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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: khalil on December 24, 2008, 10:23:03 pm

Title: complex numbers
Post by: khalil on December 24, 2008, 10:23:03 pm: has any1 seen the binomial method?
ive got the heinemann text book and they have this weird way which involves n's and r's
some1 please explain
solve (3-5i)^5 using the binomial method
thanks
Title: Re: complex numbers
Post by: hard on December 24, 2008, 10:49:14 pm: looks pretty simple, just use the binomial method then to simplify and all use the i^2= -1 etc. Not too sure though so correct me anyone?
Title: Re: complex numbers
Post by: Mao on December 25, 2008, 09:23:52 am: expand it as any polynomial, but remember

$i^2 = -1,\; i^3 = -i,\; i^4 = 1,\; i^5 = i,\; etc$
Title: Re: complex numbers
Post by: khalil on December 25, 2008, 01:35:03 pm: yer but it says to use the binomial method
Title: Re: complex numbers
Post by: Damo17 on December 25, 2008, 06:23:56 pm: Quote from: khalil on December 25, 2008, 01:35:03 pm
yer but it says to use the binomial method

Then use it:

$(x+y)^{r}= x^{r}+rx^{r-1}y+\frac{r(r-1)}{2!}x^{r-2}y^2+\frac{r(r-1)(r-2)}{3!}x^{r-3}y^3+............$

Now using that:
$(3-5i)^{5}= 3^{5}+5(3)^{5-1}(-5i)+\frac{5(5-1)}{2!}(3)^{5-2}(-5i)^{2}+\frac{5(5-1)(5-2)}{3!}(3)^{5-3}(-5i)^3+ \frac{5(5-1)(5-2)(5-3)}{4!}(3)^{5-4}(-5i)^{4}+\frac{5(5-1)(5-2)(5-3)(5-4)}{5!}(3)^{5-5}(-5i)^{5}$

$= (3)^{5}+5(3^{4})(-5i)+10(3^{3})(-5i)^{2}+10(3^{2})(-5i)^{3}+5(3)(-5i)^{4}+(-5i)^{5}$

$=243-2025i-6750+11250i+9375-3125i$

$=2868+6100i$

If your not sure on your answer straight away when doing question like this, just put it in your calculator with (a+bi) set in mode and you will see if it is right.
Title: Re: complex numbers
Post by: Damo17 on December 25, 2008, 07:05:28 pm: Khalil: Doesn't the book show the binomial method (like below) and an example?

Binomial Theorem:
$(x+y)^{r}= x^{r}+rx^{r-1}y+\frac{r(r-1)}{2!}x^{r-2}y^2+\frac{r(r-1)(r-2)}{3!}x^{r-3}y^3+............$
Title: Re: complex numbers
Post by: khalil on December 26, 2008, 04:18:48 pm: Quote from: Damo17 on December 25, 2008, 07:05:28 pm
Khalil: Doesn't the book show the binomial method (like below) and an example?

Binomial Theorem:
$(x+y)^{r}= x^{r}+rx^{r-1}y+\frac{r(r-1)}{2!}x^{r-2}y^2+\frac{r(r-1)(r-2)}{3!}x^{r-3}y^3+............$

yer it does but it contains weird things like n over 0

thanks by the way
Title: Re: complex numbers
Post by: khalil on December 26, 2008, 04:24:34 pm: Can someone do this:
State the values of |z| and Arg z when z equals
-3 cos(- π/3)+ 3i sin (- π/3)

the things that look like n's are pie
Title: Re: complex numbers
Post by: ell on December 26, 2008, 04:47:21 pm: Quote from: khalil on December 26, 2008, 04:24:34 pm
Can someone do this:
State the values of |z| and Arg z when z equals
-3 cos(- π/3)+ 3i sin (- π/3)

the things that look like n's are pie

Let $z=-3\cos\left(-\frac{\pi}{3}\right)+ 3i\sin\left(-\frac{\pi}{3}\right)$

$= -3\left(\frac{1}{2}\right)+ 3i\left(-\frac{\sqrt{3}}{2}\right)$

$= -\frac{3}{2}- \frac{3\sqrt{3}}{2}i$

To find $Arg(z)$ , first determine what quadrant the complex number is in. Picture an Argand diagram - since the real part is negative and the imaginary part is also negative it is in the 3rd quadrant.

