VCE Specialist 3/4 Question Thread!

[Sanguinne]

The set of points in the complex plane defined by |z+2i|=|z| corresponds to

A) the point given by z = -i
B) The line Im(z) = -1
C) the line Im(z) = -i
D) The line Re(z) = -1
E) The circle with the center -2i and radius 1

By subbing in x + yi, the cartesian equation i worked out was x = -1, however I am not sure what this corresponds to.
If there is any other way of doing this kind of question, what would it be?
Thanks

[drake]

Im(z) is basically the same as the y-axis while Re(z) is basically the same as the x-axis. So x=-1 corresponds to Re(z)=-1. But I think when you substituted z=x+yi, you did something wrong as the answer should be Im(z)=-1. Try it again and see if you end up with y=-1.

[RKTR]

how come i get y=-1 which is B?

|z-(-2i)|=|z|
same distance from two points which are 0 and -2i
therefore is y=-1

or you can do it as
x^2+(y+2)^2=x^2+y^2
y^2+4y+4=y^2
4y=-4
y=-1

[Sanguinne]

Im(z) is basically the same as the y-axis while Re(z) is basically the same as the x-axis. So x=-1 corresponds to Re(z)=-1. But I think when you substituted z=x+yi, you did something wrong as the answer should be Im(z)=-1. Try it again and see if you end up with y=-1.

yeah i made a mistake in my arithmetic, it should be y=-1

thanks

[M_BONG]

For circular functions graphs do you guys recommend memorising the general shapes of the graph then applying transformations?

Ie. Memorise shape of sin^-1(x) graph, then apply transformations if required? Also, is memorising the domains and ranges of that graph the only way - any way to conceptually remember this?

Also with reciprocal graphs (sort of asked this before), do you guys first draw f(x)=tan(x) then cot(x), reciprocating all key points or do you memorise general shapes then apply transformations?

I never like to memorise graph shapes because 1) I always memorise incorrectly 2) I start doubting myself under time pressure (SACs or exams). Are there any ways out of this for reciprocal and inverse circular functions?

[lzxnl]

I personally never bother with memorising graphs. I just work them out from the basic sine/cosine/tangent graphs that I DO know from memory (you should at least know the features of these graphs). But, I never bother drawing the sine/cosine/tangent graph in because I can just directly draw the graph of y = 1+ 2 csc(3x+pi/5) by plotting the important points/drawing in asymptotes of the graph (consider extrema and intercepts).

[M_BONG]

I personally never bother with memorising graphs. I just work them out from the basic sine/cosine/tangent graphs that I DO know from memory (you should at least know the features of these graphs). But, I never bother drawing the sine/cosine/tangent graph in because I can just directly draw the graph of y = 1+ 2 csc(3x+pi/5) by plotting the important points/drawing in asymptotes of the graph (consider extrema and intercepts).

Oh yeah that's what I prefer to do;

But what about  inverse graphs like sin^-1(x), do you just remember basic shape of sin (x), restrict it to a one-to-one function then reflect over y=x? Does that take longer than memorising/recalling the graph shape?

[lzxnl]

Hey An,

I need help with this question that requires knowledge on both the cross product and dot product;

Use the identity a x (b x c) = (a . c) b - (a . b)c   to prove a x (b x c) is not equal to (a x b) x c in general, that is, the vector product is not associative.

a, b and c are representative of vectors, I didn't know how to include the tildas.

Help would be appreciated, thanks

This question isn't a VCE level spesh question, just saying.
However, you would just use (a x b) x c = -c x (a x b) = -( (c.b)a - (c.a)b)) = (c.a)b - (c.b)a, which is not equal to (a.c)b - (a.b)c obviously.

Oh yeah that's what I prefer to do;

But what about  inverse graphs like sin^-1(x), do you just remember basic shape of sin (x), restrict it to a one-to-one function then reflect over y=x? Does that take longer than memorising/recalling the graph shape?

I don't even. I just mark in the endpoints, mark in the point of inflection, draw something which looks like a basic inverse sine in shape and then label the asymptotes.

[eagles]

Find (1-i)⁹ in Cartesian form.

First step shown in worked solutions is:

(1-i) = squareroot (1² + (-1)²) cis (-pi/4)

From there I know what to do, using de Moivre's theorem, but what I wish to understand is where the angle, -pi/4, came from?

Please refresh me! Cheers.

[brightsky]

okay in general, if you have some complex number raised to the power of something big, you know you need to use de moivre's theorem. de moivre's theorem can only really be applied if the complex number which is you being raised to the power of something big is written in polar form. so the first step is to convert (1-i) into complex form. now, a complex number, represented on an argand diagram, is simply a dot. picture (1 - i) in your mind. it is simply the point (1, -1). now convert to polar form. you know that the modulus (distance from origin) can be found easily using pythagoras' theorem: sqrt(1^2 + (-1)^2) = sqrt(2). you can find the Argument (capital a) by noting the angle which the complex number makes with the positive direction of the x-axis (or real axis). in this case, this angle is 45 degrees, since your right angled triangle, in this particular case, is also an isosceles triangle. what is 45 degrees in radians? pi/4. so (1-i) = sqrt(2) cis(-pi/4). raise to the power of 9. 16sqrt(2) cis(-9pi/4). now this isn't in proper polar form, since the angle should be between -pi (non-inclusive) and pi (inclusive). this can be easily fixed by adding 2pi. the answer is 16sqrt(2) cis(-9pi/4 + 8pi/4) = 16sqrt(2) cis(-pi/4).

edit: awkward error corrected.

[eagles]

Yep, thanks for the prompt reply!

But since 1-i is in 4th quadrant, will it be more correct if we wrote the answer with the angle -pi/4 instead of pi/4?

[brightsky]

oh woops awks, yeah you're right. should be -pi/4. see edit.

[eagles]

Ahaha thanks. I have another question:

x² + y² = 1/4

y² is equal or less than 1/4

Then -1/2 y is equal or less than 1/2

I'm having trouble understanding the third line; how does the second line flow to the third line?

Cheers.

[lzxnl]

Ahaha thanks. I have another question:

x² + y² = 1/4

y² is equal or less than 1/4

Then -1/2 y is equal or less than 1/2

I'm having trouble understanding the third line; how does the second line flow to the third line?

Cheers.

Have trouble understanding the third line? It's wrong.

y^2 <= 1/4
|y| <= 1/2 (think about this line carefully and why I chose |y| instead of y)
so -1/2 <= y <= 1/2

[eagles]

That makes more sense, thanks!

Just wanted to make sure whether I understood correctly:

Did you mod the y as it restricts the range to [-1/2, 1/2]?

Thank you