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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: /0 on July 21, 2009, 10:37:54 pm

Title: Spesh Qs
Post by: /0 on July 21, 2009, 10:37:54 pm: Let $p(z) = 1+z^2+z^4$ , and let $\alpha$ be a zero of $p(z)$ .

Show that $\alpha^6 =1$
Show that $\alpha^2$ is also a zero of $p(z)$ .
Title: Re: Spesh Qs
Post by: kamil9876 on July 21, 2009, 10:41:33 pm: $0=1 + a^2 + a^4$ (1)

multiply both sides by a^2:

$0=a^2 + a^4 + a^6$ (2)

(2) - (1):
$ a^6 - 1=0$

$a^6=1$
Title: Re: Spesh Qs
Post by: TrueTears on July 21, 2009, 10:43:00 pm: lol this was itute exam Q
Title: Re: Spesh Qs
Post by: kamil9876 on July 21, 2009, 10:54:30 pm: $p(a^2)=1+a^4+a^8$
$a^2 p (a^2)=a^2+a^6+a^{10}$
$=a^2+1+a^4$
$=p(a)$
$=0$

$\therefore a^2 p(a^2)=0$

$but a^2 \neq 0 because a^2=0 \implies a^6=0$

$\therefore p(a^2)=0$
Title: Re: Spesh Qs
Post by: kamil9876 on July 21, 2009, 10:56:34 pm: Quote from: TrueTears on July 21, 2009, 10:43:00 pm
lol this was itute exam Q

Orly? meh I was feeling creative today so it worked out, but under exam pressure I don't know if I would have come up with this.
Title: Re: Spesh Qs
Post by: /0 on July 21, 2009, 10:59:47 pm: Quote from: TrueTears on July 21, 2009, 10:43:00 pm
lol this was itute exam Q

Really? I got it from HSC 2008

Thanks kamil :)
Title: Re: Spesh Qs
Post by: kamil9876 on July 22, 2009, 12:39:50 am: I just realised I multipled everything by a^2 unnecesarily, look:

$p(a^2)=1+a^4+a^8=1+a^4+a^2 a^6=1+a^4+a^2=p(a)$

Oh and the schoolkid way of doing first part would probably involve the geometric sequence, but what I did is essentially derive the formula for it in disguise.
Title: Re: Spesh Qs
Post by: dcc on July 22, 2009, 05:13:59 pm: A 'lazy' way of doing this (i.e. no inspiration whatsoever) is:

$ \begin{align*} p(z) &= \left(z^2 + \frac{1}{2}\right)^2 + \frac{3}{4}\\ \implies p(\alpha) &= \left(\alpha^2 + \frac{1}{2}\right)^2 + \frac{3}{4} = 0\\ \left(\alpha^2 + \frac{1}{2}\right) &= \pm \frac{\sqrt{3}i}{2}\\ \implies \alpha^2 &= \frac{-1 \pm \sqrt{3}i}{2} = \exp\left( \pm \frac{2\pi i}{3} \right)\\ \end{align*}$ .

Notice that $\alpha^4 = \exp\left( \pm \frac{4\pi i}{3}\right) = \exp\left( \mp \frac{2 \pi i}{3} \right) = \overline{\alpha}^2$ . Therefore $p\left(\alpha^2\right) = \left(\overline{\alpha}^2 + \frac{1}{2}\right)^2 + \frac{3}{4} = 0$ (You should be able to see that without any evaluation at all, but sometimes it helps just to write everything down).

$\alpha^6 = \exp \left( \pm 2 \pi i \right) = 1$ .
Title: Re: Spesh Qs
Post by: /0 on July 22, 2009, 05:48:18 pm: Thanks, it's great to see different approaches, no matter how inspired they are ;p
Title: Re: Spesh Qs
Post by: /0 on July 22, 2009, 11:34:03 pm: 1. Use the method of cylindrical shells to find the volume of the solid formed when the shaded region bounded by

$y=0$ , $y=\frac{\log_ex}{x}$ , $x=1$ and $x=e$ is rotated about the y-axis.

