complex numbers question (1 Viewer)

[conics2008]

sketch

arg (z-2) - arg (z) = pi/2

i know arg ( z-2/z ) = pi /2
= arg ( 1 - 2/z ) = pi/2

but i cant seem to put it in drawings..

i know its a circle.. and like there is a chord from 0 to 2

but how would u approach quetions in the form of

arg ( z-w) - arg ( z-s) = theta

where w= a+ib and s= c+id and 0<theta<2pi or pi/2

Thanks

 

[conics2008]

what i get is a circle possibe a ssemi circle...

i know the point 0 and 2 and the diameter etc etc.. but how would u be confident in that ???

eg tests u can preform ??

 

[Poad]

I've seen a question like this before, and if I remember correctly, you're right about it being a semi-circle.

 

[u-borat]

this is pretty generic complex number graph question.
anything which has arg of something minus arg of something else =an angle, gives you part of a circle...and if the angle is pi/2, its a semicircle.

 

[tacogym27101990]

yeah thats a good question aye
ive always wondered how we can just assume that its a circle
has it got something to do with exterior angle of a triangle
and then the locus of that triangle creates a circle because two of the points are fixed??
prob way off track

 

[Poad]

I think its because the angle in a semicircle taken from the extremities of the diameter to the circumference is pi/2 (and the question asks
arg (z-2) - arg (z) = pi/2).

Maybe I'm wrong, I dunno.

 

[conics2008]

guys.. in general its two lines intersectin... and makin the angle which is given..

you needd ot know your exterior angle rule of a triangle.. etc.. but i just want to know how can i do test on it to prove its true ???

 

[Affinity]

sketch

arg (z-2) - arg (z) = pi/2

i know arg ( z-2/z ) = pi /2
= arg ( 1 - 2/z ) = pi/2

next line:

1-2/z = ki for some positive real k.

z = 2/(1+ki)
z = (2-2ki)/(1+k^2)

so
x = 2/(1+k^2)
y = -2k/(1+k^2)
k>0

you get something along the lines of

x^2 + y^2 = 2x
or

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

but y<0

===> the bottom half.

 

[Affinity]

in general if you get arg(EXPR) = theta you can proceed with the equivalent condition EXPR = r*(cos(theta) + isin(theta)) for some positive r.

 

[timmiitippii]

Umm i think in the Excel Study guide they provide a proof using angles theta and phi made with the real axis... and all these mubo jumbo but yeh... there is a reason why the points forms a chord and the "= angle" is the angle formed at the circumference and the chord...
Sorry if nto much help but i can't find my excel atm LOL :rofl:

 

[doink]

use the external angle of a triangle is equal to the two opposite interior angles, have teh 2 base points of triangle be the two arguments and let the angle up top be pi/2. Then arg(x+h) = arg(x) + pi/2 hence pi/2 = arg(x) - arg (x+h)

 

[Iruka]

It is basically Thales' Theorem.

http://en.wikipedia.org/wiki/Thales'_theorem

I think as far as actually demonstrating that algebraically, Affinity's method is probably the easiest.

 

[conics2008]

affinitys way is in the cambridge book aswell.. buts like how would you GENERALLY approach quetsions like this type...

because i just know how they will turn out eg

i know its a circle for a start.. i know that angle at the end if the angle at the circumference and those lines that it cuts what ever it might be is either a chord or a diameter.. etc etcc but the prob is how do i put it in writing other than affintiy way... it looks short and simple.. but i know ill stuff up using his method.

thanks

 

[vds700]

Hi i got a complex number Q, might as well tack it on this thread. Its the part in the red box i don't get.

 

[Iruka]

The n^th roots of unity form a regular n-gon. You chop the n-gon up into n isosceles triangles in which the equal sides have length one. Then you use the area of a triangle formula from trig

area = 1/2 ab sin(C)

to evaluate the area of each triangle.

 

[Saviour]

The n^th roots of unity form a regular n-gon. You chop the n-gon up into n isosceles triangles in which the equal sides have length one. Then you use the area of a triangle formula from trig

area = 1/2 ab sin(C)

to evaluate the area of each triangle.

Why didn't I think of that?
Truly throughout my engineering escapade I should have thought of this during my complex number days.

 

[conics2008]

Hi i got a complex number Q, might as well tack it on this thread. Its the part in the red box i don't get.

hey it goes like this



let z be the root

z^n=1

Mod = 1 and arg = 2pik/n

hence the point A0,A1,A2,A3 etc etc are the roots of z^n=1

if you draw a unit circle and place al these roots you will realise that the radius is 1 and its actually equally spaced at an angle of 2pi/n

hence finding an area of 1 of the triangle is

1/2 (1)(1) Sin ( 2pi/n ) = 1/2sin2pi/n

but there are n triangles

there fore n * 1/2sin2pi/n = n/2sin2pi/n


=]

t

 

[vds700]

Thanks Conics and Iruka. Area rule, dunno why i didn't think of that.