VCE Specialist 3/4 Question Thread!

[Cort]

Right, just to test you all on some theory wise ( I lie, I'm retarded and I have a memory of a gold fish) --

When it comes to complex numbers, finding the roots/solutions can be found via:
De Moivre's Law, correct? Are there other ways I can actually approach it? Pedantically, do you know which exercise in chapter 3 Maths Quest might have this? Or even essentials? I have both, but I can't seem to find relevant explanation or ideas to it.

Furthermore, to break it down (someone correct me if I'm wrong)
In complex numbers, the coordinates are mapped in an argand diagram
Z= x + yi is the Cartesian Form. You normally do addition and subtraction with these (with their respectable values) because it's easier.
Re(z) = x and Im(z) = y. When it comes to dividing, multiplying, or dealing with powers, you convert to Polar form, yes?

What about factorising? Oh god panic panic panic.

[enwiabe]

Right, just to test you all on some theory wise ( I lie, I'm retarded and I have a memory of a gold fish) --

When it comes to complex numbers, finding the roots/solutions can be found via:
De Moivre's Law, correct? Are there other ways I can actually approach it? Pedantically, do you know which exercise in chapter 3 Maths Quest might have this? Or even essentials? I have both, but I can't seem to find relevant explanation or ideas to it.

Furthermore, to break it down (someone correct me if I'm wrong)
In complex numbers, the coordinates are mapped in an argand diagram
Z= x + yi is the Cartesian Form. You normally do addition and subtraction with these (with their respectable values) because it's easier.
Re(z) = x and Im(z) = y. When it comes to dividing, multiplying, or dealing with powers, you convert to Polar form, yes?

What about factorising? Oh god panic panic panic.

Here's the vTextbook tutorial on Complex Numbers:

https://www.youtube.com/watch?v=DGEciZ9mQjw

Polar form starts roughly 7 minutes in and De Moivre's theorem shortly after around 9:20, but I recommend watching the whole video because it's a really nice explanation of everything you've been learning so far in the topic.

Hope that helps! The full vTextbook Specialist Maths course will be available in the next week, so everyone keep your eyes peeled

[Cort]

Bloody brilliant. Thanks

[Special At Specialist]

how do I differentiate sec^2(5x^2)? :x

Using the formula:

[eagles]

Hello! I have a quick question:

If a, b, c are elements of the complex plane and |a|=|b|=|c|= 1,
show that (a)(a conjugate) = (a)² = 1 and similarly for (b)(b conjugate) = (b)² = 1.

All your help is appreciated!

[Phenomenol]

Hello! I have a quick question:

If a, b, c are elements of the complex plane and |a|=|b|=|c|= 1,
show that (a)(a conjugate) = (a)² = 1 and similarly for (b)(b conjugate) = (b)² = 1.

All your help is appreciated!

Let a = x + yi and go from there.

[M_BONG]

Can anyone help me?

This question is really frustrating and I have no access to a teacher/tutor until Monday...

Ok the question is on last year's SAC.. it's a Locus question.

For those of you confused, screenshot 2 is the actually solution to the question... screenshot 3 is the consequential solution (ie. if you got the wrong region for part a). I decided to do both parts for practice.

I don't understand how my teacher derived the following:

Screenshot 2 "Solutions"
(iii) the minimum value of Arg (W)



Screenshot 3 "Part I don't Understand"
(iii) the minimum value of Arg (w) .

How did he get the pi/6 as the tangent to the circle?



Thanks so much in advance; I have a SAC on Wednesday so I desperately need your help ASAP!

[lzxnl]

Can anyone help me?

This question is really frustrating and I have no access to a teacher/tutor until Monday...

Ok the question is on last year's SAC.. it's a Locus question.

For those of you confused, screenshot 2 is the actually solution to the question... screenshot 3 is the consequential solution (ie. if you got the wrong region for part a). I decided to do both parts for practice.

I don't understand how my teacher derived the following:

Screenshot 2 "Solutions"
(iii) the minimum value of Arg (W)



Screenshot 3 "Part I don't Understand"
(iii) the minimum value of Arg (w) .

How did he get the pi/6 as the tangent to the circle?



Thanks so much in advance; I have a SAC on Wednesday so I desperately need your help ASAP!

Minimum value of Arg(w)? This essentially means we want to draw a line from the origin to the curve and want to minimise the angle it makes with the positive x-axis direction. Let's deal with the case of the full circle first. Can you see that such a line must be a tangent to the circle? If you take any other point through the origin intersecting with the circle, you'll find that you'll have two intersection points with the circle and your line will have a larger gradient.