$Arg(z)=\tan^{-1} \left(\frac{3\sqrt{3}}{2}\div \frac{3}{2}\right)=\tan^{-1} \left(\sqrt{3}\right)=-\frac{2\pi}{3}$ (since the angle must be in the third quadrant)

$|z|=\sqrt{\left(-\frac{3}{2}\right)^{2}+\left(- \frac{3\sqrt{3}}{2}\right)^{2}}=\sqrt{\frac{9}{4}+\frac{27}{4}}=3$
Title: Re: complex numbers
Post by: khalil on December 26, 2008, 07:38:59 pm: Wow thanks
the method in the book is much more complex
Title: Re: complex numbers
Post by: bigtick on December 26, 2008, 08:29:47 pm: -3 cos(- π/3)+ 3i sin (- π/3) = -3 cos(π/3)- 3i sin (π/3) = -3cis(n/3) = 3cis(-2n/3)
|z| = 3, Arg(z) = -2n/3
Title: Re: complex numbers
Post by: Damo17 on December 26, 2008, 09:46:55 pm: Quote from: khalil on December 26, 2008, 04:18:48 pm
Quote from: Damo17 on December 25, 2008, 07:05:28 pm
Khalil: Doesn't the book show the binomial method (like below) and an example?

Binomial Theorem:
$(x+y)^{r}= x^{r}+rx^{r-1}y+\frac{r(r-1)}{2!}x^{r-2}y^2+\frac{r(r-1)(r-2)}{3!}x^{r-3}y^3+.........$
yer it does but it contains weird things like n over 0

thanks by the way

Yeah, there are different ways of expressing the binomial methods. The one that I did above is Newton's generalized binomial theorem. Below is the one you most likely have is your book.

$(x+y)^n=\binom{n}{0}x^{n}y^{0}+\binom{n}{1}x^{n-1}y^{1}+\binom{n}{2}x^{n-2}{y^{2}+\binom{n}{3}x^{n-3}y^{3}+.....$

what is crucial in understanding this is:
$\binom{n}{k}=\frac{n!}{k!\,(n-k)!}$

To understand better lets use an example. The question I did above.

$(3-5i)^{5}= \binom{5}{0}3^{5}(-5i)^{0}+\binom{5}{1}3^{5-1}(-5i)^{1}+\binom{5}{2}3^{5-2}(-5i)^{2}+\binom{5}{3}3^{5-3}(-5i)^{3}+\binom{5}{4}3^{5-4}(-5i)^{4}+\binom{5}{5}3^{5-5}(-5i)^{5} $
At this point, to save alot of unnecessary working out, know that $\binom{n}{0}$ and $\binom{n}{n}$ are both equal to 1.

$= 3^{5}+\frac{5!}{1!\,(5-1)!}3^{4}(-5i)+\frac{5!}{2!\,(5-2)!}3^{3}(-5i)^{2}+\frac{5!}{3!\,(5-3)!}3^{2}(-5i)^{3}+\frac{5!}{4!\,(5-4)!}3(-5i)^{4}+(-5i)^{5}$

$=3^{5}+\frac{120}{24}(3)^{4}(-5i)+\frac{120}{12}(3)^{3}(-5i)^{2}+\frac{120}{12}(3)^{2}(-5i)^{3}+\frac{120}{24}(3)(-5i)^{4}+(-5i)^{5} $

$=243+5(81)(-5i)+10(27)(-5i)^{2}+10(9)(-5i)^{3}+5(3)(-5i)^{4}+(-5i)^{5} $

$=243-2025i-6750+11250i+9375-3125i$

$=2868+6100i$

It is rather simple once you understand:
$\binom{n}{k}=\frac{n!}{k!\,(n-k)!}$
Title: Re: complex numbers
Post by: humph on December 26, 2008, 09:52:09 pm: And of course for small numbers it's much easier to remember the coefficients via Pascal's triangle (though this becomes increasingly impractical for larger numbers).
Title: Re: complex numbers
Post by: Mao on December 26, 2008, 09:56:04 pm: combinatorics are easier to remember like this:

$\binom{9}{3}=\frac{9\times 8\times 7}{1\times 2\times 3}$

on top, starting with 9, count down 3 numbers, on the bottom, starting with 1, count up 3 numbers

$\binom{49}{5}=\frac{49\times 48 \times 47 \times 46 \times 45}{1\times 2\times 3\times 4\times 5}$

makes life so much easier :)
Title: Re: complex numbers
Post by: khalil on December 26, 2008, 09:56:36 pm: Quote from: Damo17 on December 26, 2008, 09:46:55 pm
Quote from: khalil on December 26, 2008, 04:18:48 pm
Quote from: Damo17 on December 25, 2008, 07:05:28 pm
Khalil: Doesn't the book show the binomial method (like below) and an example?

