VCE Specialist 3/4 Question Thread!

[brightsky]

Wait does z x conj(z) = mod(z)^2?

Indeed!

Proof:

Let z = a + bi. Then conj(z) = a - bi.
z*conj(z) = (a+bi)(a-bi) = a^2 - b^2 i^2 = a^2 + b^2 = mod(z)^2

Also, mod(z)^2 - 2z - 2conj(z) + 1 = 0
mod(z)^2 - 2z - 2conj(z) + 4 = 3

Where does the 3 come from?

I added 3 to both sides of the equation.

[M_BONG]

Hey

Just having some difficulties understanding the explanation (in the attachment) where Maths Quest tries to explain consideration of 'behaviour near asymptotes".

FYI, y1 in this case refers to and y2 refers to

What does it mean when it says, y approaches y1 from above the graph? What about to approach from below the graph?

I understand how to find horizontal and vertical asymptotes etc but just seem confused as to what it means to approach from below or above the graph.

[mdntimp]

Hey guys! I have a question. It is not in the text book, it is extra work i was set to do on Parametric equations.

THE QUESTION: Determine the equation (I assume it means parametric equation) for a person on the "tea-cup" ride.
Each cup revolves counterclockwise on it's own 1 metre axis 6 times while completing one full revolution on a larger 4 metre axis, also in a counterclockwise direction. Draw a sketch of the path and then determine possible equations.
Note: This problem involves addition of ordinates.

So far, I have:

Cartesian equations for the two circles:
LARGER CIRCLE: x^2+y^2=4
SMALLER CIRCLE: (x-1.5)^2+y^2=0.25

Parametric equations for the two circles:
LARGER CIRCLE: x=2cos(t) , y=2sin(t) where t is an element of [0, 2Pi]
SMALLER CIRCLE: x=0.5cos(t)+1.5 , y=0.5sin(t) where t is an element of [0, 12Pi]

Use this online graphing calculator to check what my graphs look like, if you would like: https://www.desmos.com/calculator

Is what I'm doing even correct, or am I WAAAY off the mark?

AND, as far as addition of ordinates goes... I have no idea how to do that or how exactly it applies to this question. Please help! Thank you heaps   

[Eugenet17]

Help please





for the first question, I know what the graphs look like and whatnot, but I don't understand why we're placing certain restrictions on the graph?

Also, A complex number z satisfies the inequality mod(z+2-2sqrt(3)i)≤2 , b) Find the greatest possible value of Arg(z). How do i find this?

[survivor]

Let z1= 1+2i, and z2=2-i

a) Represent on an Argand diagram:
i)z1
ii)z2
iii)2z1 + z2
iv) z1-z2

b) Verify the parts iii and iv follow the vector triangle properties


Can plot on argand diagram, but not sure how to verify. If anyone could help me it would be greatly appreciated.
Thanks

[Alwin]

Hey

Just having some difficulties understanding the explanation (in the attachment) where Maths Quest tries to explain consideration of 'behaviour near asymptotes".

FYI, y1 in this case refers to and y2 refers to

What does it mean when it says, y approaches y1 from above the graph? What about to approach from below the graph?

I understand how to find horizontal and vertical asymptotes etc but just seem confused as to what it means to approach from below or above the graph.

Approach from below: when x = some number (usually a range of numbers but I chose just one here for example's sake), say 5 then your asymptote=10 but your actual graph=9.99999999 (less than asymptote)
Approach from above: when x is say 20 asymptote=5 but the actual graph=5.0001 (bigger than asymptote)

Graphically, using your attachment, on the LHS:
When x<-0.63 the graph has a y-value less than the asymptote. You can test numbers, but for any vertical line you draw for this range of x values the graph is always smaller than the asymptote
Vice versa for x>-0.63

Let z1= 1+2i, and z2=2-i

a) Represent on an Argand diagram:
i)z1
ii)z2
iii)2z1 + z2
iv) z1-z2

b) Verify the parts iii and iv follow the vector triangle properties


Can plot on argand diagram, but not sure how to verify. If anyone could help me it would be greatly appreciated.
Thanks

So what we do is "convert" (for lack of a better word) an argand diagram into a vector plane. The Re(z) axis becomes the direction of the i vector and the Im(z) axis becomes the direction of the j vector.

