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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: BlueYoHo on April 10, 2009, 06:05:33 pm

Title: General Complex Numbers Questions.
Post by: BlueYoHo on April 10, 2009, 06:05:33 pm: Alright I really suck at complex numbers. It's mostly because I only did half of the spec course last year and this wasn't a topic that I covered.
But enough of making excuses...

Q3.
Show that if $\frac{-\pi}{2} < Arg (z1) < \frac{\pi}{2}$ and $\frac{-\pi}{2} < Arg (z2) < \frac{\pi}{2}$ then

$Arg(z1 z2) = Arg(z1) + Arg(z2)$ and $Arg(\frac{z1}{z2}) = Arg(z1) - Arg(z2)$

I don't know how to do it really... I tried letting z1 = a + bi and z2 = c + di... yeah but I'm not sure if that gets me anywhere. Please Help :)

Q5.
Show that $sin\theta + cos\theta i = cis(\frac{\pi}{2} - \theta)$

One question, with this question: When it says something like show blah = rah am I allowed, instead, to show that rah = blah if that makes sense? If so I think I might be able to do the question...
Title: Re: General Complex Numbers Questions.
Post by: TrueTears on April 10, 2009, 06:07:19 pm: Q5

$sin(\theta) = cos(\frac{\pi}{2} - \theta)$

$cos(\theta) = sin(\frac{\pi}{2} - \theta)$

subbing in yields:

$cos(\frac{\pi}{2} - \theta) + sin(\frac{\pi}{2} - \theta)i = cis(\frac{\pi}{2} - \theta)$
Title: Re: General Complex Numbers Questions.
Post by: BlueYoHo on April 10, 2009, 06:16:38 pm: Quote from: TrueTears on April 10, 2009, 06:07:19 pm
$sin(\theta) = cos(\frac{\pi}{2} - \theta)$

$cos(\theta) = sin(\frac{\pi}{2} - \theta)$

Are these like things that we just have to know? Or is there a way you got there?

Also, the question says
$sin\theta + cos\theta i$
Does the $i$ multiplied by the $cos\theta$ do a difference? Because in " $cis$ " isn't it $cos\theta + i sin\theta$ ? With the $i$ multiplied with the $sin\theta$ instead.
Title: Re: General Complex Numbers Questions.
Post by: TrueTears on April 10, 2009, 06:19:26 pm: Yes you should know those complementary formulas.

Think about the question this way:

Let $\frac{\pi}{2} - \theta = X$

$sin(\theta) = cos(X)$

and $cos(\theta) = sin(X)$

subbing this in $sin(\theta) + cos(\theta)i$ yields

$cos(X) + sin(X)i = cis(X)$

what is X? $X = \frac{\pi}{2} - \theta$

sub this in yields $cis(\frac{\pi}{2} - \theta)$
Title: Re: General Complex Numbers Questions.
Post by: BlueYoHo on April 10, 2009, 06:20:29 pm: Oh lol no shit. thanks.
Title: Re: General Complex Numbers Questions.
Post by: BlueYoHo on April 10, 2009, 06:24:51 pm: Ok what about these one:

Simplify
a) $(sin\theta + cos\theta i)^7$
b) $(sin\theta + cos\theta i)(sin\phi + cos\phi i)$

Just want to see what you do and I'll try do the rest myself.
Title: Re: General Complex Numbers Questions.
Post by: TrueTears on April 10, 2009, 06:28:06 pm: a) since $sin(\theta) + cos(\theta)i = cis(\frac{\pi}{2} - \theta)$

$(cis(\frac{\pi}{2} - \theta))^7 = cis(7 \times (\frac{\pi}{2} - \theta))$

$= cis(\frac{7\pi}{2} - 7\theta) = cis(\frac{3\pi}{2} - 7\theta)$ ( $\frac{7\pi}{2}$ is the same as $\frac{3\pi}{2}$ just minus $2\pi$ )
Title: Re: General Complex Numbers Questions.
Post by: BlueYoHo on April 10, 2009, 06:28:58 pm: Dam you rock so hard. Thanks man. Ima try the other one now.
Title: Re: General Complex Numbers Questions.
Post by: TrueTears on April 10, 2009, 06:31:57 pm: b) $sin(\phi) + cos(\phi)i = cis(\frac{\pi}{2} - \phi)$ (same principles as how to derive the other one)

so $(cis(\frac{\pi}{2} - \theta)) \times (cis(\frac{\pi}{2} - \phi))$ $= Cis( \frac{\pi}{2} - \theta + \frac{\pi}{2} - \phi) = Cis(\pi - \theta - \phi)$
Title: Re: General Complex Numbers Questions.
Post by: BlueYoHo on April 10, 2009, 06:32:38 pm: Yeah. Mad thanks a lot man.
Title: Re: General Complex Numbers Questions.
Post by: Mao on April 10, 2009, 07:23:20 pm: Quote from: BlueYoHo on April 10, 2009, 06:16:38 pm
Quote from: TrueTears on April 10, 2009, 06:07:19 pm
$sin(\theta) = cos(\frac{\pi}{2} - \theta)$

$cos(\theta) = sin(\frac{\pi}{2} - \theta)$

Are these like things that we just have to know? Or is there a way you got there?

