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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Damo17 on August 07, 2008, 08:15:32 pm

Title: Complex Number Help!
Post by: Damo17 on August 07, 2008, 08:15:32 pm: I'm having trouble working out one question:

If n is an even natural number show that (-1)^n/2 = i^n

(Don't know how to use latex)
Title: Re: Complex Number Help!
Post by: perfectscore on August 07, 2008, 08:17:58 pm: y are u doing complex no.s now?
Title: Re: Complex Number Help!
Post by: Mao on August 07, 2008, 08:20:04 pm: Quote from: perfectscore on August 07, 2008, 08:17:58 pm
y are u doing complex no.s now?

because he's doing unit 1/2
Title: Re: Complex Number Help!
Post by: Damo17 on August 07, 2008, 08:20:39 pm: We just started it in General Maths Advanced, but it's so basic and i've finished the chapter already so I've gone onto next years (Specialist) complex numbers chapter to pose a challenge to me.
Title: Re: Complex Number Help!
Post by: Mao on August 07, 2008, 08:21:02 pm: Quote from: Damo17 on August 07, 2008, 08:15:32 pm
I'm having trouble working out one question:

If n is an even natural number show that (-1)^n/2 = i^n

(Don't know how to use latex)

did you mean:
$(-1)^{\frac{n}{2}}=i^{n}$
Code: [Select]
[tex](-1)^{\frac{n}{2}}=i^{n}[/tex]
or
$\frac{(-1)^{n}}{2}=i^{n}$
Code: [Select]
[tex]\frac{(-1)^{n}}{2}=i^{n}[/tex]
Title: Re: Complex Number Help!
Post by: enwiabe on August 07, 2008, 08:23:06 pm: i = sqrt(-1)
i^2 = (sqrt(-1))^2 = -1

Therefore,
(-1)^(n/2) = (i^2)^(n/2) = i^(2*n/2) = i^n as required.

I, too, am latex-challenged. :P
Title: Re: Complex Number Help!
Post by: Collin Li on August 07, 2008, 08:24:32 pm: He would have meant the first one, Mao.
Title: Re: Complex Number Help!
Post by: Mao on August 07, 2008, 08:25:27 pm: Quote from: enwiabe on August 07, 2008, 08:23:06 pm
i = sqrt(-1)
i^2 = (sqrt(-1))^2 = -1

Therefore,
(-1)^(n/2) = (i^2)^(n/2) = i^(2*n/2) = i^n as required.

I, too, am latex-challenged. :P

you are a disgrace :P

he meant:
$i = \sqrt{-1}$
$i^2 = \left(\sqrt{-1}\right)^2 = -1$

Therefore,
$(-1)^{n/2} = \left(i^2\right)^{n/2} = i^{2\cdot (n/2)} = i^n$ as required.
Title: Re: Complex Number Help!
Post by: Damo17 on August 07, 2008, 08:27:00 pm: Quote from: coblin on August 07, 2008, 08:24:32 pm
He would have meant the first one, Mao.
Yeah, the first one.

Thanks, I get it now.
Title: Re: Complex Number Help!
Post by: Mao on August 07, 2008, 08:28:33 pm: on a side note, it is true that $(-1)^{n/2}=(i)^n,\; n\in \mathbb{R}$ , not only limited to even integers, since the definition of the imaginary number is $(-1)^{1/2}$
Title: Re: Complex Number Help!
Post by: Damo17 on August 07, 2008, 08:33:29 pm: Quote from: Mao on August 07, 2008, 08:28:33 pm
on a side note, it is true that $(-1)^{n/2}=(i)^n,\; n\in \mathbb{R}$ , not only limited to even integers, since the definition of the imaginary number is $(-1)^{1/2}$

Great, I'm loving learning this, it's a bit harder than unit 1/2 but so much more interesting.
Title: Re: Complex Number Help!
Post by: Damo17 on August 09, 2008, 06:38:18 pm: Another question i'm finding hard.

