Author Topic: General Complex Numbers Questions. (Read 7822 times) Tweet Share

BlueYoHo · « **Reply #15 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...?

ok thanks lol.

pHysiX · « **Reply #16 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 =]

TrueTears · « **Reply #17 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...?

ok thanks lol.

Top of page 150 of your essentials book. Check it out

BlueYoHo · « **Reply #18 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...?

ok thanks lol.
Top of page 150 of your essentials book. Check it out

Omg lol, yeah there it is.

Thanks.

EDIT:and you too physix

kamil9876 · « **Reply #19 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 $-\theta$

BlueYoHo · « **Reply #20 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 $3 + i$

TrueTears · « **Reply #21 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.

BlueYoHo · « **Reply #22 on:** April 12, 2009, 09:40:14 pm »

swweet thanks

BlueYoHo · « **Reply #23 on:** April 12, 2009, 09:42:01 pm »

By the way, please tell me I'm not hallucinating, but isn't $\sqrt{-4} = 2i$ ?

TrueTears · « **Reply #24 on:** April 12, 2009, 09:43:11 pm »

BlueYoHo · « **Reply #25 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 $2$ ?

TrueTears · « **Reply #26 on:** April 12, 2009, 09:58:27 pm »

BlueYoHo · « **Reply #27 on:** April 12, 2009, 10:01:25 pm »

omG no shit, u clever person. Thanks man.

Pizza Break now :laugh:

TrueTears · « **Reply #28 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.

Over9000 · « **Reply #29 on:** April 12, 2009, 10:03:40 pm »

Funny you should say that as I have crust cravings

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Author Topic: General Complex Numbers Questions. (Read 7822 times) Tweet Share

BlueYoHo

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