Construct a right angle triangle. One of the angle is $\theta$ , since the sum of angles add up to 180 degrees, the other angle must be $90^o - \theta = \frac{\pi}{2}-\theta$

e.g. for sine, $\sin \theta = \frac{O}{H}$ , the opposide side to $\theta$ is adjacent to $\frac{\pi}{2}-\theta$ , i.e. $\cos \left( \frac{\pi}{2} - \theta \right) = \frac{O}{H}$ , hence $\sin (\theta) = \cos \left( \frac{\pi}{2} - \theta \right)$

similarly, $\cos (\theta) = \sin \left( \frac{\pi}{2} - \theta \right)$ , $\tan (\theta) = \cot \left( \frac{\pi}{2} - \theta \right)$ , $\cot (\theta) = \tan \left( \frac{\pi}{2} - \theta \right)$
Title: Re: General Complex Numbers Questions.
Post by: BlueYoHo on April 10, 2009, 07:25:14 pm: Sorry, I got one more question...

I got the $sin\theta + icos\theta = cis(\frac{\pi}{2} - \theta)$ part, BUT:
Prove
$sin\theta - icos\theta = cis(\theta - \frac{\pi}{2})$ ???

$sin\theta = cos(\frac{\pi}{2} - \theta)$
$cos\theta = sin(\frac{\pi}{2} - \theta)$

$cos(\frac{\pi}{2} - \theta) - sin(\frac{\pi}{2} - \theta)$

And I don't know how to go further
Title: Re: General Complex Numbers Questions.
Post by: BlueYoHo on April 10, 2009, 07:26:45 pm: Quote from: Mao on April 10, 2009, 07:23:20 pm
Quote from: BlueYoHo on April 10, 2009, 06:16:38 pm
Quote from: TrueTears on April 10, 2009, 06:07:19 pm
$sin(\theta) = cos(\frac{\pi}{2} - \theta)$

$cos(\theta) = sin(\frac{\pi}{2} - \theta)$

Are these like things that we just have to know? Or is there a way you got there?

Construct a right angle triangle. One of the angle is $\theta$ , since the sum of angles add up to 180 degrees, the other angle must be $90^o - \theta = \frac{\pi}{2}-\theta$

e.g. for sine, $\sin \theta = \frac{O}{H}$ , the opposide side to $\theta$ is adjacent to $\frac{\pi}{2}-\theta$ , i.e. $\cos \left( \frac{\pi}{2} - \theta \right) = \frac{O}{H}$ , hence $\sin (\theta) = \cos \left( \frac{\pi}{2} - \theta \right)$

similarly, $\cos (\theta) = \sin \left( \frac{\pi}{2} - \theta \right)$ , $\tan (\theta) = \cot \left( \frac{\pi}{2} - \theta \right)$ , $\cot (\theta) = \tan \left( \frac{\pi}{2} - \theta \right)$

Oh ok I see that. Clever. Thanks Mao.
Title: Re: General Complex Numbers Questions.
Post by: pHysiX on April 10, 2009, 07:30:26 pm: Use a quadrant one "theta" for simplicity:

$- \sin (\frac{\pi }{2} - \theta )$ is equivalent to $\sin (\theta - \frac{\pi }{2})$ since you'll move into a negative quadrant

Also, $\cos (\theta - \frac{\pi }{2})$ is a quadrant 4, hence it's the same as $\cos (\frac{\pi }{2} - \theta )$ [both positive]

Hence, $cis(\theta - \frac{\pi }{2})$
=]
Title: Re: General Complex Numbers Questions.
Post by: TrueTears on April 10, 2009, 07:32:26 pm: Quote from: BlueYoHo on April 10, 2009, 07:25:14 pm
Sorry, I got one more question...

I got the $sin\theta + icos\theta = cis(\frac{\pi}{2} - \theta)$ part, BUT:
Prove
$sin\theta - icos\theta = cis(\theta - \frac{\pi}{2})$ ???

