Author Topic: Complex Numbers (Read 21513 times) Tweet Share

/0 · « **Reply #75 on:** April 06, 2009, 02:55:32 pm »

After much wearisome simplification,

$\frac{z+1}{z-1} = \frac{x^2+y^2-1}{(x-1)^2+y^2}-\frac{2y}{(x-1)^2+y^2}i$

If $Re(z) =0$ , then

$\frac{x^2+y^2-1}{(x-1)^2+y^2}=0$

$\boxed{\therefore x^2+y^2=1\ , \ x \neq 1}$

d0minicz · « **Reply #76 on:** April 06, 2009, 03:09:00 pm »

If, with an Argand diagram with the origin O, the point P represents z and Q represents $\frac {1}{z}$ , prove that O,P and Q are collinear and find the ratio OP:OQ in terms of |z|.
thanks guys

TonyHem · « **Reply #77 on:** April 06, 2009, 03:31:54 pm »

Collinear = In a straight line.

OP = z
Q = 1/Z
Q = 1/OP
Say you let Z = 1+i
Then OP = 1+i, and Q = 1/1+i which equals 1+i/2
(1,i) and (1/2,i/2) = on the same line

So Z = Double of 1/Z
|Z| = Sqrt{2}
so |Z|^2 = 2
so |Z|^2:1 (Since 2 is double 1 :])

kamil9876 · « **Reply #78 on:** April 06, 2009, 08:59:58 pm »

All you have done is proven this for a particular case. you shouldve let z=x+yi and proved it for all complex numbers other than zero.

P=x+yi

Q=1/(x+yi)
=(x-yi)/(x^2+y^2)

But these aren't collinear :S And in you're case of z=1 + i its not collinear either because
1/z=(1-i)/(1^2 - i^2)
=(1-i)/2

Dominicz, can you recheck the question? are you sure you didn't miss some conjugate sign in the expression of P or Q?

TonyHem · « **Reply #79 on:** April 06, 2009, 11:00:53 pm »

Its 1/ conjugate of Z ( for the question bit)
I missed it too... its co-linear for the 1+1i bit if I use conjugate of Z.
Missed that bit... and yeah I know it's just for that question :S

TonyHem · « **Reply #80 on:** April 07, 2009, 10:20:56 am »

Quote from: d0minicz on April 07, 2009, 09:57:57 am

That's the question that's written in the book.

4H of essentials? Question 13.
Its got Q represents 1/conjugateZ

Damo17 · « **Reply #81 on:** April 07, 2009, 10:22:29 am »

Quote from: d0minicz on April 07, 2009, 09:57:57 am

That's the question that's written in the book.

You have stated the question slightly wrong.

Instead of it being $\frac{1}{z}$ it is $\frac{1}{\tilde{z}}$ which changes everything and the question can be done.

Quote from: TonyHem on April 06, 2009, 03:31:54 pm

Collinear = In a straight line.

OP = z
Q = 1/Z
Q = 1/OP
Say you let Z = 1+i
Then OP = 1+i, and Q = 1/1+i which equals 1+i/2
(1,i) and (1/2,i/2) = on the same line

So Z = Double of 1/Z
|Z| = Sqrt{2}
so |Z|^2 = 2
so |Z|^2:1 (Since 2 is double 1 :])

If you have $Q=\frac{1}{z}$ using $z=1+i$ :

$Q=\frac{1-i}{2}$
And this point does not lie on the line of $1+i$ connected to the origin but it does lie on the line of $1-i$ connected to the origin. So if $OP=z$ and $Q=\frac{1}{z}$ then $OP$ and $Q$ can not be co-linear.

That is why for questions like this it is very important to read carefully as sometimes in the book it does not look like $\tilde{z}$ .

d0minicz · « **Reply #82 on:** April 07, 2009, 10:28:32 am »

for fucks sakkkkkkkkkkeeeeeeeeeeeeeeeeeeeeee
sorry guys
thanks damo.

d0minicz · « **Reply #83 on:** April 07, 2009, 01:42:51 pm »

so how do i prove that it's collinear andd how do i find the zatio using the conjugate

TrueTears · « **Reply #84 on:** April 07, 2009, 01:49:57 pm »

$P = x+yi$

$Q = \frac{1}{x-yi} = \frac{x}{x^2+y^2} + \frac{y}{x^2+y^2}i$

for collinear: $OP = kOQ$ for some constant k

$x+yi = k(\frac{x+yi}{x^2+y^2}$

$k = x^2+y^2$ so they are collinear.

$|OP| = \sqrt{x^2+y^2}$

$|OQ| = \frac{1}{\sqrt{x^2+y^2}}$

$\frac{|OP|}{|OQ|} = \frac{\sqrt{x^2+y^2}}{\frac{1}{\sqrt{x^2+y^2}}} = \frac{x^2+y^2}{1}$

so $|OP| : |OQ| = (x^2+y^2) : 1 = |z|^2 : 1$

TrueTears · « **Reply #85 on:** April 07, 2009, 02:40:58 pm »

Solution for Question asked:

$|z-1-2i|=3$

$|\frac{w}{2} -1-2i|=3$

$w = x+yi$

$|\frac{x+yi}{2} - 1 - 2i| =3$

$(\frac{x}{2} - 1)^2 + (\frac{y}{2} - 2)^2 = 9$

$x^2 - 4x + y^2 - 8y + 20 = 36$

$x^2 - 4x + (2)^2 - (2)^2 + y^2 - 8y + (4)^2 - (4)^2 = 16$

$(x-2)^2 + (y-4)^2 = 36$

centre (2,4) radius 6

TrueTears · « **Reply #86 on:** April 08, 2009, 08:33:15 pm »

Quote from: d0minicz on April 06, 2009, 12:17:19 pm

What about this one : $z: Arg(z-1) = \frac {\pi}{3}$

let $z = x+yi$

$Arg((x-1)+yi) = \frac{\pi}{3}$

$tan^{-1}(\frac{y}{x-1}) = \frac{\pi}{3}$

tan both sides yields $\frac{y}{x-1} = \sqrt{3}$ (Note $x>1$ )

$y = (x-1)\sqrt{3}$

$y = \sqrt{3}x - \sqrt{3} , x>1$ is the Cartesian equation.

d0minicz · « **Reply #87 on:** April 11, 2009, 05:47:45 pm »

Express in the modulus argument form, $0 < \theta < \frac {\pi}{2}$ .
$1+ cot\theta i$
thanks

TrueTears · « **Reply #88 on:** April 11, 2009, 05:54:06 pm »

$|z| = \sqrt{1+cot^2\theta} = cosec\theta$

to work out Arg
$<br />\theta = tan^{-1}(\frac{cot\theta}{1})$

$tan\theta = \frac{cot\theta}{1}$

but we know $cot\theta = tan(\frac{\pi}{2} - \theta)$
$<br />tan\theta = tan(\frac{\pi}{2} - \theta)$

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

so $Arg(z) = \frac{\pi}{2} - \theta$

therefore $1+cot\thetai = cosec(\theta) Cis(\frac{\pi}{2} - \theta)$

d0minicz · « **Reply #89 on:** August 28, 2009, 10:23:51 pm »

How do you factorise things like $z^3 + 64$
theres 2 formulas but i forgot them lol

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