It is given that $\int \log_exdx = x\log_ex-x+C$
Title: Re: Spesh Qs
Post by: /0 on July 27, 2009, 09:50:59 pm: Describe the key features of $\{z:|z-i|=5\}$

I said "Circle center (0,1), radius 5 in the complex plane"

The answers said "Circle center 0+i, radius 5"

Would my answer be incorrect?
Title: Re: Spesh Qs
Post by: TrueTears on July 27, 2009, 09:56:51 pm: Remember I asked this in class once? Back in good old term 2, and Mr B said we could use Cartesian notation in the Argand diagram.
Title: Re: Spesh Qs
Post by: /0 on July 27, 2009, 10:00:15 pm: Quote from: TrueTears on July 27, 2009, 09:56:51 pm
Remember I asked this in class once? Back in good old term 2, and Mr B said we could use Cartesian notation in the Argand diagram.

Oh yeah true!
Title: Re: Spesh Qs
Post by: Mao on July 27, 2009, 10:49:17 pm: Quote from: TrueTears on July 27, 2009, 09:56:51 pm
Remember I asked this in class once? Back in good old term 2, and Mr B said we could use Cartesian notation in the Argand diagram.

It is preferable that you don't, since the axis's of the Argand diagram are defined as Re(z) and Im(z). :)
Title: Re: Spesh Qs
Post by: /0 on July 28, 2009, 06:42:52 pm: ah i see...

Also: Find $\frac{dy}{dx}$ for $xy +\frac{y^2}{x}=2$

If I do implicit diff straight away I end up with $\frac{y^2-x^2y}{x^3+2xy}$

If I rearrange so that x is not in the denominator, then implicit diff, I get $\frac{2(1-xy)}{x^2+2xy}$

Can someone please check this?
Title: Re: Spesh Qs
Post by: TrueTears on July 28, 2009, 06:43:27 pm: Haha I just did this Q /0, yeah their both right, I think the Q asked you to find the equation of the tangents? Both expressions for the gradient function give the same gradient at those points.
Title: Re: Spesh Qs
Post by: /0 on July 28, 2009, 08:44:23 pm: Quote from: TrueTears on July 28, 2009, 06:43:27 pm
Haha I just did this Q /0, yeah their both right, I think the Q asked you to find the equation of the tangents? Both expressions for the gradient function give the same gradient at those points.

Yeah that's right, but what I'm more worried about is that the expressions are not the same for other points.

i.e. for $x=2$ , solving for y you get $y=-1 \pm 2\sqrt{2}$

Let's take the positive value of y...

With $\frac{y^2-x^2y}{x^3+2xy}$ the gradient is $\frac{-2}{51}(113\sqrt{2}-151)$

With $\frac{2(1-xy)}{x^2+2xy}$ the gradient is $\frac{4}{93}(11\sqrt{2}-26)$

... which leads me to believe that somehow, one of these expressions is wrong.
Title: Re: Spesh Qs
Post by: TrueTears on July 28, 2009, 08:48:17 pm: Yeah that JUST happened to me on Kilbaha exam Q 1... the expressions don't work for some points but they do for others... I'm gonna ask kamil about it later and see what he says on it.
Title: Re: Spesh Qs
Post by: /0 on July 28, 2009, 08:53:50 pm: hmm interesting
Also another question, this time from MAV 2007 Exam 2, picture attached

This might be due to my lack of understanding with solids of revolutions, but I was under the impression that A would be the correct expression, but it is not.
Title: Re: Spesh Qs
Post by: Flaming_Arrow on July 28, 2009, 08:57:41 pm: is it B?
Title: Re: Spesh Qs
Post by: kamil9876 on July 28, 2009, 09:01:06 pm: Quote from: Flaming_Arrow on July 28, 2009, 08:57:41 pm
is it B?

Yep
Title: Re: Spesh Qs
Post by: /0 on July 28, 2009, 09:03:00 pm: ok don't worry i think i see why now
Title: Re: Spesh Qs
Post by: Damo17 on July 28, 2009, 09:39:28 pm: Quote from: /0 on July 28, 2009, 08:53:50 pm
hmm interesting
Also another question, this time from MAV 2007 Exam 2, picture attached

This might be due to my lack of understanding with solids of revolutions, but I was under the impression that A would be the correct expression, but it is not.

it can't be A because it is not in the form $\pi \int_a^b ((y_1)^2 - (y_2)^2) dx$ . where $y_1$ is upper function and $y_2$ is lower function.

you can not combine to two functions.
$\pi \int_a^b ((y_1)^2 - (y_2)^2) dx \ne \pi \int_a^b (y_1 - y_2)^2 dx$
Title: Re: Spesh Qs
Post by: Mao on July 28, 2009, 09:44:49 pm: Replying to the question relating implicit diff, $xy + \frac{y^2}{x} = 2$ , the solutions should be the same. It appears you have made a small error in solving the relation for y when x=2, as well as an error in implicit differentiation.

