Author Topic: General Complex Numbers Questions. (Read 7826 times) Tweet Share

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...

TrueTears · « **Reply #1 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:

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

BlueYoHo · « **Reply #2 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.

TrueTears · « **Reply #3 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 $cis(\frac{\pi}{2} - \theta)$

BlueYoHo · « **Reply #4 on:** April 10, 2009, 06:20:29 pm »

Oh lol no shit. thanks.

BlueYoHo · « **Reply #5 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.

TrueTears · « **Reply #6 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 $2\pi$ )

BlueYoHo · « **Reply #7 on:** April 10, 2009, 06:28:58 pm »

Dam you rock so hard. Thanks man. Ima try the other one now.

TrueTears · « **Reply #8 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 $(cis(\frac{\pi}{2} - \theta)) \times (cis(\frac{\pi}{2} - \phi))$ $= Cis( \frac{\pi}{2} - \theta + \frac{\pi}{2} - \phi) = Cis(\pi - \theta - \phi)$

BlueYoHo · « **Reply #9 on:** April 10, 2009, 06:32:38 pm »

Yeah. Mad thanks a lot man.

Mao · « **Reply #10 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, $\cos (\theta) = \sin \left( \frac{\pi}{2} - \theta \right)$ , $\tan (\theta) = \cot \left( \frac{\pi}{2} - \theta \right)$ , $\cot (\theta) = \tan \left( \frac{\pi}{2} - \theta \right)$

BlueYoHo · « **Reply #11 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

BlueYoHo · « **Reply #12 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.

pHysiX · « **Reply #13 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, $cis(\theta - \frac{\pi }{2})$
=]

TrueTears · « **Reply #14 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

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

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Author Topic: General Complex Numbers Questions. (Read 7826 times) Tweet Share

BlueYoHo

General Complex Numbers Questions.

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