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Pretty pictures in complex (1 Viewer)

friction

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Any one have any tips or anything in this part of 4u for me. I can never seem to get any of them right and there has been one in most the papers that i have done. These are things like parralellograms etc. Or stuff between abs values = other stuff. Any tips.
 

Timothy.Siu

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Js^-1 said:
http://upload.wikimedia.org/wikipedia/commons/d/d3/Color_complex_plot.jpg

Not exactly relevant, but...It's a pretty picture :)

(Color plot of complex function (x^2-1) * (x-2-I)^2 / (x^2+2+2I), hue represents the argument, sat and value represents the modulo, sourced from Wikipedia.org)
lol wouldn't never thought.

uhh i'm not sure wat the OP wants help on, are u talking about vectors? well just draw it and remember if its 2 imaginary numbers z and w, z+w=the longer diagonal and z-w= shorter one, also, notice if its a square/rhombus/rectangle as they all have special stuff associated with them
and for absolute stuff u mean the modulus? if its |z|=2 then u draw a circle with radius 2 basically |z-1|=2 u draw with 1 as its centre,
i'm not too sure about this stuff so correct me if i'm wrong....so anyone else wanna help him?
 

Js^-1

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If you have the |z-w| = r or any other number, then that means the distance of z, from the point w, is always equal to r. Note w is complex, of the form w=x+iy.
Also note, the modulus can be an inequality, such as |z+i|< 2. This means that the modulus of (distance from) -i (since |z-w|, then |z-(-i)| ) is always less than 2. This then represents the locus of the interior of a circle, radius 2, and centre w=0-i

As for things like angles and stuff, remember that multiplication of a complex number z by another complex number w = rcis(a), is equivalent to enlarging the vector of z by a factor of r, and adding the arguments of z and w. Hence if P = zw, and z=Rcis(A) and w=rcis(a), the P = R*rcis(a+A).
Multiplication by i is a special form of this. Since |i|=1 and arg(i) = pi/2
then multiplication by i involves a rotation of the complex number by pi/2 in an anticlockwise direction.

Also, you need to remember properties of quadrilaterals and other geometrical properties. Things like the diagonals of a square are equal and other things like that.

Other than that, just the basic things like vector addition and subtraction and the parralellograms that are formed.
Hope this helped, not quite sure what you were asking though...
 

friction

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Im not sure what i am really asking either :)

But yea that pretty much it. I dont think i know my properties of quadrilaterals well enough.

I know the basic |z-w| = r but one was a question like

3. |z-8| = 2 Re |z-2|

a) Sketch the curve (+ important features) and write its equation. State the type of the curve.

b) Write down values for |z+8| - |z-8|

c) find possible values for arg z
 

Js^-1

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Have you tried writing it down algebraically?
i.e. |z| = (x<sup>2</sup>+y<sup>2</sup>)<sup>1/2</sup>
I'm not sure how that would work with |z-8| though...
would that mean |z-8| = (x<sup>2</sup>-8+y<sup>2</sup>)<sup>1/2</sup> ?
 

tommykins

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i use a different method.

say |z-7+2i| = 9
vector wise, i change it to |z-(7-2i)| = 9
using the brackets, i know the centre is at (7, -2i)
 

u-borat

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^^same as what i do

friction, with that question, try subbing in z=x+iy then use your normal modulus and re(z) rules?

((x-8)^2 + y^2)^1/2 = 2 (x-2)

square both sides

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

expand and simplify. you get a hyperbola.
 

friction

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Yea i can do the basic ones but sones more difficult like the one i showed i cannot get. And its only question 3.
 

tommykins

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hahaha, reminds me our teacher had an evil one in our trial -

|z-1| ≤ 3 and Im(z) ≥ 3
 

u-borat

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lol is that like the point 1, 3?

edit; i always get confused with the ones that turn out as parabolas.
 

tommykins

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u-borat said:
lol is that like the point 1, 3?

edit; i always get confused with the ones that turn out as parabolas.
yeah, in the heat of the moment i was liek OMG WTFUX WHAR IS THE LOCUS (because my circle wasn't to scale) then looking back im like LOLWUT .
 

adnan91

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tommykins said:
hahaha, reminds me our teacher had an evil one in our trial -

|z-1| ≤ 3 and Im(z) ≥ 3
its just a shaded in circle with center (1,0) and radius 3
 

adnan91

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tommykins said:
the point of intersection between the circle (x-1)²+y² = 9 and the line y = 3.
u mean evrything inside the circle and y is greater than or equal to 3
 

tommykins

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adnan91 said:
u mean evrything inside the circle and y is greater than or equal to 3
equality only holds for both locus's at the point of intersection.
 

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