implicit differentiation:
$\implies x \frac{dy}{dx} + y + \frac{2y}{x}\frac{dy}{dx} - \frac{y^2}{x^2} = 0$

$\implies \frac{dy}{dx}\cdot \left( \frac{2y + x^2}{x}\right) = \frac{y^2 - x^2 y}{x^2}$

$\implies \frac{dy}{dx} = \frac{y^2 - x^2 y }{x^3 + 2xy}$

The other method:

$x^2 y + y^2 = 2x$

implicit diff:

$2xy + x^2 \frac{dy}{dx} + 2y\frac{dy}{dx} = 2$

$\implies \frac{dy}{dx} \cdot (x^2 + 2y) = 2 - 2xy$

$\implies \frac{dy}{dx} = \frac{2(1-xy)}{x^2 + 2y}$

When x=2, $2y + \frac{1}{2}y^2 = 2 \implies y = -2 \pm 2\sqrt{2}$

for the positive value, $\frac{dy}{dx}_1 =-2 + \frac{5}{2\sqrt{2}}$
$\frac{dy}{dx}_2 = -2 + \frac{5}{2\sqrt{2}}$
Title: Re: Spesh Qs
Post by: /0 on August 09, 2009, 01:25:18 am: Apparently the answer for this is 100N but i get 41.2N... what does everyone else get?

(highlight to uncover text)
Title: Re: Spesh Qs
Post by: unknown id on August 09, 2009, 01:40:50 am: I get 100N.

$F = ma$

$P - 10gsin30^\circ - \frac{\sqrt 3}{5}gcos30^\circ = 10 \times 2.16$
Title: Re: Spesh Qs
Post by: TrueTears on August 09, 2009, 01:42:40 am: Yeah I get 100.

EDIT: too fast unknown id lol was just about to post that.
Title: Re: Spesh Qs
Post by: TonyHem on August 09, 2009, 01:48:42 am: He forgot the 10
Title: Re: Spesh Qs
Post by: /0 on August 09, 2009, 03:10:57 am: thanks
Title: Re: Spesh Qs
Post by: /0 on August 09, 2009, 09:09:29 pm: Is it true/false that if you take the average of the three coordinates of a triangle you will get the coordinates of the centre of the circumcircle?
I've heard that it can be used to find the centroid but i'm not sure if that's the same thing.
Title: Re: Spesh Qs
Post by: TrueTears on August 09, 2009, 09:15:13 pm: Quote from: /0 on August 09, 2009, 09:09:29 pm
Is it true/false that if you take the average of the three coordinates of a triangle you will get the coordinates of the centre of the circumcircle?
I've heard that it can be used to find the centroid but i'm not sure if that's the same thing.
Is this that Kilbaha question? |z-c| = r shit?

I just made 3 simultaneous equations to find the centroid.
Title: Re: Spesh Qs
Post by: /0 on August 16, 2009, 09:06:19 pm: true true

Another Q

If you have a length in $m$ and you take the log base 10 of it then do the units stay the same or wut? :P
Title: Re: Spesh Qs
Post by: kamil9876 on August 16, 2009, 09:15:05 pm: Can you provide an example of the context?
Title: Re: Spesh Qs
Post by: TrueTears on August 16, 2009, 09:17:54 pm: I will provide the example, it is in fact this question:
http://vcenotes.com/forum/index.php/topic,9192.msg167561.html#msg167561
Title: Re: Spesh Qs
Post by: kamil9876 on August 16, 2009, 09:26:46 pm: ok well because the question says "x cm". x=10 for 10cm, not x=10cm otherwise "x cm" would imply "10cm cm".

==============================================================================

Difference between a maths and physics question:

maths:

Displacement of a particle x m from the origin at time t seconds after blah blah is given by:

$x=sin(\omega t)$ where $\omega=2\pi$

physics:

Displacement of a particle,x , at a time t after blah blah is given by:

$x=sin(\omega t) m$ where $\omega=2\pi ^c s^{-1}$
Title: Re: Spesh Qs
Post by: /0 on August 16, 2009, 09:33:16 pm: thanks kamil... it was just a general question
Title: Re: Spesh Qs
Post by: kamil9876 on August 16, 2009, 09:36:06 pm: yeah, in general what i tried to say (without explicitly saying it) is that mathematical functions take unitless numbers as arguments so really there should be some constant with a unit to cancel out the unit of the quantity, or the quantity can just be defined by a real number as 'how many units'(latter being the case in most applied math questions)
Title: Re: Spesh Qs
Post by: /0 on September 08, 2009, 10:17:30 pm: If a question asks whether the 'path' of one vector function crosses the path of another vector, what does it mean by 'cross'?