$sin\theta = cos(\frac{\pi}{2} - \theta)$
$cos\theta = sin(\frac{\pi}{2} - \theta)$

$cos(\frac{\pi}{2} - \theta) - sin(\frac{\pi}{2} - \theta)$

And I don't know how to go further

$sin(\theta) -icos(\theta)$ is just $sin(\theta) + icos(\theta)$ conjugate.

so we know the conjugate of $Cis(X) = Cis(-X)$

$sin(\theta) +icos(\theta) = Cis(\frac{\pi}{2} - \theta)$

so $Cis(-X) = Cis(-(\frac{\pi}{2} - \theta)) = Cis(\theta - \frac{\pi}{2})$

Think of it like... let $a = sin(\theta)$ and $b = cos(\theta)$

a + ib conjugate is a - ib

sub a and b back in yields

$sin(\theta) -icos(\theta)$
Title: Re: General Complex Numbers Questions.
Post by: BlueYoHo on April 10, 2009, 07:37:28 pm: Quote from: TrueTears on April 10, 2009, 07:32:26 pm
so we know the conjugate of $Cis(X) = Cis(-X)$

:-[ Oh did we know that...? :P

ok thanks lol.
Title: Re: General Complex Numbers Questions.
Post by: pHysiX on April 10, 2009, 07:37:35 pm: Quote from: TrueTears on April 10, 2009, 07:32:26 pm
Quote from: BlueYoHo on April 10, 2009, 07:25:14 pm
Sorry, I got one more question...

I got the $sin\theta + icos\theta = cis(\frac{\pi}{2} - \theta)$ part, BUT:
Prove
$sin\theta - icos\theta = cis(\theta - \frac{\pi}{2})$ ???

$sin\theta = cos(\frac{\pi}{2} - \theta)$
$cos\theta = sin(\frac{\pi}{2} - \theta)$

$cos(\frac{\pi}{2} - \theta) - sin(\frac{\pi}{2} - \theta)$

And I don't know how to go further

$sin(\theta) -icos(\theta)$ is just $sin(\theta) + icos(\theta)$ conjugate.

so we know the conjugate of $Cis(X) = Cis(-X)$

$sin(\theta) +icos(\theta) = Cis(\frac{\pi}{2} - \theta)$

so $Cis(-X) = Cis(-(\frac{\pi}{2} - \theta)) = Cis(\theta - \frac{\pi}{2})$

Think of it like... let $a = sin(\theta)$ and $b = cos(\theta)$

a + ib conjugate is a - ib

sub a and b back in yields

$sin(\theta) -icos(\theta)$

hmm that's pro but thing is, u are saying that "the conjugate of cis(theta)" is equal to cis(theta)

haha is that allowed? just wondering =]
Title: Re: General Complex Numbers Questions.
Post by: TrueTears on April 10, 2009, 07:39:12 pm: Quote from: BlueYoHo on April 10, 2009, 07:37:28 pm
Quote from: TrueTears on April 10, 2009, 07:32:26 pm
so we know the conjugate of $Cis(X) = Cis(-X)$

:-[ Oh did we know that...? :P

ok thanks lol.
Top of page 150 of your essentials book. Check it out :P
Title: Re: General Complex Numbers Questions.
Post by: BlueYoHo on April 10, 2009, 07:47:01 pm: Quote from: TrueTears on April 10, 2009, 07:39:12 pm
Quote from: BlueYoHo on April 10, 2009, 07:37:28 pm
Quote from: TrueTears on April 10, 2009, 07:32:26 pm
so we know the conjugate of $Cis(X) = Cis(-X)$

:-[ Oh did we know that...? :P

ok thanks lol.
Top of page 150 of your essentials book. Check it out :P
Omg lol, yeah there it is. :P Thanks.

EDIT:and you too physix :)
Title: Re: General Complex Numbers Questions.
Post by: kamil9876 on April 10, 2009, 08:59:56 pm: for Q5 in first post you could also say:

$RHS= cis(\frac{\pi}{2}) cis (-\theta)$
$= i cis(-\theta)$
$= i (cos(-\theta) + i sin(-\theta))$
$=i (cos(\theta) -i sin(\theta))$
$=i cos(\theta) -i^2 sin(\theta)$
$=LHS$

Which is pretty neat since it allows u to solve a larger class of problems. i.e: if there was a pi/4 instead of pi/2. or if it was $+\theta$ rather than $-\theta$
Title: Re: General Complex Numbers Questions.
Post by: BlueYoHo on April 12, 2009, 09:33:01 pm: This question isn't working for me. I'm not sure if I'm approaching it right...:
Solve for z
$z^2 - (6 + 2i)z + (8 + 6i) = 0$

I first tried the quadratic formula which got really messy and confusing, so then I thought maybe I should complete the square?
I have a feeling my working out is wrong but here it is anyway:

$z^2 - (6 + 2i)z + (8 + 6i) = 0$

$[z^2 - (6 + 2i)z + (3 + i)^2] - (3 + i)^2 + (8 + 6i) = 0$

$[(z - 3 + i)^2] - (9 - 2i + i^2) + (8 + 6i) = 0$

$[(z - 3 + i)^2] + 16 = 0$

$[(z - 3 + i)^2] - (4i)^2 = 0$

So should the solution be $3 - i \pm 4i$ or something like that?