If $z= x+yi$ find the values of $x$ and $y$ such that $\frac{z-1}{z+1} = z+2$
Title: Re: Complex Number Help!
Post by: Mao on August 09, 2008, 06:45:57 pm: $z=x+yi$

$\implies \frac{(x+yi)-1}{(x+yi)+1}=(x+yi)+2$

$\implies (x-1)+yi=(x+1+yi)\cdot (x+2+yi)$

$\implies (x-1)+yi=(x+1)(x+2)-y^2 +(2x+3)yi$

so, equating real/imag parts:

imaginary:
$y=(2x+3)y$

$\implies 2x+3=1,\; y\neq 0$

$x=-1$

real:
$x-1=x^2+3x+2-y^2$

$\implies y^2=x^2+2x+3$

$\implies y^2=2$

$\implies y=\pm \sqrt{2}$

$\boxed{x=-1,\; y=\pm \sqrt{2}}$

$z=\left{ -1+\sqrt{2}i,\; -1-\sqrt{2}i\right}$
Title: Re: Complex Number Help!
Post by: dcc on August 09, 2008, 06:52:27 pm: Quote from: Mao on August 07, 2008, 08:25:27 pm
$i = \sqrt{-1}$
$i^2 = \left(\sqrt{-1}\right)^2 = -1$

Therefore,
$(-1)^{n/2} = \left(i^2\right)^{n/2} = i^{2\cdot (n/2)} = i^n$ as required.

$(-1)^{n/2} = \left((-i)^2\right)^{n/2} = \left(-i\right)^{2\cdot (n/2)} = (-i)^n$

Quote from: Mao on August 07, 2008, 08:28:33 pm
on a side note, it is true that $(-1)^{n/2}=(i)^n,\; n\in \mathbb{R}$ , not only limited to even integers, since the definition of the imaginary number is $(-1)^{1/2}$

$\left(-1\right)^{5/2} = i^{5} = i \neq (-i)^{5} = -i$
Title: Re: Complex Number Help!
Post by: Damo17 on August 09, 2008, 06:54:40 pm: Thanks Mao, got it now.
It all seems so easy when you show it like that, doing it so quickly.
Title: Re: Complex Number Help!
Post by: Damo17 on August 09, 2008, 07:07:33 pm: One more for the night that I gave up on yesterday.

If ${z= x+yi}$ , determine the values of $x$ and $y$ such that ${z=\sqrt{3+4i}}$ .
Title: Re: Complex Number Help!
Post by: Mao on August 09, 2008, 07:08:52 pm: Quote from: dcc on August 09, 2008, 06:52:27 pm
Quote from: Mao on August 07, 2008, 08:25:27 pm
$i = \sqrt{-1}$
$i^2 = \left(\sqrt{-1}\right)^2 = -1$

Therefore,
$(-1)^{n/2} = \left(i^2\right)^{n/2} = i^{2\cdot (n/2)} = i^n$ as required.

$(-1)^{n/2} = \left((-i)^2\right)^{n/2} = \left(-i\right)^{2\cdot (n/2)} = (-i)^n$

Quote from: Mao on August 07, 2008, 08:28:33 pm
on a side note, it is true that $(-1)^{n/2}=(i)^n,\; n\in \mathbb{R}$ , not only limited to even integers, since the definition of the imaginary number is $(-1)^{1/2}$

$\left(-1\right)^{5/2} = i^{5} = i \neq (-i)^{5} = -i$

that was enwiabe's reasoning. I personally don't agree with it that much, as $\left(a^b\right)^c=a^{(bc)}$ when b or c is a fraction is generally valid only if a is positive.

it is analogous to this: http://en.wikipedia.org/wiki/False_proof#Version_3

i would probably have done this:

$\left(-1\right)^{n/2}=\left((i)^{2}\right)^{n/2}=i^n$

but for Damo's sake, it was logical enough :P
Title: Re: Complex Number Help!
Post by: Mao on August 09, 2008, 07:09:47 pm: Quote from: Damo17 on August 09, 2008, 07:07:33 pm
One more for the night that I gave up on yesterday.