For example, if you had the vector (1,1) and the 'path' r(t) = (t,-t), would that 'cross' the (1,1) vector?
Title: Re: Spesh Qs
Post by: TrueTears on September 08, 2009, 10:21:01 pm: I'd assume cross is just that the path of both the vector functions will intersect each other. The particles themselves don't have to intersect at the same time but the paths they travel on will intersect.

r(t) = (t,-t) what notation is that?
Title: Re: Spesh Qs
Post by: Damo17 on September 08, 2009, 10:22:15 pm: Quote from: /0 on September 08, 2009, 10:17:30 pm
If a question asks whether the 'path' of one vector function crosses the path of another vector, what does it mean by 'cross'?

For example, if you had the vector (1,1) and the 'path' r(t) = (t,-t), would that 'cross' the path of the vector?

I believe if it says 'crosses the path' it means that at some point in time, one vector is at a position where the other is (if collision), or has previously been there.

In your example, no they wouldn't cross, at t=1, one vector would be at (1,1) and the other at (1,-1).
Title: Re: Spesh Qs
Post by: /0 on September 08, 2009, 10:23:13 pm: lazy notation

according to MAV 2008 in that particular example the paths wouldn't cross... so i wonder if the domain for 'crossing' is something like (0,1), not [0, 1]? or something or rather...
Quote from: TrueTears on September 08, 2009, 10:21:01 pm
I'd assume cross is just that the path of both the vector functions will intersect each other. The particles themselves don't have to intersect at the same time but the paths they travel on will intersect.

r(t) = (t,-t) what notation is that?
Title: Re: Spesh Qs
Post by: TrueTears on September 08, 2009, 10:23:52 pm: Wait r(t) = t i -t j?
Title: Re: Spesh Qs
Post by: /0 on September 08, 2009, 10:24:20 pm: Quote from: Damo17 on September 08, 2009, 10:22:15 pm
Quote from: /0 on September 08, 2009, 10:17:30 pm
If a question asks whether the 'path' of one vector function crosses the path of another vector, what does it mean by 'cross'?

For example, if you had the vector (1,1) and the 'path' r(t) = (t,-t), would that 'cross' the path of the vector?

I believe if it says 'crosses the path' it means that at some point in time, one vector is at a position where the other is (if collision), or has previously been there.

In your example, no they wouldn't cross, at t=1, one vector would be at (1,1) and the other at (1,-1).

oh so do you think the path has to go THROUGH the point (1,1), not just through the 'vector' pointing there?

Quote from: TrueTears on September 08, 2009, 10:23:52 pm
Wait r(t) = t i -t j?

yer
Title: Re: Spesh Qs
Post by: TrueTears on September 08, 2009, 10:24:57 pm: Quote from: /0 on September 08, 2009, 10:24:20 pm
Quote from: Damo17 on September 08, 2009, 10:22:15 pm
Quote from: /0 on September 08, 2009, 10:17:30 pm
If a question asks whether the 'path' of one vector function crosses the path of another vector, what does it mean by 'cross'?

For example, if you had the vector (1,1) and the 'path' r(t) = (t,-t), would that 'cross' the path of the vector?

I believe if it says 'crosses the path' it means that at some point in time, one vector is at a position where the other is (if collision), or has previously been there.

In your example, no they wouldn't cross, at t=1, one vector would be at (1,1) and the other at (1,-1).

oh so do you think the path has to go THROUGH the point (1,1), not just through the 'vector' pointing there?

Quote from: TrueTears on September 08, 2009, 10:23:52 pm
Wait r(t) = t i -t j?

yer
Yes that what it means by crossing doesn't it? The paths literally have to MEET. [But doesn't have to be at same time]

Quote from: /0
Quote from: TrueTears on Today at 10:23:52 PM
Wait r(t) = t i -t j?

yer
Okay well x = t and y = -t

so y = -x is the cartesian equation. Since there is no restriction on t, then domain and range of the function is R.