Binomial Theorem:
$(x+y)^{r}= x^{r}+rx^{r-1}y+\frac{r(r-1)}{2!}x^{r-2}y^2+\frac{r(r-1)(r-2)}{3!}x^{r-3}y^3+.........$
yer it does but it contains weird things like n over 0

thanks by the way

Yeah, there are different ways of expressing the binomial methods. The one that I did above is Newton's generalized binomial theorem. Below is the one you most likely have is your book.

$(x+y)^n=\binom{n}{0}x^{n}y^{0}+\binom{n}{1}x^{n-1}y^{1}+\binom{n}{2}x^{n-2}{y^{2}+\binom{n}{3}x^{n-3}y^{3}+.....$

what is crucial in understanding this is:
$\binom{n}{k}=\frac{n!}{k!\,(n-k)!}$

To understand better lets use an example. The question I did above.

$(3-5i)^{5}= \binom{5}{0}3^{5}(-5i)^{0}+\binom{5}{1}3^{5-1}(-5i)^{1}+\binom{5}{2}3^{5-2}(-5i)^{2}+\binom{5}{3}3^{5-3}(-5i)^{3}+\binom{5}{4}3^{5-4}(-5i)^{4}+\binom{5}{5}3^{5-5}(-5i)^{5} $
At this point, to save alot of unnecessary working out, know that $\binom{n}{0}$ and $\binom{n}{n}$ are both equal to 1.

$= 3^{5}+\frac{5!}{1!\,(5-1)!}3^{4}(-5i)+\frac{5!}{2!\,(5-2)!}3^{3}(-5i)^{2}+\frac{5!}{3!\,(5-3)!}3^{2}(-5i)^{3}+\frac{5!}{4!\,(5-4)!}3(-5i)^{4}+(-5i)^{5}$

$=3^{5}+\frac{120}{24}(3)^{4}(-5i)+\frac{120}{12}(3)^{3}(-5i)^{2}+\frac{120}{12}(3)^{2}(-5i)^{3}+\frac{120}{24}(3)(-5i)^{4}+(-5i)^{5} $

$=243+5(81)(-5i)+10(27)(-5i)^{2}+10(9)(-5i)^{3}+5(3)(-5i)^{4}+(-5i)^{5} $

$=243-2025i-6750+11250i+9375-3125i$

$=2868+6100i$

It is rather simple once you understand:
$\binom{n}{k}=\frac{n!}{k!\,(n-k)!}$
thanks for ur help....i havent done factorials in yr 11 so my understanding of them isnt so great
do u have to use the binomial method a lot ?
Title: Re: complex numbers
Post by: humph on December 26, 2008, 09:58:11 pm: Not much in spec, but moreso in methods. But it's a technique you're expected to know.
Title: Re: complex numbers
Post by: Damo17 on December 26, 2008, 09:59:11 pm: Quote from: humph on December 26, 2008, 09:52:09 pm
And of course for small numbers it's much easier to remember the coefficients via Pascal's triangle (though this becomes increasingly impractical for larger numbers).

Well yes, but in a way if it was even $(x+y)^{25}$ I would think that pascal's triangle would be quicker. It would maybe take 5 min to draw it on another page. However it would depend on whether you are using a calculator or not. If so then using $\binom{n}{k}=\frac{n!}{k!\,(n-k)!}$ with a calculator would be quicker.