If ${z= x+yi}$ , determine the values of x and y such that ${z=\sqrt{3+4i}}$ .

$\implies z^2=3+4i$

$\implies (x+yi)^2=3+4i$

$\implies (x^2-y^2)+2xyi=3+4i$

equating real and imaginary:

$2xy=4,\; \implies y=\frac{2}{x}$

$x^2-y^2=3$

$\implies x^2-\frac{4}{x^2}=3$

$\implies x^4-3x^2-4=0,\; x\neq 0$

$\implies x^2=\frac{3\pm\sqrt{9+16}}{2}=\frac{1}{2}(3\pm 5)=4$

$\implies x=\pm 2$

$\implies y=\pm 1$
Title: Re: Complex Number Help!
Post by: Collin Li on August 09, 2008, 07:13:26 pm: $(x+yi)^2 = 3+4i$

$x^2 - y^2 + 2xyi = 3+4i$

Equating real and imaginary parts:

$x^2 - y^2 = 3$ --{1}

$2xy = 4$ --{2}

From {2}: $x = \frac{2}{y}$

Substituting into {1}: $\frac{4}{y^2} - y^2 = 3$

$\implies y^4 + 3y^2 - 4 = 0$

Using the quadratic formula with respect to $y^2$ :

$\implies y^2 = \frac{-3 \pm \sqrt{9 + 16}}{2}$

$\implies y^2 = \frac{-3 \pm 5}{2}$

$\implies y^2 = 1$ (rejecting the negative solution -- since $x$ and $y$ are real numbers)

$\implies y = \pm 1$

$\implies x = \pm 2$

$z = 2 + i$ or $z = -2 - i$
Title: Re: Complex Number Help!
Post by: Collin Li on August 09, 2008, 07:20:19 pm: When using the polar form method to obtain roots is ugly and cumbersome, sometimes cartesian expansion will do the job more efficiently (especially if its just square roots).

You can do it with higher powers to obtain some fairly cool trigonometric equalities too.
Title: Re: Complex Number Help!
Post by: Damo17 on August 09, 2008, 07:24:38 pm: Thanks Mao and Coblin. What great, helpful and kind people we have on here. Thanks alot!

Quote from: coblin on August 09, 2008, 07:20:19 pm
When using the polar form method to obtain roots is ugly and cumbersome, sometimes cartesian expansion will do the job more efficiently (especially if its just square roots).

You can do it with higher powers to obtain some fairly cool trigonometric equalities too.

I've just finished Conjugates and division of complex numbers.
I'm going to learn Polar form of Complex Numbers tomorrow and the basic operations of them.
Title: Re: Complex Number Help!
Post by: Ahmad on August 09, 2008, 08:10:46 pm: Here is another way to do it,

(http://i202.photobucket.com/albums/aa132/ahmad0/geo1-2.png)

If I can find the complex form of point E, I can readjust the length of this complex number so that it's $\sqrt{5}$ and I'll be done.

AB was extended so that angle ABE = ADC. Simple arithmetic shows that $\angle\mbox{ DBC} = 180 - 2\theta$ . Therefore $\angle\mbox{ BCD} = \theta \implies \triangle BCD$ is isosceles therefore, $\mbox{BC = BD = 3}$ .