Therefore it won't ever intersect with the point (1,1)
Title: Re: Spesh Qs
Post by: /0 on September 08, 2009, 10:27:07 pm: mmm cool thanks, it just seems weird seeing as in the question they have OA = 500i+450j which is just a 'point', and yet they ask whether the vector function crosses the 'path' of OA. Wouldn't normally call a point a path rite?
Title: Re: Spesh Qs
Post by: Damo17 on September 08, 2009, 10:27:21 pm: Quote from: /0 on September 08, 2009, 10:24:20 pm
Quote from: Damo17 on September 08, 2009, 10:22:15 pm
Quote from: /0 on September 08, 2009, 10:17:30 pm
If a question asks whether the 'path' of one vector function crosses the path of another vector, what does it mean by 'cross'?

For example, if you had the vector (1,1) and the 'path' r(t) = (t,-t), would that 'cross' the path of the vector?

I believe if it says 'crosses the path' it means that at some point in time, one vector is at a position where the other is (if collision), or has previously been there.

In your example, no they wouldn't cross, at t=1, one vector would be at (1,1) and the other at (1,-1).

oh so do you think the path has to go THROUGH the point (1,1), not just through the 'vector' pointing there?

My limited knowledge says yes.
Title: Re: Spesh Qs
Post by: TrueTears on September 08, 2009, 10:28:20 pm: Quote from: /0 on September 08, 2009, 10:27:07 pm
mmm cool thanks, it just seems weird seeing as in the question they have OA = 500i+450j which is just a 'point', and yet they ask whether the vector function crosses the 'path' of OA. Wouldn't normally call a point a path rite?
LOL I remember that dodgy plane question!

(wait didn't they say the plane landed at (500,450) or something? soz memory is a bit weary here lol)
Title: Re: Spesh Qs
Post by: /0 on September 08, 2009, 10:30:12 pm: Quote from: TrueTears on September 08, 2009, 10:28:20 pm
Quote from: /0 on September 08, 2009, 10:27:07 pm
mmm cool thanks, it just seems weird seeing as in the question they have OA = 500i+450j which is just a 'point', and yet they ask whether the vector function crosses the 'path' of OA. Wouldn't normally call a point a path rite?
LOL I remember that dodgy plane question!

yeah true lol
anyway thanks TT and damo XD
Title: Re: Spesh Qs
Post by: Mao on September 08, 2009, 10:30:43 pm: For two vector functions $r_1 (t)$ and $r_2(t)$

Cross: for some arbitrary $t_1$ and $t_2$ , $r_1(t_1) = r_2(t_2)$ (i.e. they cross the same point)

Collide: for some $t_1$ , $r_1(t_1) = r_2(t_1)$ (i.e. they cross the same point at the same time)

A path defined by the tip of the position vector described by r(t). Hence, when two paths cross, we're saying that the tips of two position vectors coincide. Since all position vectors originate from O, thus the two position vectors must coincide exactly for a 'cross'. For a collision, the two position vectors must coincide exactly at the time.
Title: Re: Spesh Qs
Post by: /0 on September 08, 2009, 10:32:01 pm: Quote from: TrueTears on September 08, 2009, 10:28:20 pm
Quote from: /0 on September 08, 2009, 10:27:07 pm
mmm cool thanks, it just seems weird seeing as in the question they have OA = 500i+450j which is just a 'point', and yet they ask whether the vector function crosses the 'path' of OA. Wouldn't normally call a point a path rite?
LOL I remember that dodgy plane question!

(wait didn't they say the plane landed at (500,450) or something? soz memory is a bit weary here lol)
Here's the question

Relative to a fixed origin, O, the position of the plane, P, at time $t \geq 0$ is given by $r(t) = (15t^2+30t)i+(10t^2-20t)j$ . An airport officilal is standing at A(500,450) and observes the plane landing, worried that he may be hit by the plane. State, giving reasons, whether the plane's path crosses the path of vector OA.

Quote from: Mao on September 08, 2009, 10:30:43 pm
For two vector functions $r_1 (t)$ and $r_2(t)$

Cross: for some arbitrary $t_1$ and $t_2$ , $r_1(t_1) = r_2(t_2)$ (i.e. they cross the same point)

Collide: for some $t_1$ , $r_1(t_1) = r_2(t_1)$ (i.e. they cross the same point at the same time)

A path defined by the tip of the position vector described by r(t). Hence, when two paths cross, we're saying that the tips of two position vectors coincide. Since all position vectors originate from O, thus the two position vectors must coincide exactly for a 'cross'. For a collision, the two position vectors must coincide exactly at the time.

cool thanks mao! hate those tricky linguistics
Title: Re: Spesh Qs
Post by: /0 on September 08, 2009, 10:36:03 pm: Lol i think MAV might have had other ideas.

