complex nuuuuumbers (1 Viewer)

[ashokkumar]

If p is real, and -2<p<2, show that the roots of the equation x^2 + px + 1 = 0 are non-real complex numbers with modulus 1.

 

[jet]

 

[Cazic]

Supposing p < 2, we start by completing the square:



Since p is real and not equal to 2, the first equality tells us that x is non-real.

 

[jet]

Supposing p < 2, we start by completing the square:



Since p is real and not equal to 2, the first equality tells us that x is non-real.

Euler's formula is not taught in the 4 unit HSC, so I'm sure many people will have trouble understanding your working (excellent as it is).

 

[Cazic]

Ack. Why would they talk about complex numbers and not introduce the relationship between the trig and exponential functions

 

[jet]

Ack. Why would they talk about complex numbers and not introduce the relationship between the trig and exponential functions

Because, to be quite honest, they are tools. It isn't even considered in the new syllabus either.

 

[adomad]

because, to be quite honest, they are tools. It isn't even considered in the new syllabus either.

+1. new syllabus??

 

[jet]

+1. new syllabus??

The maths syllabi are changing quite soon. I think in 2011, though a teacher would have to correct me. They can be found on the Board of Studies Website. There are some new, very interesting topics which I would have loved to have completed.

http://www.boardofstudies.nsw.edu.au/syllabus_hsc/mathematics-advanced.html

 

[Cazic]

+1. new syllabus??

I'd rather learn Euler's formula on my own than have to sit through a statistics topic *shudder*. I feel dirty just typing that word.

 

[ashokkumar]

how did you change it so that the discriminant became 4 - p^2?

 

[Cazic]

He took out a factor of i².

 

[jet]

how did you change it so that the discriminant became 4 - p^2?

Factor p^2 - 4 into -1(4 - p^2) = i^2(4 - p^2)

Taking the square root of this gives i x sqrt(4 - p^2)

 

[adomad]

they look the same?? minor differences. the removed conics from 4U lol....

 

[jet]

they look the same?? minor differences. the removed conics from 4U lol....

There is a COMPLETELY new differential equations unit. There is no harder 3 unit, only inequalities.

 

[adomad]

There is a COMPLETELY new differential equations unit. There is no harder 3 unit, only inequalities.

true that

 

[ninetypercent]

such a shame that they had to remove harder 3 unit.

 

[Trebla]

I'd rather learn Euler's formula on my own than have to sit through a statistics topic *shudder*. I feel dirty just typing that word.

Statistics would be interesting if statistical modelling was introduced in the HSC course such as probability distribution and density functions (e.g. normal, gamma, poisson, geometric densities) which is much more theoretical (uses some calculus) and interesting than the tedious number crunching stereotype that statistics is often labelled with. At the moment high school statistics is boring.

Also, Euler's formula is actually in the proposed new course. I think it is in the second order ODEs topic in the Extension 2 draft.

 

[jet]

Statistics would be interesting if statistical modelling was introduced in the HSC course such as probability distribution and density functions (e.g. normal, gamma, poisson, geometric densities) which is much more theoretical (uses some calculus) and interesting than the tedious number crunching stereotype that statistics is often labelled with. At the moment high school statistics is boring.

Also, Euler's formula is actually in the proposed new course. I think it is in the second order ODEs topic in the Extension 2 draft.

On second read, you're right. I thought I had seen it there, though I couldn't find it in the complex numbers section, so I thought it wasn't there.

That does seem slightly confusing though, rather than teaching it in the context of complex numbers.

 

[elv09]

to the OP, an easier way to show that the roots have unit modulus (i.e. a modulus of 1) would be to use the relationship between roots and coefficients (sum and product of roots).

assume that one of the roots is alpha (A as i do not know how to use Latex or whatever you call it), then the other root of the quadratic would be conjugate alpha, (A~) (conjugate root theorem)

therefore, using the product of roots, A x A~ = 1.
hence using the theorem that (zz~ = |z| squared) |A|^2 = 1
and as by definition, modulus is >0. |A| = 1 ....

i could be wrong ....

 

[jet]

to the OP, an easier way to show that the roots have unit modulus (i.e. a modulus of 1) would be to use the relationship between roots and coefficients (sum and product of roots).

assume that one of the roots is alpha (A as i do not know how to use Latex or whatever you call it), then the other root of the quadratic would be conjugate alpha, (A~) (conjugate root theorem)

therefore, using the product of roots, A x A~ = 1.
hence using the theorem that (zz~ = |z| squared) |A|^2 = 1
and as by definition, modulus is >0. |A| = 1 ....

i could be wrong ....

That's perfectly right