Therefore, $\triangle\mbox{ABE}$ is similar to $\triangle\mbox{ADC}$ , this means,

$\frac{y}{y+x} = \frac{5}{5+BD} = \frac{5}{8}$ , and since $x + y = 4$ we have $x = \frac{3}{2}$

Therefore $E = 3 + \frac{3}{2} i$ adjusting to make it have length $\sqrt{5}$ gives $2 + i$ , and we mustn't forget the other negative solution, therefore $z = \pm (2 + i)$ .
Title: Re: Complex Number Help!
Post by: Umesh on August 09, 2008, 08:15:38 pm: wow
Title: Re: Complex Number Help!
Post by: bigtick on August 09, 2008, 09:54:33 pm: Quote from: Mao on August 07, 2008, 08:28:33 pm
on a side note, it is true that $(-1)^{n/2}=(i)^n,\; n\in \mathbb{R}$ , not only limited to even integers, since the definition of the imaginary number is $(-1)^{1/2}$

It is true only for even natural numbers.
Title: Re: Complex Number Help!
Post by: Mao on August 09, 2008, 10:25:10 pm: did you try using any non-even decimal numbers?

n=0.24692 works
n=-19.6 also works...
$n=2\sqrt{2}$ works also...
Title: Re: Complex Number Help!
Post by: bigtick on August 10, 2008, 08:55:26 am: I was referring to the original question. What I meant to say was, it is not true for odd natural numbers. You said it is not only limited to even integers. You have to think complex consistently. I think you have missed the essence of the question.
Title: Re: Complex Number Help!
Post by: Mao on August 10, 2008, 09:32:57 am: Quote from: bigtick on August 10, 2008, 08:55:26 am
I was referring to the original question. What I meant to say was, it is not true for odd natural numbers. You said it is not only limited to even integers. You have to think complex consistently. I think you have missed the essence of the question.

please explain how it is not true for odd natural numbers.
Title: Re: Complex Number Help!
Post by: bigtick on August 10, 2008, 10:03:51 am: There is another value for (-1)^(1/2) besides i. Think complex.
Title: Re: Complex Number Help!
Post by: Mao on August 10, 2008, 10:44:39 am: $z=(-1)^{1/2} \not\equiv z^2=-1$

the notation of $x^{1/2}$ is equivalent to $\sqrt{x}$ , which is the principle square root.

if the question did not mean the principle square root, it would have been presented as $(-1)^n=(i)^{2n}$
Title: Re: Complex Number Help!
Post by: Collin Li on August 10, 2008, 11:21:00 am: $(-1)^\frac{1}{2} = \pm i$

Problems arise for the odd powers when taking $(-1)^\frac{1}{2} = -i$
Title: Re: Complex Number Help!
Post by: dcc on August 10, 2008, 11:26:30 am: Quote from: bigtick on August 10, 2008, 10:03:51 am
There is another value for (-1)^(1/2) besides i. Think complex.

Let $a = (-1)^{\frac{n}{2}} = \exp \left(i \left( \pi + 2 \pi m \right) \right)^{\frac{n}{2}} = \exp \left( i \left(\frac{\pi n}{2} + \pi m n \right) \right),\ m \in \mathbb{Z}$

$a = \exp \left(\frac {i \pi n}{2} \right) \cdot \exp \left(i \pi m n\right) = \exp \left(i \pi k\right) \cdot \exp \left( 2 \pi m k \right)$ (since $n = 2k,\ k \in \mathbb{Z}$ )

$\begin{align*} \implies a &= \left( \cos \pi k + i \sin \pi k \right) \cdot \left( \cos 2 \pi m k + i \sin 2 \pi m k \right)\\ a &= \left(\cos \pi k + 0\right) \cdot \left(1 + 0i \right)\\ a &= \cos \pi k = \boxed{ \cos \frac{\pi n}{2} } \end{align*}$

Now, for the other side:

Let $b = i^n = \exp \left( i \left(\frac{\pi}{2} + 2 \pi j\right) \right)^n,\ j \in \mathbb{Z}$

$b = \exp \left( i \left( \frac{\pi n}{2} + 2 \pi n j \right) \right) = \exp \left( \frac{i \pi n}{2} \right) \cdot \exp \left( 2 \pi i n j \right)$

Since $n = 2t,\ t \in \mathbb{Z}$ :

