Author Topic: TrueTears question thread (Read 67351 times) Tweet Share

TrueTears · « **Reply #300 on:** March 10, 2009, 08:20:14 pm »

$sin(t + g) = k \times cos(t - g)$

show that $tan(t) = \frac{k - tan(g)}{1 - k \times tan(g)}$

Mao · « **Reply #301 on:** March 10, 2009, 08:39:46 pm »

$\sin (t+g) = k \cos (t-g)$

$\sin t \cos g + \sin g \cos t = k (\cos t \cos g + \sin t \sin g)$

$\sin t (\cos g - k \sin g) = \cos t (k\cos g - \sin g)$

$\frac{\sin t}{\cos t} = \frac{k \cos g - \sin g}{\cos g - k \sin g}$

$\tan t = \frac{k \cos g - \sin g}{\cos g - k\sin g} \times \frac{\sec g}{\sec g}$

$\tan t = \frac{k - \tan g}{1 - k\tan g}$

as required

TrueTears · « **Reply #302 on:** March 10, 2009, 08:42:57 pm »

nice haha

TrueTears · « **Reply #303 on:** March 16, 2009, 06:57:21 pm »

Just a quick question regarding complex conjugates

say you have $z = -7 + 3i$

now the conjugate here would be $-7 - 3i$ right?

but if you have $+7 + 3i$ , when this is multiplied with $z$ it also gives a real number

ie, $(-7+3i)(7+3i) = -49 - 21i + 21i + 9i^2 = -49-9 = -58$

however... the conjugate i always thought would just be -7-3i,

ie $(-7+3i)(-7-3i) = 49 +21i - 21i-9i^2 = 49+9 = 58$

Now I'm wondering, does this mean $z$ has 2 conjugates? I mean one yields a positive answer and the other is a negative, but they are still different numbers...

Thanks guys

EDIT: I always thought each complex number had only ONE conjugate?

Damo17 · « **Reply #304 on:** March 16, 2009, 07:10:12 pm »

Quote from: TrueTears on March 16, 2009, 06:57:21 pm

Just a quick question regarding complex conjugates

say you have $z = -7 + 3i$

now the conjugate here would be $-7 - 3i$ right?

but if you have $+7 + 3i$ , when this is multiplied with $z$ it also gives a real number

ie, $(-7+3i)(7+3i) = -49 - 21i + 21i + 9i^2 = -49-9 = -58$

however... the conjugate i always thought would just be -7-3i,

ie $(-7+3i)(-7-3i) = 49 +21i - 21i-9i^2 = 49+9 = 58$

Now I'm wondering, does this mean $z$ has 2 conjugates? I mean one yields a positive answer and the other is a negative, but they are still different numbers...

Thanks guys

EDIT: I always thought each complex number had only ONE conjugate?

Well no, z can only have 1 conjugate as the conjugate signifies that there is "imaginary refection" meaning the same size but opposite angle.

Well this is what I think.

TrueTears · « **Reply #305 on:** March 16, 2009, 07:13:11 pm »

ohh nice thanks Damo17, so basically a conjugate is just the imaginary part reflected in the x axis?

I always thought it was just you multiply the conjugate by the original complex number and as long as it yields a real number then it is a conjugate.

But now that you've cleared it up thanks

Damo17 · « **Reply #306 on:** March 16, 2009, 07:19:10 pm »

Quote from: TrueTears on March 16, 2009, 07:13:11 pm

ohh nice thanks Damo17, so basically a conjugate is just the imaginary part reflected in the x axis?

I always thought it was just you multiply the conjugate by the original complex number and as long as it yields a real number then it is a conjugate.

But now that you've cleared it up thanks

No problem. That's right, the conjugate is just the imaginary part reflected in the x-axis. Remember if $z= a cis \theta$ then $\tilde{z}=a cis (-\theta)$
When you multiply a complex number by its complex conjugate, you always get a real positive number.

