VCE Specialist 3/4 Question Thread!

[cosine]

Please refer to the attachment


 
Why is it that they only took the value of a from the first bracket and sub it into the equation to solve for b?
I mean, the second one can also be solved over the complex system, so why didn't they take the a values of that? 

[DSubShell]

Please refer to the attachment


 
Why is it that they only took the value of a from the first bracket and sub it into the equation to solve for b?
I mean, the second one can also be solved over the complex system, so why didn't they take the a values of that?

When you have a complex number of the form:


While the whole number is Complex, the coefficients and are actually only real. It makes sense. Try substitue a & b with complex numbers... and after simplifying you'll just end up with a+bi again, albeit a different a and b.

So with these types of questions, we only find the real solutions for a & b. in this case, solving for only gives imaginary answers... so we only use

[flares]

What is the first topic you do in Specialist Mathematics when you return back to school in year 12?

Thanks,

[appleandbee]

What is the first topic you do in Specialist Mathematics when you return back to school in year 12?

Thanks,

Conic functions :Ellipses, Hyperbolas, Parametric Funtions

[flares]

Conic functions :Ellipses, Hyperbolas, Parametric Funtions

eh, gross....

Thanks so much for that
Also, what textbook would you guys recommend?
I've got the Heinemann VCE SPecialist mathematics enhanced textbook (its on the booklist).
Thanks again

[keltingmeith]

Conic functions :Ellipses, Hyperbolas, Parametric Funtions

Hate to jump in here, but just for anybody not doing these first:
Don't stress, you DON'T have to start with these topics. Different teachers will choose different topics to start with, and there is no wrong order to learn specialist (hell, you could quite easily start with mechanics, although most schools don't). So don't worry if you're not doing these now/won't do them when you get back.

eh, gross....

Thanks so much for that
Also, what textbook would you guys recommend?
I've got the Heinemann VCE SPecialist mathematics enhanced textbook (its on the booklist).
Thanks again

Our favourite mod pi has written an amazing review of each VCE textbook (plus others!) here, which I highly recommend reading.

[cosine]

Solve



If z =rCis,  then

Okay, so convert -1 into polar form:

-1 = Cis 

Why does the text book say -1 in polar form is equal to Cis -pi?

[keltingmeith]

Solve



If z =rCis,  then

Okay, so convert -1 into polar form:

-1 = Cis 

Why does the text book say -1 in polar form is equal to Cis -pi?

Probably just mucked up their domain - however, they are the same angle, and if enough people made the mistake in an exam, I reckon VCAA probably would've disregarded it as long as the other 7 answers were in the appropriate domain.

[brightsky]

Solve



If z =rCis,  then

Okay, so convert -1 into polar form:

-1 = Cis 

Why does the text book say -1 in polar form is equal to Cis -pi?

I'd suggest setting out your working in the following way:

z^8 + 1 = 0
z^8 = -1
z^8 = cis(pi)
z = (cis(pi + 2k*pi))^1/8, where k E Z
z = cis(pi/8(1 + 2k))

Now, we require:

-pi < pi/8(1+2k) ≤ pi
-9/2 < k ≤ 7/2

Since k E Z, we know that k = -4, -3, -2, -1, 0, 1, 2, 3.

Let k = -4. Then z = cis(-7pi/8).
Let k = -3. Then z = cis(-5pi/8).
Let k = -2. Then z = cis(-3pi/8).
Let k = -1. Then z = cis(-pi/8).
Let k = 0. Then z = cis(pi/8).
Let k = 1. Then z = cis(3pi/8).
Let k = 2. Then z = cis(5pi/8).
Let k = 3. Then z = cis(7pi/8).

Note the following:

- There are 8 solutions in total.
- The modulus of each solution is 1.
- The argument of each solution differs from the next by 2pi/8 = pi/4.

This means that the 8 solutions to the equation z^8 + 1 = 0 all lie equally spaced along the circumference of the circle centred at the origin with a radius of 1.

[cosine]

Probably just mucked up their domain - however, they are the same angle, and if enough people made the mistake in an exam, I reckon VCAA probably would've disregarded it as long as the other 7 answers were in the appropriate domain.

Thanks for the reassurance buddy. By the way, I usually dont like to 'just' do math questions, I usually like to know why and such.. lol (i know, nerdy)

So, there was a question asking to solve , I mean we can easily solve it by factoring the cube roots using the formula, but we can also do the above method. Whats the difference between the both and when do we know what to use what?? Thanks

[cosine]

I'd suggest setting out your working in the following way:

z^8 + 1 = 0
z^8 = -1
z^8 = cis(pi)
z = (cis(pi + 2k*pi))^1/8, where k E Z
z = cis(pi/8(1 + 2k))

What does k E Z refer to? k is an element of???
Also, how did z = cis(pi/8(1 + 2k)) come about? Where did the pi in the (cis(pi + 2k*pi))^1/8 go? And were did the 1 come from?

[keltingmeith]

Thanks for the reassurance buddy. By the way, I usually dont like to 'just' do math questions, I usually like to know why and such.. lol (i know, nerdy)

Honestly, I don't care if it's nerdy, I respect you way more for this sentence. :')

So, there was a question asking to solve , I mean we can easily solve it by factoring the cube roots using the formula, but we can also do the above method. Whats the difference between the both and when do we know what to use what?? Thanks

Both are acceptable - one way is just breaking it up to linear factors and solving it over the Real number field before continuing on to the complex number field.

For example, by the factorisation method:



From this, we have one real solution, which is the first. For the second equation, we must solve it over the complex field, so we'd apply the quadratic formula and get:



Now, by the mod-arg method:



By converting from cartesian to polar or vice-versa, we can see that both methods work and give us the correct answer.

[brightsky]

What does k E Z refer to? k is an element of???
Also, how did z = cis(pi/8(1 + 2k)) come about? Where did the pi in the (cis(pi + 2k*pi))^1/8 go? And were did the 1 come from?

k E Z means k is an element of Z, which is the symbol used to denote the set of all integers. Ensure that you are familiar with what the symbols N, Z, Q, R and - seeing as you are doing Spesh next year - C, represent.

z = (cis(pi + 2k*pi))^1/8 = cis(1/8*(pi + 2k*pi)) = cis(pi/8(1 + 2k)) by De Moivre's Theorem, which states that if z = cis(θ), then z^n = cis(nθ), where n E Q (read: n is an element of the set of all rational numbers). The pi and the 1 emerged as a result of factorisation.

[cosine]

Honestly, I don't care if it's nerdy, I respect you way more for this sentence. :')

Both are acceptable - one way is just breaking it up to linear factors and solving it over the Real number field before continuing on to the complex number field.

For example, by the factorisation method:



From this, we have one real solution, which is the first. For the second equation, we must solve it over the complex field, so we'd apply the quadratic formula and get:



Now, by the mod-arg method:



By converting from cartesian to polar or vice-versa, we can see that both methods work and give us the correct answer.

Thanks brother, it makes sense now (:

[cosine]

k E Z means k is an element of Z, which is the symbol used to denote the set of all integers. Ensure that you are familiar with what the symbols N, Z, Q, R and - seeing as you are doing Spesh next year - C, represent.

z = (cis(pi + 2k*pi))^1/8 = cis(1/8*(pi + 2k*pi)) = cis(pi/8(1 + 2k)) by De Moivre's Theorem, which states that if z = cis(θ), then z^n = cis(nθ), where n E Q (read: n is an element of the set of all rational numbers). The pi and the 1 emerged as a result of factorisation.

Ahh totally see where it comes from now! Thank you!