

At this point, we have to see if there's any special angle of theta such that when you take the tangent of it, you get 2/3

[plush perch]

then tan^-1 (2/3)

[uneven talon]

Yes

And we have 1/2 theta

[plush perch]

which is 33.69

[uneven talon]

On the left side

So then we multiply by 2 on both sides to isolate theta

[plush perch]

so i’d multiply all my values by 2?

[plush perch]

ohh i see i can do that instead

but what i was gonna do is correct yes?

[uneven talon]

Yep

You were doing the right algebra

[plush perch]

okay so if i do it my way (sorry it’s just easier)

i do 33.69 -180 which is -146.3

but then i multiply that by 2 and its outside the range

and that goes for all my other values i only get one value for that question

which is 33.69

okay wait i did it your way

it’s easier

i got this are these all the values?

@uneven talon

[uneven talon]

I'm not sure about the exact values

As in idk what they are

[plush perch]

okay this doesn’t work out i’m so confused

can you please work it out and show me???

[uneven talon]

But it should just be $2\arctan{\frac23} + 2\pi n$

[woven radish]

hiidostuff

[plush perch]

that’s what you do to your values last

otherwise it’s wrong

[uneven talon]

Well yes

But let's go through it slowly

We know $\tan{\frac12 \theta} = \frac23$

[woven radish]

hiidostuff

[uneven talon]

So then

$\frac12 \theta = \arctan{\frac23}$

[woven radish]

hiidostuff

[uneven talon]

And finally we get theta = 2arctan(2/3)

But remember that tangent is periodic every pi units

So we have to add pi * n

Where n is an integer

[devout snow]

@plush perch Has your question been resolved?

[plush perch]

i’m really confused

can you tell me what i’ve done wrong?

[devout snow]

@plush perch Has your question been resolved?

[uneven talon]

Hello, sorry I got distracted

Ah I see

You multiplied tan(1/2 theta) by 2 to get tan theta

But first you have to take the inverse tangent

Then you multiply by 2

[plush perch]

i see

what values do you get when you do it?

within the range of course

[plush perch]

@uneven talon

[uneven talon]

Uhh lemme see

-112.62 and 67.38

[plush perch]

okok thank you

[sinful rover]

!done

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[simple mauve]

Hi i need help for that question please help

[devout snow]

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[wise storm]

I would try drawing a picture and labeling it

[simple mauve]

ok

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[solid osprey]

how many 2 digits number is divisible by the sum of digits

[solid osprey]

is there a non brureforce way to do this?

[olive snow]

So 10a + b is divisible by a+b ?

[solid osprey]

the only way i could think of is to split into the sum being 9, 6, 3 since their divisibility rule is the sum, so 18 27 36 45 54 63 72 81 90, 24 42 60, 21 12 30

[solid osprey]

yeah

so a+b|9a

then say gcd(a,b)=c then c(x+y)|9cx=>x+y|9x

[olive snow]

Is a+b and a coprim ?

[solid osprey]

no

yes if a,b is coprime

[solid osprey]

this implies x+y|9 or x+y=1 or 3 or 9

eh wait

x+y=1 means x=1 y=0, so all multiples of 10

[solid osprey]

3 and 9 was this

is that all

12+9=21

ans key is 23 so missing 2 :/

[alpine python]

did you get 48 and 84?

[solid osprey]

no

ok i see

thanks those are the missing ones

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[wheat bough]

Can someone please help me on this question:

[winter patrol]

what have you tried

[wheat bough]

nvm i got the answer

i was just confused by the wording at first

[minor skiff]

(: glad its solved

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[tropic nebula]

Helpp

[tropic nebula]

what the heck is this a thing

[solid osprey]

whats the question?

[tropic nebula]

In a pool at an aqueous, a dolphin jumps out of the Wota travelling at 20cm per second it's height above wota level after t seconds is given by h=20t - 16t

@solid osprey

[solid osprey]

???

what is it asking

[tropic nebula]

that's the question

oh wait a sec

this

i haven't saw the other page sry

[solid osprey]

ok which are you stuck on

[tropic nebula]

the full question i never got something like that on my entire schools while it came up on exams from somewhere none knows

[solid osprey]

have you not learned quadratics?