Mao · « **Reply #307 on:** March 16, 2009, 10:12:04 pm »

the conjugate of -7+3i is -7-3i

if you happen to multiply it by -1:

$(-7+3i)\cdot (7+3i) = (-7+3i)\cdot (-1\cdot (-7-3i)) = -|-7+3i|^2$

it's not another conjugate, it's just the negative of the conjugate.

TrueTears · « **Reply #308 on:** March 16, 2009, 10:17:26 pm »

Quote from: Mao on March 16, 2009, 10:12:04 pm

the conjugate of -7+3i is -7-3i

if you happen to multiply it by -1:

$(-7+3i)\cdot (7+3i) = (-7+3i)\cdot (-1\cdot (-7-3i)) = -|-7+3i|^2$

it's not another conjugate, it's just the negative of the conjugate.

oh i get it now, yeah factorising out that -1 haha

Mao · « **Reply #309 on:** March 16, 2009, 10:26:17 pm »

A complex conjugate is defined such that:
$z \cdot \tilde{z} = |z|^2$

There is an infinite set of complex numbers that can multiply a complex number to get a real number (or positive real number), any scalar multiple works. (z=-7+3i, another vector could be 14+6i, or -21-9i, etc)

Damo17 · « **Reply #310 on:** March 16, 2009, 10:28:02 pm »

And also, the reason why $z\tilde{z}$ cannot be negative is because:
$z \tilde{z} = (x + iy)(x - iy) = x^2 + iyx - ixy -(i^2)y^2 = x^2 + y^2$
Now as $x^2+y^2 = r^2$ , $z\tilde{z}$ has to always be positive.

TrueTears · « **Reply #311 on:** March 16, 2009, 10:34:42 pm »

Thanks mao and damo appreciate all your help

TrueTears · « **Reply #312 on:** March 28, 2009, 06:26:12 pm »

Find real values of 'a' for which 'ai' is a solution of the polynomial equation: $z^4 - 2z^3 +7z^2 -4z + 10 = 0$

Over9000 · « **Reply #313 on:** March 28, 2009, 06:41:16 pm »

With that question, we are basically trying to find a value of z that makes the equation = 0
Let z =xi

$(xi)^{4} - 2(xi)^{3} + 7(xi)^{2} - 4xi + 10=0$
$x^{4} + 2x^{3}i - 7x^{2} - 4xi + 10 =0$
$(2x^{3}-4x)i=0$ and $(x^{4} - 7x^{2} +10)=0$
solve this $x = \pm{\sqrt{2}}$

TrueTears · « **Reply #314 on:** March 28, 2009, 06:42:44 pm »

Quote from: Over9000 on March 28, 2009, 06:41:16 pm

With that question, we are basically trying to find a value of z that makes the equation = 0
Let z =xi

$(xi)^{4} - 2(xi)^{3} + 7(xi)^{2} - 4xi + 10=0$
$x^{4} + 2x^{3}i - 7x^{2} - 4xi + 10 =0$
$(2x^{3}-4x)i=0$ and $(x^{4} - 7x^{2} +10)=0$
solve this $x = \pm{\sqrt{2}}$

Oh yeah that was my thought process, but I expanded that $(xi)^{4} - 2(xi)^{3} + 7(xi)^{2} - 4xi + 10=0$ out wrong, damn!

Thanks >9000

Search the forums now!

We have moved!

Author Topic: TrueTears question thread (Read 67351 times) Tweet Share

TrueTears

Re: TrueTears question thread

Mao

Re: TrueTears question thread

TrueTears

Re: TrueTears question thread

TrueTears

Re: TrueTears question thread

Damo17

Re: TrueTears question thread

TrueTears

Re: TrueTears question thread

Damo17

Re: TrueTears question thread

Mao

Re: TrueTears question thread

TrueTears

Re: TrueTears question thread

Mao

Re: TrueTears question thread

Damo17

Re: TrueTears question thread

TrueTears

Re: TrueTears question thread

TrueTears

Re: TrueTears question thread

Over9000

Re: TrueTears question thread

TrueTears

Re: TrueTears question thread

Recent Posts

User:
Password: