Complex Numbers/Polynomials (1 Viewer)

[cutemouse]

Hi,

I was wondering, are solving equations of the form zⁿ=a+ib (where a,b are real and =/= 0) in the course?

Thanks

 

[Shadowdude]

I would assume so, using De Moivre's theorem I think. Although I'm not sure.

 

[gurmies]

Probably, but for n > 2 - you'd assume arctan(b/a) was workable.

 

[jet]

Probably, but for n > 2 - you'd assume arctan(b/a) was workable.

Yup. Also, for n > 2 you'd convert to mod-arg form.

 

[GUSSSSSSSSSSSSS]

a question of that form was in my trial paper this year, so i wud assume so

the method to solve has already been stated above

 

[cutemouse]

Well, just because it appeared in a trial it doesn't necessarily mean that it's in the HSC course... I'll look through past HSC papers from 1990 and see if I find such questions.

I know of course solving z^n = +/- 1 is certainly in the course though. Just not sure about z^n = a+ib.

 

[GUSSSSSSSSSSSSS]

Well, just because it appeared in a trial it doesn't necessarily mean that it's in the HSC course... I'll look through past HSC papers from 1990 and see if I find such questions.

I know of course solving z^n = +/- 1 is certainly in the course though. Just not sure about z^n = a+ib.

yerrrrr gud point

well just look thru the past papers then ..

 

[khorne]

It's not very different from solving z^2 = +/-a, so I believe it's still probably in the course.

 

[h3ll h0und]

Hi,

I was wondering, are solving equations of the form zⁿ=a+ib (where a,b are real and =/= 0) in the course?

Thanks

yep came up in my 4u assessment task today

 

[addikaye03]

yep came up in my 4u assessment task today

what was the exact Q?

 

[cutemouse]

yep came up in my 4u assessment task today

As I said, just because they put it in an assessment task, it doesn't necessarily mean it is in the course.

And yes I did go through HSC past papers from 1990 and found one (1994 or 1995 IIRC). But it had a lead in part.

So it probably is in the course. But funnily enough I don't think Cambridge goes through it, and I thought that they'd at least cover syllabus material...

 

[untouchablecuz]

As I said, just because they put it in an assessment task, it doesn't necessarily mean it is in the course.

And yes I did go through HSC past papers from 1990 and found one (1994 or 1995 IIRC). But it had a lead in part.

So it probably is in the course. But funnily enough I don't think Cambridge goes through it, and I thought that they'd at least cover syllabus material...

they do, in the polynomials topic (i think)

anyway, its just an application of de moivre's; i would think that it's an implied part of the syllabus

 

[cutemouse]

they do, in the polynomials topic (i think)

I must've missed it then... will have another look tomorrow.

anyway, its just an application of de moivre's; i would think that it's an implied part of the syllabus

Yeah but some stuff are not in the course. Like for instance partial fractions with repeated factors.

 

[Trebla]

Yeah but some stuff are not in the course. Like for instance partial fractions with repeated factors.

Even though it is not explicitly stated in the syllabus, it is still assessible in the HSC because you have all the appropriate tools to solve it, without needing anything new or learning about it explicitly. This comes under the "skills" section of the course where you are expected to apply your knowledge of a topic to unfamiliar questions. Generally these types of questions will have a lead in from different parts.

Using your example of partial fractions with repeated factors, it is actually commonly seen in HSC exams, even though the syllabus explicitly states that they are not in the course. The tricky little loophole around this is that they can ask people to do this for example:

"Find a and b such that: 2/(x + 1)(x + 2)² = a/(x + 1) + b/(x + 2) + c/(x + 2)²"

But they shouldn't ask:

"Decompose 2/(x + 1)(x + 2)² into partial fractions"

In the first question, you are GIVEN the form of the decomposition and it's merely a matter of algebra and general partial fraction skills to solve it which is perfectly doable given we already have general tools to solve partial fractions.
However, the second question requires you to actually KNOW the form of the decomposition in order to do it. This is the bit that is 'not in the syllabus' whereas the first question doesn't require you to recall such a thing.

 

[cutemouse]

Yeah thanks...

I was looking over some examples and came across one. I think the book is wrong though. Here's the question, could someone please have a look?

Solve z⁵=-4+4i

Answer: Roots of z⁵=-4+4i are where k=3, 11, 19, 27, and 35.

I reckon it should be just instead of . Could someone please confirm?

 

[jet]

Shouldn't it just be 2?

|LHS| = 4² + 4² = 16 + 16 = 32

Therefore |z| = fifth root of 32 = 2

 

[cutemouse]

Shouldn't it just be 2?

|LHS| = 4² + 4² = 16 + 16 = 32

Therefore |z| = fifth root of 32 = 2

Umm



Isn't that right? :S

Reason I'm asking is that it's from Coroneos... which is usually very correct. This would be the first error that I have picked up, that is, if it is actually an error, which is what I'm trying to confirm.

 

[jet]

OHHHHHHHH. Lol, I forgot the square root. I'm such an idiot.

You are right.

 

[h3ll h0und]

what was the exact Q?

i cant remember wat the question was exactly but it was like solve z^6 = a +ib ... a and b were integers and i cant remember them...there were much more harder questions i had to worry about =(