diagram of arg(z-1)=2argz (1 Viewer)

[totallybord]

hi,
and other questions does anyone no?

 

[nick90]

if mod z = 1(or any other constant for that matter), then to prove that arg (z-1)=2argz you first draw the unit circle in the Argand diagram. The vector representing z-1 is the line drawn from 1 to any point on the unit circle, while the vector representing z is the line drawn from 0 to the same point on the unit circle. Using circle geometry, the aangle at the centre of the circle is double the angle at the circumference on the same arc. This means that the angle given by arg(z-1) is twice the size of the angle made by argz.
i.e arg(z-1)=2argz

Hope that helps

 

[Affinity]

Concise answer:

circle of radius 1 centered at (1,0), but excluding point (0,0) PLUS the real axis excluding the closed interval from (0,0) to (1,0)

Intuition:

The circle part could guessed from the fact that the paralellogram 0,z-1,z,1 is infact a rhombus because of the angles. the other part was actually not anticipated but discovered when ploughing through the dirty equations

sketch of proof:

arg(z-1) = 2arg(z)
arg(z-1) = arg(z^2)
so k(z-1) = z^2 for some k>0
z^2 - kz + k = 0

so z = (k +/- sqrt(k^2 - 4k))/2

then for k in (0,4) you get the circle (just check that |z-1| =1 by calculation, and that every point is infact attained. for k > 4 you get the part of the real line

yeah this one is a bit nasty, but the technique used here is quite straight forward

Note: The geometric approach can give you good intuition, but you should always confirm your guesses analytically, most who do this one geometrically would miss the part on the real line.

Note: I used arg(a) + arg(b) = arg(ab) which might have problems (not here though) unless it's interpreted in the general sense not as numbers (-pi,pi] or [0,2pi), but as values of R mod 2pi (you can also think of the arg function giving you not the angle theta but the complex number cos(theta+isin(theta) ) don't worry too much if you don't get this paragraph

 

[totallybord]

thanks nick90 and affinity
i dun really get affinity's method whats k stand for? is this a formula?
and can you just draw the circle anywhere you like?
nick90: is this what you meant by it? (attachment)
also, how do i find the mod-arg form of this q:
1+cos(pi/4) +isin(pi/4) => i found the modulus by half angle or double angle formula whatever its called by i didnt get the rite answer for the arg which is spose to be 3pi/8
any help is appreciated

 

[chousta]

K is a constant. affinity is showing you how to show it in a more analytical approach

 

[totallybord]

Re: diagram of arg(z-1)=2argz AND COMPLEX NO qs

also on diagrams how o i no which ones r open circles n why r there open circles ?
and how do i figure out the arg of z=(1 + i)/(root3 - i)?
q: sketch and describe the locus of 2|z| = z+(conj)z +4
i got y^2 = 4x +4 is that rite?

 

[totallybord]

another question: (sorry exams on mon):
if u times z by i, it turns anticlockwise 90degrees but does z have to be at the origin? or can it be anywhere?

 

[totallybord]

another q: z1=4-i and z2=2i are the vertices of an iscosceles right angle triangle whose right angel is at z3, find z3. that was in relation to my previous q... i got 2 answers bt i dunno if they are rite
also : find the locus os z if arg (z-i/z+1)= pi/2 => how do i draw that diagram? i no its a semi circle rite? but whats the centre and all that? or do i need to find the locus first?

 

[totallybord]

hey 3unitz, |z| = sqrt(x^2+y^2) not x^2+y^2

 

[totallybord]

ok how do i shade -pi < argz< pi
i thought that wld be the whole graph since pi is the same as -pi rite?