More technically, 1 + 0i represents i  and 0 + 1i represents j

so now, z₁ = 1 + 2i becomes i + 2j  and  z₂ = 2 - i becomes 2i - j
I'll only do (iii) for you so you can try (iv) yourself

2z₁ + z₂ = 4 + 3 i
equivalent to 4i + 3j

Now, i would say the "vector triangle properties" just means that they sum (head to tail) to 0 or something along those lines.
So: 2(i + 2j) + 2i - j - 4i + 3j = 0

That's the only triangle property of vectors I can think of... although there might be another

EDIT did realise at first there were multiple unanswered qs, will work my way through them from the earliest adding them in unless someone else answers first

[~T]

Hey

Just having some difficulties understanding the explanation (in the attachment) where Maths Quest tries to explain consideration of 'behaviour near asymptotes".

FYI, y1 in this case refers to and y2 refers to

What does it mean when it says, y approaches y1 from above the graph? What about to approach from below the graph?

I understand how to find horizontal and vertical asymptotes etc but just seem confused as to what it means to approach from below or above the graph.

When we observe asymptotic behaviour, we are observing a situation in which the distance between two graphs approaches zero. This can occur in two ways: when one graph approaches another from above, or when it approaches from below. This is literally what it sounds like. When it says "y approaches y1 from above the graph", the whole graph of is approaching the graph of from above - in essence, y is slightly larger in value than y1, and this gap is getting smaller. If y was slightly smaller but still approaching, then y would be approaching y1 from below.

Hey guys! I have a question. It is not in the text book, it is extra work i was set to do on Parametric equations.

THE QUESTION: Determine the equation (I assume it means parametric equation) for a person on the "tea-cup" ride.
Each cup revolves counterclockwise on it's own 1 metre axis 6 times while completing one full revolution on a larger 4 metre axis, also in a counterclockwise direction. Draw a sketch of the path and then determine possible equations.
Note: This problem involves addition of ordinates.

Hey, I'm not entirely sure where you've gone with this, nor exactly what the question is asking. However, this is what I think it means by addition of ordinates.

We have one larger circle on which the tea-cup is revolving, which would be x^2+y^2=4 as you said. Indeed, the parametric equations are x1 =2cos(t) and y1 =2sin(t) (among many other possibilities of course).

We also have the smaller circle on which the tea-cup is revolving relative to the larger circle, which would be x^2+y^2=1/4. The parametric equations are of a similar form to x=1/2 cos(t) and y=1/2 sin(t). HOWEVER, we know that this revolution occurs "6 times while completing one full revolution on a larger 4 metre axis", so the period is one sixth of the larger one. So, we write the parametric equations here as x2 =1/2 cos(6t) and y2 =1/2 sin(6t).

From here, the "addition of ordinates" refers to adding the sets of parametric equations together - we essentially have smaller, faster revolutions happening in relation to a larger, slower revolution, and we can write the final set of parametric equations as:

Help please

for the first question, I know what the graphs look like and whatnot, but I don't understand why we're placing certain restrictions on the graph?

Also, A complex number z satisfies the inequality mod(z+2-2sqrt(3)i)≤2 , b) Find the greatest possible value of Arg(z). How do i find this?

Well, there isn't really a reason why there are restrictions put on the graph. That's just because the textbook wants to test your understanding of domain, I guess The set of S has all the points with modulus 2 and arguments between 0 and pi/2 inclusive. I guess the point of the question lies in the following questions, however. For the set T, what is the domain of the argument? If T is composed of all points w where w=z^2, then we have to look back at what the set of S was.







