VCE Specialist 3/4 Question Thread!

[cosine]

i know how to solve this question. My question is, how did we know we had to use de moivres theorem. Besides the  Arg (z^5) , this can easily be mis interpreted (in my case) to raise the z value to the fifth power (this isnt an option in the MC). Most questions of this type, usually have the power outside the actual expression e.g. (2cis(pi/2)) ^2

or if z =......, what is z^2

how would you know to actually use the theorem for the whole thing? e.g. 4^5 cis (5  x ... ) ?

thanks

Good question tbh, but the question clearly states and we know that
 
So, if we have then Arg(z) =

Now, the question asks you to raise Arg(z^5) and we know from De Moivre's theorem,

 

= but Arg(z) is ALWAYS between (

So subtract 4pi and your answer is there.

[keltingmeith]

Dam EurlerFan beat me to it -.-

It's okay, young padawan - you'll get your turn someday. ;P

[Chang Feng]

EularFan101:
sorry again but could you please further clarify the questions on the attachment below relating your explanations? thank you- btw your explanation is really good. thanks agian

[Chang Feng]

and also if you don't mind. whats the difference between scalar resolute and vector resolute. is it just that scalar resolute has magnitude, whilst vector resolute has magnitude and direction. and according to textbook it seems that scalar resolute is just a further simplified version of vector resolute (regarding the formula)- is there any difference her. thanks again,

[keltingmeith]

EularFan101:
sorry again but could you please further clarify the questions on the attachment below relating your explanations? thank you- btw your explanation is really good. thanks agian

The reason for the 1/|v| is because I multipled by |v| in there so I could create a dot product. However, I can't just multiply by random numbers because I feel like it. So, I also divided by |v|, which meant that essentially all I did was multiply by 1 - which we know won't change anything.

As for why , we know that . Moving that |u| from the RHS to the LHS gives us

For your other question, the scalar resolute is simply the SIGNED magnitude of the vector resolute. By signed, I mean that if the scalar resolute is positive, the vector resolute points in the same direction as the vector you're projecting onto. If it's negative, it points in the opposite direction.

Since the vector projection () can be split into a direction () and signed magnitude (), the scalar resolute is just that signed magnitude.

[Zues]

how would i do "a" ?

thanks

[keltingmeith]

how would i do "a" ?

thanks

Consider what the angle would be at each of those points, considering that they each lie on either the positive or negative side of the axis.

[IndefatigableLover]

Can someone please let me know WHY Arg(z) must be in the range of ?
I know how to apply it, but I dont see why we are required to keep the Argument between that range and please explain it in the dumbest way, no mathematical jargon please lol. Also, if the Arg(z) is not in the range , why do we have to minus/add , is it because its our domain?

I'll let someone explain the first part since it's a bit iffy to me as well but you add/subtract because on the unit circle, a full revolution is right? So if you add/subtract by that amount, you end up in the same position making the value you had before (which was not in the given range of Arg(z)) now in the given range

[IndefatigableLover]

Yeah i understand that, but why 2pi? why not pi

If you used then it wouldn't be in the same position as your original answer making it totally different value since it's only half a revolution around the unit circle. Using keeps it in it's original position along the unit circle and gives you a new answer which is valid unlike if you add/subtract .

Hope that made sense LOL

[Zues]

Yeah i understand that, but why 2pi? why not pi

Why would you add pi?

[keltingmeith]

Can someone please let me know WHY Arg(z) must be in the range of ?

Because we defined it that way.

[IndefatigableLover]

So theres no need to understand the theory behind it bro?

Nope since if you think about it, there's nothing really to "test" about it apart from knowing the range of Arg(z) and putting your final answer in the correct form.

[mikehepro]

It's also worthwhile to note that Arg and arg are different. I'm not sure if you will lose marks if you left it in principle form when it asked for arg but just something to be aware of.

[IndefatigableLover]

It's also worthwhile to note that Arg and arg are different. I'm not sure if you will lose marks if you left it in principle form when it asked for arg but just something to be aware of.

Well Arg(z) is simply whilst (aka making sure that it's in the desired range) but yes definitely do know the difference between the two as mikehepro has pointed out!

[Conic]

So theres no need to understand the theory behind it bro?

There are reasons for defining it this way. Each point on the complex plane (other than the origin) can only be written in the form with and in one way. It would be inconvenient if people got different correct answers for questions. Also, if , then (when ), so there are other reasons for defining it that particular way.