[tropic nebula]

i did

But there were no graphs

only graph i got was from polynomial x and y

And the height and time i never got anything like that on my book

[solid osprey]

this maybe wierd but try playing around with desmos, try to find the quirks of quadratics and their graphs

but basically, the "zeroes" of a quadratic is where the graph touches the x axis

[tropic nebula]

what is desmos

[solid osprey]

so say, a function has a zero at 1, so when x=1 it would touch the x axis

[solid osprey]

its a graphing tool

really useful ngl, if your too lazy to solve system of equations desmos can basically do it for you assuming you can use it

[tropic nebula]

is that some high level things

[tropic nebula]

how long it takes to learn desmos

[solid osprey]

its not hard

[tropic nebula]

Is it's easy i could learn it and use it up

Only if my teacher doesn't cut my marks

[solid osprey]

yeah basically

[tropic nebula]

once I used some goofy ahh Chinese Rational formula and he cooked me

[solid osprey]

lmao

https://www.desmos.com/calculator (if your marks are cut dont blame me :p)

try just typing "ax^2+bx+c" and there should be a button to add sliders, click add all and try to play around with the sliders and see the graph

[tropic nebula]

nah I'll bring this link to my teacher 🗿

Dude what is this now

Why the slope ain't sloping

[frozen aurora]

what are you trying to do?

this?

write "p(t) = 20t - 16t^2"

https://www.desmos.com/calculator/2wesiavwg2

[devout snow]

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[tropic nebula]

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[surreal gorge]

I have to integrate this, but i can't find any way to match the general cases i learned.

[surreal gorge]

Here is the usual things i work with

[fossil locust]

try completing the square of -x^2 + 5x - 6

then the integral of $\frac{1}{\sqrt{a^2 - x^2}}$ is a standard integral

[woven radish]

south

[surreal gorge]

I'm sorry, what do you mean by "completing"? English is not my native language

[fossil locust]

it's complétion du carré in French

[fossil locust]

it matches with $\frac{u'}{\sqrt{1 - u^2}}$ if you take $u = x/a$

cause you then have $\frac{1}{a} \frac{1}{\sqrt{1 - (x/a)^2}} = \frac{1}{\sqrt{a^2 - a^2 (x/a)^2}} = \frac{1}{\sqrt{a^2 - x^2}}$

[woven radish]

south

[surreal gorge]

Ohh, i see

Thank you!

[fossil locust]

no worries!

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[inner imp]

n.263
solutions are: 2kπ and 3π/2 + 2kπ
I can only find one solution

[tardy edge]

I think it's because cos(π-π/4) isn't cos(π/4) but -cos(π/4)

Cause that's the only mistake I see

[inner imp]

let me see

[inner imp]

Where is it tho

Oh

I just realized too

Let me see if i can solve it now

[slender mirage]

Umm, I think you might be doing the exercise wrong tho (@_@;)

[inner imp]

Nope, it’s correct.

[inner imp]

^ this was the mistake

Thank you Alex 👍👍👍

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[inner imp]

.reopen

[devout snow]

✅

[slender mirage]

Ah not that. I meant expanding and solving

[inner imp]

No it’s correct

[tardy edge]

yeah so you should get sin(π/4 - x) = √2/2 and from there 2 solutions

[inner imp]

i got cos x - sin x = 1

[tardy edge]

yes and go from there

[slender mirage]

Do you mind if I suggest something?

[inner imp]

Ya

What’s up?

[tardy edge]

√2(cosx-sinx) = √2

divide this by 2

and go from there

[slender mirage]

Consider

sin(x + π - π/4) = - sin(x - π/4)
cos(x - 7π/4) = cos(x + π/4 - 2π) = cos(x + π/4) = sin(π/4 - x)

So basically your equation is 2sin(π/4 - x) = √2 or sin(π/4 - x) = sin(π/4)

The same is true for the other exercises. I believe the point of this exercise is to make you familiar with manipulating arguments within sine / cosine functions, i.e., sin(π/2 - x) = cos x, cos(π - x) = -cos x, etc.

Q 264 will help understand this better

[inner imp]

Oh that makes sense

Thank you so much for helping

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[tranquil grail]

how should I go about solving the second part?

(the hence or otherwise part)

[fossil locust]

try expanding (z - 1/z)^5

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[fossil locust]

and then pairing up terms in the form z^n - 1/z^n

[tranquil grail]

will do, thanks

[fossil locust]

np!

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[regal berry]

$2x^2 +7x + 11 >0 => 2(x^2 + 3x + 5) +2 >0$ now what? i need to complete (a+b)^2

[woven radish]

Simon James B
Compile Error! Click the reaction for more information.
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[regal berry]

a is x but what is b

2ab = 2xb

but i can't find what b is

[iron kindle]

what are you referencing?