This is the "interesting" point of the question I assume; we now have points with argument up to pi, rather than pi/2. Do a similar thing for the set of U.

With what in particular are you have trouble regarding question 23?

As for your last question, I always completed these types of questions geometrically. Firstly, we work out what the equation means graphically. gives us a circle (filled in) of centre and radius 2. If we plot this on a cartesian plane (or argand diagram) and we think about what the "greatest possible value of Arg(z)" means, we might get ourselves somewhere.

Essentially, the Argument is a measure of the angle that a ray from the origin to a point makes with the positive direction of the x-axis or real axis. So, if we look at our graph, we can see that the maximum Argument is going to occur when this "ray" is at a tangent to the circle, yeah? We also know that a tangent to a circle is at right angles to the radius, yeah? So in this case, if we draw a triangle formed by the points at the origin, the point of tangency, and the centre of the circle, we have ourselves a right angled triangle. You should be able to work out some side lengths (two in fact) and from there you can work out the angle required Good luck!

Let z1= 1+2i, and z2=2-i

a) Represent on an Argand diagram:
i)z1
ii)z2
iii)2z1 + z2
iv) z1-z2

b) Verify the parts iii and iv follow the vector triangle properties


Can plot on argand diagram, but not sure how to verify. If anyone could help me it would be greatly appreciated.
Thanks

EDIT: looks like Alwin has already helped you I remember that question being a little odd, because of the strong relationship between vectors and complex numbers. I would also do the sum head-to-tail is 0 (even though in "adding" these vectors you're pretty much just subtracting the complex numbers you found in the first place). Perhaps this relationship is what the textbook is trying to highlight, however.

[Eugenet17]

Thank you so much, oh and don't worry about the second question, I've already done it

[M_BONG]

Is it appropriate to use co-ordinates in an Argand diagram?

For example, if I was trying to plot the line z: Im (Z) = 2.    Is it wrong if I wrote in (0,2)?  Would it be better if I did (0 , 2i) ?

Thanks!

[brightsky]

(0,2i) is actually wrong; the correct representation is (0,2). I would suggest that you use Cartesian coordinates when labelling any complex number. (You can technically write 0 + 2i, but such a representation is often very inconvenient.)

[mdntimp]

Hey, I'm not entirely sure where you've gone with this, nor exactly what the question is asking. However, this is what I think it means by addition of ordinates.

We have one larger circle on which the tea-cup is revolving, which would be x^2+y^2=4 as you said. Indeed, the parametric equations are x1 =2cos(t) and y1 =2sin(t) (among many other possibilities of course).

We also have the smaller circle on which the tea-cup is revolving relative to the larger circle, which would be x^2+y^2=1/4. The parametric equations are of a similar form to x=1/2 cos(t) and y=1/2 sin(t). HOWEVER, we know that this revolution occurs "6 times while completing one full revolution on a larger 4 metre axis", so the period is one sixth of the larger one. So, we write the parametric equations here as x2 =1/2 cos(6t) and y2 =1/2 sin(6t).

From here, the "addition of ordinates" refers to adding the sets of parametric equations together - we essentially have smaller, faster revolutions happening in relation to a larger, slower revolution, and we can write the final set of parametric equations as:

THANK YOU T, YOU ARE FANTASTIC   

[Irving4Prez]

Hey,

Could anyone help me with this problem?

sec^2(tan^-1 (x))

Cheers in advance

[Stick]

That's not a question. :S

[Irving4Prez]

That's not a question. :S

It's a multiple choice asking, "... is equal to"

A) 1 + x^2
B) 1 - x^2
C) x^2 - 1
D) 1/(1+x^2)

[b^3]

Now we can draw a little right angled triangle out with , so and , which will allow us to find the third side.
.
So we next need to find , which we can from our triangle.


Hence option A.