SPOILERS; THESE ARE SOLUTIONS.
Title: Re: Spesh Qs
Post by: Mao on September 08, 2009, 10:37:41 pm: It sounds like they defined 'the path of the vector OA' like a line segment, and any crossing of the plane with this line segment counts as 'crossing the path of the vector OA'.

Tricky wording
Title: Re: Spesh Qs
Post by: /0 on September 08, 2009, 10:39:38 pm: Quote from: Mao on September 08, 2009, 10:37:41 pm
It sounds like they defined 'the path of the vector OA' like a line segment, and any crossing of the plane with this line segment counts as 'crossing the path of the vector OA'.

Tricky wording

Would you say that MAV is unreliable in their definitions in this case?
Title: Re: Spesh Qs
Post by: Mao on September 08, 2009, 10:52:04 pm: Quote from: /0 on September 08, 2009, 10:39:38 pm
Quote from: Mao on September 08, 2009, 10:37:41 pm
It sounds like they defined 'the path of the vector OA' like a line segment, and any crossing of the plane with this line segment counts as 'crossing the path of the vector OA'.

Tricky wording

Would you say that MAV is unreliable in their definitions in this case?

mm not really. As you've said, the 'path' of a single point is not really logical. Since OA is just a single vector, the path of OA can be interpreted as the line segment OA.

Note the difference between a path of a vector function and the path of a single vector.
Title: Re: Spesh Qs
Post by: TrueTears on September 08, 2009, 11:03:45 pm: That's what I thought, the path of OA as the line segment OA, maybe it's just a interpretation thing.
Title: Re: Spesh Qs
Post by: /0 on September 18, 2009, 01:55:43 am: "Express the roots of $z^2-2\sqrt{3}z+4=0$ in terms of $z_1$ , where $z_1 = -\sqrt{3}+i$ "

(The roots are $\sqrt{3}\pm i$ )

I did "roots are $2\sqrt{3}+z_1$ , $2\sqrt{3}+z_1-2i$ "

Answers got " $\sqrt{3}-i = -z_1$ , $\sqrt{3}+i = -\bar{z}_1$ "

Is that alright? I reckon they weren't really clear about what 'express' meant so I just took each root and expressed them in terms of $z_1$ XD
Title: Re: Spesh Qs
Post by: kamil9876 on September 18, 2009, 02:39:11 am: haha LOLcat LOL'ed

edit: 666 posts. Will not post anymore so any queries you want to ask me please do so in this thread and I will edit the post accordingly kthx

TT: 3a. Note that B is constant hence you can differentiate flux quite easily in terms of area and speed of bar. 3b. Equate F=ma to F=IDb and use the expression from part a to get rid of I. 3c. Solve the differential equation or be creative like me and find the definite integral of both sides from 0 to T (watch out for change in terminals :))
Title: Re: Spesh Qs
Post by: TrueTears on September 18, 2009, 02:43:25 am: Kamil how do you solve Q 3 a c b)
Title: Re: Spesh Qs
Post by: dekoyl on September 18, 2009, 02:58:05 am: Quote from: /0 on September 18, 2009, 01:55:43 am
Answers got " $\sqrt{3}-i = -z_1$ , $\sqrt{3}+i = -\bar{z}_1$ "

Is that alright? I reckon they weren't really clear about what 'express' meant so I just took each root and expressed them in terms of $z_1$ XD
Hmm
I just wrote $roots: -z_1, -\bar{z}_1$ for my answers and just ticked it. :S
Title: Re: Spesh Qs
Post by: /0 on September 18, 2009, 04:30:54 am: Quote from: kamil9876 on September 18, 2009, 02:39:11 am
haha LOLcat LOL'ed

edit: 666 posts. Will not post anymore so any queries you want to ask me please do so in this thread and I will edit the post accordingly kthx

TT: 3a. Note that B is constant hence you can differentiate flux quite easily in terms of area and speed of bar. 3b. Equate F=ma to F=IDb and use the expression from part a to get rid of I. 3c. Solve the differential equation or be creative like me and find the definite integral of both sides from 0 to T (watch out for change in terminals :))

you can have this thread i'll make a new one
Title: Re: Spesh Qs
Post by: kamil9876 on September 18, 2009, 10:59:33 am: LOL i was just kidding. There, 667 posts :P