$\begin{align*} b &= \left( \cos \pi t + i \sin \pi t \right) \cdot \left( \cos 4 \pi j t + i \sin 4 \pi j t \right)\\ b &= \left( \cos \pi t + 0 \right) \cdot \left(1 + 0i \right)\\ b &= \cos \pi t = \boxed{\cos \frac{\pi n}{2}}\\ \end{align*} $

So $\boxed{ \left(-1\right)^{\frac{n}{2}} = i^{n} = \cos \frac{\pi n}{2} }$ for even integers, regardless of which branch you select.
Title: Re: Complex Number Help!
Post by: Mao on August 10, 2008, 12:05:48 pm: i see four problems with this current debate:

1) we're stuck on the definition of $a^{1/2}$ , which I believe represents the principle branch, but interpreted differently by others.
the habitual thing I have learnt to do is to apply the $\pm$ when a square is removed: $a^2=b\implies a=\pm (b)^{1/2}$
as opposed to counting two branches in a square root [a principle branch evaluation in my terms]: $a=b^{1/2}\implies a=\{-\sqrt{b},\sqrt{b}\}$
*this is also convention used in mathematical computing solutions

2) failing that, $(-1)^{n/2}=(i)^n,\; n\in \mathbb{R}$ is not completely wrong, it represents a branch of solutions. the statement "only for even natural numbers" is right to the same degree as it is wrong.

3) this discussion is way beyond the scope of the question itself. one might argue that the even integers restriction is applied strictly for mathematical correctness, but I would argue that given the context, it was meant simply so that year 11 students can recognise $i^2=-1$

4) [personal opinion] I do not find the attitude and manner in which the disagreement has been voiced to be appropriate nor acceptable. In an intellectual environment such as this, it will be more constructive to refute a statement with reasoning rather than simply state "you are wrong". It does not come friendly, and I have had complaints from others regarding this. Please heed this advice.
Title: Re: Complex Number Help!
Post by: Collin Li on August 10, 2008, 04:12:22 pm: If it is to teach $i^2 = -1$ , then it can only be justified for even numbers.

There is no need to be so defensive about what you are doing. It is just worth pointing out that the definition $i$ should not be thought of as $i = \sqrt{-1}$ explicitly, but rather: $i^2 = -1$ (an implicit definition).
Title: Re: Complex Number Help!
Post by: Damo17 on August 11, 2008, 09:21:17 pm: Need help with another question my teacher gave me. Tried working out for ages, just can't get it tonight.

Find values for a and b so that $z=a+bi$ satisfies $\frac{z+i}{z+2} = i$
Title: Re: Complex Number Help!
Post by: shinny on August 11, 2008, 09:40:13 pm: $\frac{z+i} {z+2}=i$

$\Rightarrow z+i=iz+2i$

$\Rightarrow z-iz=i$

$\Rightarrow z(1-i)=i$

$\Rightarrow z=\frac{i} {1-i}$

$\Rightarrow z=\frac{i(1+i)} {(1-i)(1+i)}$

$\Rightarrow z=\frac{i+i^2} {2}$

$\Rightarrow z=\frac{i} {2} - \frac{1} {2}$

$\Rightarrow a+ib=\frac{i} {2} - \frac{1} {2}$

$\boxed{answer: a=-\frac{1} {2}\; and\; b=\frac{1} {2}}$

first time using latex so hopefully its formatted alright. someone confirm this answer?
Title: Re: Complex Number Help!
Post by: dcc on August 11, 2008, 09:42:00 pm: Quote from: Damo17 on August 11, 2008, 09:21:17 pm
Find values for a and b so that $z=a+bi$ satisfies $\frac{z+i}{z+2} = i$

$a + (b + 1)i = - b + (a + 2)i$

So clearly $a = -b$ and $b + 1 = a + 2$

combining these: $b + 1 = -b + 2 \implies \boxed{b = \frac{1}{2},\ a = - \frac{1}{2}}$

edit: whoops lol too late, but same thing
Title: Re: Complex Number Help!
Post by: Damo17 on August 11, 2008, 09:48:53 pm: Thanks, fantastic again!! :)