arg(z-1)/arg(z+1) (1 Viewer)

[Teoh]

Can someone please explain to me why

arg(z-1)/arg(z+1) = pi/2

Is below the x axis, rather than above?

I mean the semicircle/triangle it makes

 

[steverulz55]

do you mean arg [(z-1)/(z+1)] = pi/2 ??

in that case it is

arg (z-1) - arg (z+1) = pi/2
<=> arg (z+1) - arg (z-1) = -pi/2

which is a semicircle below the x axis, if you use sum of interior angles equals opposite exterior in the triangle if you join up z with -1 and 1

see also http://www.boredofstudies.org/community/showthread.php?t=38614

 

[Teoh]

But why do u have to re-arrange the equation?

 

[ngai]

below is it?
hmm, when z = i
(i-1)/(i+1)
= [(i-1)^2]/(i^2 - 1^2)
= (i^2 - 2i + 1)/(-1-1)
= (-1+1-2i)/(-2)
= -2i/-2
= i
and arg i = pi/2
hence we have a contradiction, in that it cannot be just below the real axis

its actually above the real axis
draw the complex nubmers z-1, z+1
the full circle is arg(...) = |pi/2|
for the z-1 to have bigger argument, can't be below real axis
for the z+1 to have bigger argument, can't be above real axis

 

[Teoh]

Oh whoops...
z+1/z-1

So...should i just sub in z=i, and work through it by simplyfying the arguments, or what?

OR, should i just compare which one has to be the bigger angle, and draw from there?

 

[McLake]

Oh whoops...
z+1/z-1

So...should i just sub in z=i, and work through it by simplyfying the arguments, or what?

OR, should i just compare which one has to be the bigger angle, and draw from there?

You can take either approach. By the second one I take it that you mean that you would simplify the LHS and RHS and show a relation ...