But the answer is just $3 + i$ ???
Title: Re: General Complex Numbers Questions.
Post by: TrueTears on April 12, 2009, 09:35:45 pm: $z^2 - (6+2i)z + (8+6i) = 0$

$z^2 - (6+2i)z + (\frac{(6+2i)}{2})^2 - (\frac{(6+2i)}{2})^2 + (8+6i) = 0$

$(z - (\frac{(6+2i)}{2}))^2 = 0$ (Note $-(\frac{(6+2i)}{2})^2 + (8+6i) = 0$ , which is handy :) )

$z = \frac{6+2i}{2}$

$z = 3 + i$

Note: Your mistake lies in 2 places, first it is $(z - 3 - i)^2$

and when you expanded $-(3+i)^2$ it is meant to be $-(9+6i+i^2) = (8+6i)$ so then $-(8+6i) + (8+6i)$ cancels out.

Your method is perfect.
Title: Re: General Complex Numbers Questions.
Post by: BlueYoHo on April 12, 2009, 09:40:14 pm: swweet thanks
Title: Re: General Complex Numbers Questions.
Post by: BlueYoHo on April 12, 2009, 09:42:01 pm: By the way, please tell me I'm not hallucinating, but isn't $\sqrt{-4} = 2i$ ?
Title: Re: General Complex Numbers Questions.
Post by: TrueTears on April 12, 2009, 09:43:11 pm: $\sqrt{-4} = \sqrt{-1} \times \sqrt{4} = i \times 2 = 2i$
Title: Re: General Complex Numbers Questions.
Post by: BlueYoHo on April 12, 2009, 09:55:45 pm: Yeah well then I have no idea why the following question isn't working for me.

Given that $1 - i$ is a solution to the equation $z^3 - 4z^2 + 6z - 4 = 0$ find the other two solutions
One of them is the conjugate thing so its just $1 + i$
And to find the other two you go:

$(z + 1 + i)(z + 1 - i) = z^2 + z - zi + z + 1 - i + zi + i - i^2$
$= z^2 + 2z + 2$

Then completing the square:
$(z^2 + 2z + 2)$
$(z + 1)^2 + 1$
$(z + 1)^2 - (i)^2$

So the three solutions should be: $(1 + i)$ , $(1 - i)$ , $(-1 \pm i)$

But for some reason B.O.B says its $2$ ?
Title: Re: General Complex Numbers Questions.
Post by: TrueTears on April 12, 2009, 09:58:27 pm: $(z - (1-i))(z-(1+i)) = (z - 1 + i)(z - 1 - i)$
Title: Re: General Complex Numbers Questions.
Post by: BlueYoHo on April 12, 2009, 10:01:25 pm: omG no shit, u clever person. Thanks man.

Pizza Break now :laugh:
Title: Re: General Complex Numbers Questions.
Post by: TrueTears on April 12, 2009, 10:03:03 pm: hahaha pizza break, what kind you having? Ever tried Crust pizzas omfg they are the best pizza ever.
Title: Re: General Complex Numbers Questions.
Post by: Over9000 on April 12, 2009, 10:03:40 pm: Funny you should say that as I have crust cravings
Title: Re: General Complex Numbers Questions.
Post by: BlueYoHo on April 12, 2009, 10:09:32 pm: Lol nah haven't. Dad owns pizza shop and he's come home early with some pizza :D

I'm sort of sick of it but I haven't had it for nearly 4 days now :P so i felt like some today.
Title: Re: General Complex Numbers Questions.
Post by: BlueYoHo on April 12, 2009, 10:10:56 pm: It's a pepperoni today. :)
Title: Re: General Complex Numbers Questions.
Post by: TrueTears on April 12, 2009, 10:11:25 pm: Quote from: BlueYoHo on April 12, 2009, 10:10:56 pm
It's a pepperoni today. :)
yummmm you sure are making me hungry all of a sudden Lols
Title: Re: General Complex Numbers Questions.
Post by: bigtick on April 12, 2009, 11:07:17 pm: (http://img15.imageshack.us/img15/8901/missingpiece.gif)
Title: Re: General Complex Numbers Questions.
Post by: Mao on April 13, 2009, 12:24:37 am: Quote from: bigtick on April 12, 2009, 11:07:17 pm
(http://img15.imageshack.us/img15/8901/missingpiece.gif)
There it is:

$\frac{\pi}{4} = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}$