I know I'll regret this, but... (1 Viewer)

[spice girl]

Now, for any 2003 year 12's who are doing extension 2 next year:

From my experience, a lot of ppl get stumped on one or two particular questions and then they get discouraged and enter some chronic depression state.

So, if that happens to any of you, feel free to post any extension 2 (or probability) questions here as single threads, beginning with "Q:", so we know it's a question.

If med studies allow, I'll answer them ASAP. I'm sure other ex-HSC ppl won't mind having a crack at them too.

Knowing more than one way to solve a problem is good extension 2 study technique.

Make sure it's sufficiently non-trivial otherwise you'll be wasting everyone's time.

 

[gronkboyslim]

cool

 

[Ozz^E]

Originally posted by gronkboyslim
cool

Yea...dont hesitate to ask questions. Though I wont gaurantee knowing how to answer everything..well give a sincere attempt.

 

[kini mini]

Good idea, I'll have a go if anyone wants me to as well

 

[Dashpreet]

Try this one...

Cant seem to figure this one out...

Q: Express [1 + i] and [1 - i] in modulus/argument form. Use Moivre's theorem to evaluate --->

(1 + i) to the power of 20 + (1 - i) to the power of 20.

Take a crack at it...

 

[spice girl]

Re: Try this one...

Originally posted by Dashpreet
Cant seem to figure this one out...

Q: Express [1 + i] and [1 - i] in modulus/argument form. Use Moivre's theorem to evaluate --->

(1 + i) to the power of 20 + (1 - i) to the power of 20.

Take a crack at it...

firstly, it's De-moivre's thm, not Moivre's

mod-arg form means find modulus and argument

by plotting 1+i and 1-i (there's heaps of other ways to figure argument out, the most common being figuring out the quadrant, and then applying tan-1), you get arg(1+i) = pi/4, arg(1-i) = -pi/4

|x + iy| = sqrt(x^2 + y^2)
so |1 + i| = |1 - i| = sqrt(2)

so (1 + i) = sqrt(2)cis(pi/4)
(1 - i) = sqrt(2)cis(-pi/4)

using de-moivre's, (1 + i)^20 = sqrt(2)^20 cis (20pi/4) = 2^10 cis(pi) = -2^10

(1-i)^20 = sqrt(2)^20 cis (-20pi/4) = -2^10 cis(pi) = -2^10

(1 + i)^20 + (1-i)^20 = -2^10 + -2^10 = -2^11 = -2048

 

[spice girl]

and btw, use separate threads, not posts to submit new questions

i dun want this thread clogged with several questions at a time

 

[flyin']

wow free tutoring ... oh wait ive done ext2 maths already (2002) ... oh well ...

 

[BlackJack]

spice girl and the rest of us have already given free tutoring all the way since this forum started...
An alternative to paying tutors.

I'll get my mind working if I haven't already forgotten all of them by next year...

 

[spice girl]

hmm...passive tutoring maybe

we dun go out of the way to deliberately teach something

 

[drolle]

Originally posted by spice girl
...I'm sure other ex-HSC ppl won't mind having a crack at them too...

Yes please! I live on 4u maths questions!

well.... maybe not... but I'd be happy to help answer problems, and helping other people is pretty much all I ever did in 2u and about 50% of 3u anyway, and I find I quite enjoy it.

I doubt there'll be any problems that BJ or spice girl or the other super-people can't handle, so I'll just watch and point out alternative solutions if I see any.

I'm inclined to think that there's always a faster solution, it just takes twice as long to find!

 

[McLake]

Originally posted by spice girl
hmm...passive tutoring maybe

we dun go out of the way to deliberately teach something

Yes, try not to put teachers out of a job.