[radiant dune]

how did that 11 turn into a 12

and 7x to 6x

and also whats the original problem, are we trying to express this in a form something times (x+k)^2 + something else

[regal berry]

11 did not turn in 12? i am factoring 2 out of the original. 2x^2 = 2 * x^2
7x = 2 * 3x and we have 1 left
11 = 2*5 and one left

1+1 left = 2

[regal berry]

it's pinned

$2x^2 +7x + 11 >0 => 2(x^2 + 3x + 5) +2 >0$

[woven radish]

Simon James B

[iron kindle]

well, this is wrong

[regal berry]

why

[winter patrol]

7x = 2 * 3x and we have 1 left
you'd have 1x left

[radiant dune]

and 2*5 + 2 is 12

not 11

[winter patrol]

you'd have 1x+1 leftover overall

[iron kindle]

$2(x^2+3x+5)+2=2x^2+2\cdot 3x+2\cdot 5+2=2x^2+6x+12\neq 2x^2+7x+11$

[woven radish]

Bonk

[winter patrol]

but not ideal to factor 2 out like that anyway

[regal berry]

what do we do then

[winter patrol]

just factor out 2 directly,
don't bother with trying to get nice integers

[regal berry]

i need to show that the original is >0

[winter patrol]

as that wouldn't be completing the square here

[radiant dune]

if that is the only task, they eaiser way is to use the fact that the coefficient of x^2 is positive and the parabola is never crossing the x axis

by showing the b^2-4ac<0

symbols denote usual meaning

[regal berry]

Maybe don't assume i even know what a parabola is 😭

I am on algbera 1 only help

[iron kindle]

a parabola is a second order polynomial

[regal berry]

I also don't know what a second order plynomial is 😄

[iron kindle]

i.e. largest exponent is x^2

[regal berry]

my chapter is about factoring using the formulas given

[winter patrol]

if you want to go completing the square route
factor out 2 from the whole of 2x^2 +7x + 11
or the first two terms 2x^2 + 7x
either is fine, its up to personal preference

[iron kindle]

okay, do that first then

[regal berry]

$2(x^2 + 3x + 5) +1x +1 >0$ this is where i am left

[woven radish]

Simon James B

[winter patrol]

no

as said before, that doesn't help

[regal berry]

what now

[winter patrol]

you don't want to have any x terms outside

don't bother with trying to get nice integers

[iron kindle]

but this doesnt do anything

[winter patrol]

just factor out 2 directly,

[radiant dune]

you want something like this essentially

[regal berry]

Then what does anything i don't know to solve this things

[radiant dune]

and see what remains from the 11/2

[regal berry]

no idea how to do that or what that means?!!!

[iron kindle]

note that $k(x+a)^2=kx^2+2akx+ka^2$

[radiant dune]

if 7/4 is 'eaten up' by the square completion

[woven radish]

Bonk

[winter patrol]

$2x^2 + 7x + 11 = 2 \cdot \what$

[woven radish]

ℝαμOmeganato5

[iron kindle]

so you need to find $k$ and $a$ such that $kx^2+2akx = 2x^2+7x$

[woven radish]

Bonk

[regal berry]

2 * (x^2 + 7x/2 + 11/2)?!

[winter patrol]

yes

[regal berry]

$2 * (x^2 + 7x/2 + 11/2)$

[woven radish]

Simon James B

[winter patrol]

and then focus on completing the square for the stuff within the parentheses

by introducing
+(b/2)^2 - (b/2)^2

[radiant dune]

think like the x^2 + 7x/2 thingy has appeared a result of expanding some (x+k)^2

so what would k be

[regal berry]

we have x^2 + 2ab + b^2 i see the expanded formula

in our case a = x right

7x/2 is our 2ab and 11/2 is our b^2

[radiant dune]

11/2 is not b^2

7x/2 is 2ab yes

then get b from here if a is x

you need to 'split' 11/2

to make b^2 and some other positive stuff

[regal berry]

11/2 is 5.5

[winter patrol]

forget about the 11/2 for now

that's not the main focus here

and try not to use b like that here, as it conflicts with what b is normally used to represent

[regal berry]

i have no clue what to use what do to how to do anything past this point

[iron kindle]

try this

[winter patrol]

i'll give you a side example with nicer numbers \
would you be able to complete the square for
$$x^2 + 4x$$

[woven radish]

ℝαμOmeganato5

[regal berry]

x^2 + 4x = $x^2 + 4x + 4- 4$

[woven radish]

Simon James B
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[winter patrol]

yes

apply the same idea to what you have

[regal berry]

the fractions confuse me alot that's why i can't do it