So, how to find this tangent? Let's use some geometry. From circle theorems, the line from the centre of the circle to where the tangent meets the circle is perpendicular to the tangent. Thus, the circle centre, origin and the point of intersection of the tangent and circle form a right angled triangle. The tangent is NOT the hypotenuse; the hypotenuse is the line from the origin to the centre of the circle. Draw it out and you'll see. Now, let's find the complementary angle to the angle marked theta in your diagram. The radius of your circle is 1 and the distance from the centre of the circle to the origin is 2. The hypotenuse is thus 2 and the opposite angle is 1. The angle is 30 degrees and your argument is 60 degrees. This is for the "erroneous" case where the bottom semicircle is considered.

Now if you're just considering the top semicircle, you should be able to see that along the semicircle, the angle made by the line from the origin to the semicircle is always increasing, so you take the bottom right corner of the semicircle. Hopefully you can see that the point has coordinates (1,2) as it's horizontal from the centre of the circle at (0,2) and the radius is 1. So, we just need to find the angle a line connecting (0,0) and (1,2) makes with the positive x axis direction. Trig ratios yield tan theta = 2 => theta = inv tan(2)

[M_BONG]

Minimum value of Arg(w)? This essentially means we want to draw a line from the origin to the curve and want to minimise the angle it makes with the positive x-axis direction. Let's deal with the case of the full circle first. Can you see that such a line must be a tangent to the circle? If you take any other point through the origin intersecting with the circle, you'll find that you'll have two intersection points with the circle and your line will have a larger gradient.

So, how to find this tangent? Let's use some geometry. From circle theorems, the line from the centre of the circle to where the tangent meets the circle is perpendicular to the tangent. Thus, the circle centre, origin and the point of intersection of the tangent and circle form a right angled triangle. The tangent is NOT the hypotenuse; the hypotenuse is the line from the origin to the centre of the circle. Draw it out and you'll see. Now, let's find the complementary angle to the angle marked theta in your diagram. The radius of your circle is 1 and the distance from the centre of the circle to the origin is 2. The hypotenuse is thus 2 and the opposite angle is 1. The angle is 30 degrees and your argument is 60 degrees. This is for the "erroneous" case where the bottom semicircle is considered.

Now if you're just considering the top semicircle, you should be able to see that along the semicircle, the angle made by the line from the origin to the semicircle is always increasing, so you take the bottom right corner of the semicircle. Hopefully you can see that the point has coordinates (1,2) as it's horizontal from the centre of the circle at (0,2) and the radius is 1. So, we just need to find the angle a line connecting (0,0) and (1,2) makes with the positive x axis direction. Trig ratios yield tan theta = 2 => theta = inv tan(2)

Thank you!
I realised my mistake; I mistook the tangent as the hypothenuse... since I always thought the hypothenuse was just the length directly opposite the right angle.

BUT, I still don't understand why the tangent is not the hypothenuse. Can you explain to me why? From what I can see, the tangent's length is root 5 (from pythagoras)... doesn't that make it the hypothenuse?

[lzxnl]

The tangent's length isn't root 5. It's actually root 3. Think about where the right angle is. By the sine rule, the largest angle is opposite the largest side length. In the right angled triangle drawn, the hypotenuse must be opposite the right angle. The right angle is between the circle radius and the tangent, so the tangent cannot possibly be the hypotenuse.

[Thorium]

I was looking at a worked example on complex numbers, where we have to find loci. The last part of this example says:

"-8x-9=6sqrt((6x+2)^2+y^2). Note: This implies -8x-9>0. Therefore x<-9/8. Squaring both sides and adding like terms:
28x^2-36y^2=63
x^2/36-y^2/28=1/16
x^2/9-y^2/7=1/4
The locus is a hyperbola with x=<-3/2"

Now my question is 1. How the equation started to have fractions? 2. Why is it said that x<-9/8 and not x=<-9/8? 3. Is x defined for only the left branch of the hyperbola because of the previous statement?

EDIT: Corrected a typo
Hope My questions make sense.
Thanks

[Cort]

Just touching up on some revision..but I'm not too sure about these two equations. I've tried substituting them in to make 0, but I'm not getting the right answer.

[RKTR]

if z+2+2i is a factor, then  P( -2-2i)=0

[Cort]

if z+2+2i is a factor, then  P( -2-2i)=0

Huh, looks like my alegra at hand skills are absolutely terrible. Thanks for that!

[rhinwarr]

You can use combinations (eg. 5C1) instead of the numbers from pascals triangle but that is even more annoying.

For TI-nspire the syntax would be: (the x between i and tan(x) is a multiply sign)
expand((1 + i x tan(x))^5)