VCE Specialist 3/4 Question Thread!

[Willba99]

On the Casio Classpad,
Main -> interactive -> complex
It has three options down the bottom:
CompToPol
CompToTrig (what does this do??)
CompToRect

So if I want to convert polar to Cartesian youI just use the 3rd one, and for Cartesian to polar just use the first one? And never use the 2nd one?
Thanks

sounds about right

[TheAspiringDoc]

sounds about right

Wait. Hold on..
I just tried to convert a Cartesian complex number to polar form on my Casio Classpad.
CompToPol gave 6e^(11pi*i/12) which is trippy eulers identity stuff (not what I wanted!)
CompToTrig gave 6(cos(11pi/12)+sin(11pi/12)i) from which you can easily figure out the abbreviated polar form to be 6cis(11pi/12)
So it seems as though CompToTrig is better than CompToPol for converting Cartesian to polar. Is that correct?

[RuiAce]

Wait. Hold on..
I just tried to convert a Cartesian complex number to polar form on my Casio Classpad.
CompToPol gave 6e^(11pi*i/12) which is grippy eulers identity stuff (not what I wanted!)
CompToTrig gave 6(cos(11pi/12)+sin(11pi/12)i) from which you can easily figure out the abbreviated polar form to be 6cis(11pi/12)
So it seems as though CompToTrig is better than CompToPol for converting Cartesian to polar. Is that correct?

In that case, the alleged "polar form" is the exponential form \(z = re^{i\theta} \), which takes advantage of Euler's formula \( e^{ix} = \cos x + i\sin x\).

Both the polar and the trig ones should be equally viable (but if you're indecisive then maybe just use Trig as you said)

[TheAspiringDoc]

^^Thanks for your help Rui
I know this is really old, but

z = w + 1/w = 2cist + 1/(2cist) = 2cist + 1/(2(cost + isint)) = 2cist + (cost - i sint)/(2) = 5/2 cost + 3/2 i sint
parameters:
x = 5/2cost
y =3/2 sint
convert into cartesian form and you get the equation of the ellipse.

Isn't the bolded bit above implying that
Since when was this true? Why isn't it in my textbook? Why is it true?

EDIT: I've just realised multiplying both sides of the above by (cost+isint) gives LHS = 1 and RHS = cos^2t+sin^2t, which is the Pythagorean identity ... so it must be true!
But how did brightsky know this weird aforementioned formula??

[Syndicate]

^^Thanks for your help Rui
I know this is really old, butIsn't the bolded bit above implying that
Since when was this true? Why isn't it in my textbook? Why is it true?

I am not sure which textbook you use, but if I remember correctly such property is not listed in the Cambridge specialist 3/4 as well.


In the last line, what I have done is just put cos(-t) as cos(t) and isin(-t) as -isin(t).

[TheAspiringDoc]

I am not sure which textbook you use, but if I remember correctly such property is not listed in the Cambridge specialist 3/4 as well.

Oops, I haven't covered de moivre's yet!
Yep I use Cambridge.
Still, I don't get why what you've written implies why what I was asking was true ... or maybe your LaTex is just having problems and not showing some bits that I'm now seeing when quoting your post..

[Unsplash]

Hi guys,

Can someone explain to me how you would go about sketching this modulus function?

EDIT: Hold on, I think i get it. Split it up into a piecewise function, where for x ≥ 0, the graph is the same as x^2 - 4x, and it is just reflected across the y axis when x < 0. Is this correct?

[VanillaRice]

Hi guys,

Can someone explain to me how you would go about sketching this modulus function?

EDIT: Hold on, I think i get it. Split it up into a piecewise function, where for x ≥ 0, the graph is the same as x^2 - 4x, and it is just reflected across the y axis when x < 0. Is this correct?

That's right - the modulus around the x means that the y-values of corresponding positive and negative values of x are the same e.g. f(2) = f(-2)

[Vaike]

EDIT: Hold on, I think i get it. Split it up into a piecewise function, where for x ≥ 0, the graph is the same as x^2 - 4x, and it is just reflected across the y axis when x < 0. Is this correct?

Hey! Yep, you're on the money. Whenever modulus is used around each individual term, the function essentially converts all negative x inputs into their corresponding positive forms, resulting in reflections in the y axis, as shown.

Spoiler

A reflection is also seen when a function is entirely contained within modulus. However, instead of a reflection across the y axis, all outputs that would normally fall below the x-axis (that is, they result in a negative y value) are reflected in the x axis.

Spoiler

[Unsplash]

That's right - the modulus around the x means that the y-values of corresponding positive and negative values of x are the same e.g. f(2) = f(-2)

Hey! Yep, you're on the money. Whenever modulus is used around each individual term, the function essentially converts all negative x inputs into their corresponding positive forms, resulting in reflections in the y axis, as shown.

Spoiler

A reflection is also seen when a function is entirely contained within modulus. However, instead of a reflection across the y axis, all outputs that would normally fall below the x-axis (that is, they result in a negative y value) are reflected in the x axis.

Spoiler

Thanks guys! Really appreciate your help.

[TheAspiringDoc]

If z^n=complex number, what are the general steps in solving for z?
So far I know There will be n solutions, all equally spaced around a circle with centre 0+0i
But how do you find the principle solution?

[LifeisaConstantStruggle]

If z^n=complex number, what are the general steps in solving for z?
So far I know There will be n solutions, all equally spaced around a circle with centre 0+0i
But how do you find the principle solution?

The easiest way (depends on the question tho) is to convert the complex number into polar form if you haven't already, and use your knowledge on the De Moivre's Theorem to solve the question. (I'll give you an example)

Let's say we have z³=1+i

Converting it into polar form will give you z³=2^1/2cis(pi/4)

Applying De Moivre's Theorem [rcis(x)]ⁿ=rⁿcis(nx)

z=2^(1/2)*(1/3)cis[(1/3)[(pi/4)+2kpi]

the 2kpi is fairly important here because you will get a few solutions (in this case it depends on the power of the polynomial, if it is 3 then you will get 3 solutions).

Substituting 0,1,2 will give you the three solutions you need (it will revolve around 0+0i yes), if you get z⁴ you have to substitute 0, 1, 2, 3 and so on.

[A TART]

How do I do part b of this question?

I've done a and my answer is 1/2(a+b)

I've drawn a couple of funny looking trapeziums and found that the when you connect the midpoints, the resulting line is parallel, but how do I show it? (Is there like some kind of rule or an actual mathematical way to prove it?)

[_______]

How do I do part b of this question?

I've done a and my answer is 1/2(a+b)

I've drawn a couple of funny looking trapeziums and found that the when you connect the midpoints, the resulting line is parallel, but how do I show it? (Is there like some kind of rule or an actual mathematical way to prove it?)

This is interesting. I think you can just say that since it's a trapezium, you know that AB and DC are parallel. So the vector (a + b) which is AB + DC is parallel to AB and DC, so XY = (a + b)/2 is also parallel to AB.

[RuiAce]

How do I do part b of this question?

I've done a and my answer is 1/2(a+b)

This is interesting. I think you can just say that since it's a trapezium, you know that AB and DC are parallel. So the vector (a + b) which is AB + DC is parallel to AB and DC, so XY = (a + b)/2 is also parallel to AB.

Basically this. I just add more reasoning