Quote from: Mao on December 26, 2008, 09:56:04 pm
combinatorics are easier to remember like this:

$\binom{9}{3}=\frac{9\times 8\times 7}{1\times 2\times 3}$

on top, starting with 9, count down 3 numbers, on the bottom, starting with 1, count up 3 numbers

$\binom{49}{5}=\frac{49\times 48 \times 47 \times 46 \times 45}{1\times 2\times 3\times 4\times 5}$

makes life so much easier :)

That is sooo much quicker. I don't think we did that this year.
Title: Re: complex numbers
Post by: hard on December 26, 2008, 10:00:53 pm: Quote from: humph on December 26, 2008, 09:52:09 pm
And of course for small numbers it's much easier to remember the coefficients via Pascal's triangle (though this becomes increasingly impractical for larger numbers).
i still don't know any other method other than pascal's triangle
Title: Re: complex numbers
Post by: humph on December 26, 2008, 10:01:48 pm: Quote from: Damo17 on December 26, 2008, 09:59:11 pm
Quote from: humph on December 26, 2008, 09:52:09 pm
And of course for small numbers it's much easier to remember the coefficients via Pascal's triangle (though this becomes increasingly impractical for larger numbers).

Well yes, but in a way if it was even $(x+y)^{25}$ I would think that pascal's triangle would be quicker. It would maybe take 5 min to draw it on another page. However it would depend on whether you are using a calculator or not. If so then using $\binom{n}{k}=\frac{n!}{k!\,(n-k)!}$ with a calculator would be quicker.

Quote from: Mao on December 26, 2008, 09:56:04 pm
combinatorics are easier to remember like this:

$\binom{9}{3}=\frac{9\times 8\times 7}{1\times 2\times 3}$

on top, starting with 9, count down 3 numbers, on the bottom, starting with 1, count up 3 numbers

$\binom{49}{5}=\frac{49\times 48 \times 47 \times 46 \times 45}{1\times 2\times 3\times 4\times 5}$

makes life so much easier :)

That is sooo much quicker. I don't think we did that this year.
Yeah I meant that even with a calculator I find it easier just to use Pascal's triangle for polynomials up to degree six or so (mainly because I pretty much remember all of the coefficients off the top of my head).
Title: Re: complex numbers
Post by: hard on December 26, 2008, 10:03:03 pm: Quote from: Damo17 on December 26, 2008, 09:59:11 pm
Quote from: humph on December 26, 2008, 09:52:09 pm
And of course for small numbers it's much easier to remember the coefficients via Pascal's triangle (though this becomes increasingly impractical for larger numbers).

Well yes, but in a way if it was even $(x+y)^{25}$ I would think that pascal's triangle would be quicker. It would maybe take 5 min to draw it on another page. However it would depend on whether you are using a calculator or not. If so then using $\binom{n}{k}=\frac{n!}{k!\,(n-k)!}$ with a calculator would be quicker.

Quote from: Mao on December 26, 2008, 09:56:04 pm
combinatorics are easier to remember like this:

$\binom{9}{3}=\frac{9\times 8\times 7}{1\times 2\times 3}$

on top, starting with 9, count down 3 numbers, on the bottom, starting with 1, count up 3 numbers

$\binom{49}{5}=\frac{49\times 48 \times 47 \times 46 \times 45}{1\times 2\times 3\times 4\times 5}$

makes life so much easier :)

That is sooo much quicker. I don't think we did that this year.
would this be another method as opposed to pascal's triangle? This looks quite simple and quick to use.
Title: Re: complex numbers
Post by: Damo17 on December 26, 2008, 10:05:39 pm: ^ you would use that when using binomial theorem. you use that instead of big factorials.
Title: Re: complex numbers
Post by: humph on December 26, 2008, 10:09:47 pm: Quote from: Damo17 on December 26, 2008, 10:05:39 pm
^ you would use that when using binomial theorem. you use that instead of big factorials.
I think hard is saying that he only ever learnt how to calculate combinatorics via Pascal's triangle, as opposed to its formal definition via factorials.

I usually just use factorials as I usually only ever need binomials when dealing with polynomials of degree n, in which case you can't really use Pascal's triangle, and Mao's method is actually less concise.
Title: Re: complex numbers
Post by: hard on December 26, 2008, 10:13:16 pm: Quote from: humph on December 26, 2008, 10:09:47 pm
Quote from: Damo17 on December 26, 2008, 10:05:39 pm
^ you would use that when using binomial theorem. you use that instead of big factorials.
I think hard is saying that he only ever learnt how to calculate combinatorics via Pascal's triangle, as opposed to its formal definition via factorials.

I usually just use factorials as I usually only ever need binomials when dealing with polynomials of degree n, in which case you can't really use Pascal's triangle, and Mao's method is actually less concise.
oh yer fair enough.