ARG() values in complex numbers (1 Viewer)

[underthesun]

Just curious.

arg(z) = tan^-1(^im(z)<i>/</i>_re(z))

right? This is an assumed thing right? No textbooks mentioned anything like this so far...

Because this is so damn useful in finding the centres of complex number loci in the form of circles. Example arg(z-1) - arg(z+1) = pi/4...

 

[jm1234567890]

i don't know about assumed.....
you shouldn't assume too many things

but, yes it is true!

 

[McLake]

Originally posted by underthesun
Just curious.

arg(z) = tan^-1(^im(z)<i>/</i>_re(z))

right? This is an assumed thing right? No textbooks mentioned anything like this so far...

Because this is so damn useful in finding the centres of complex number loci in the form of circles. Example arg(z-1) - arg(z+1) = pi/4...

I "think" it's assuamble (pretty basic fact)

 

[wogboy]

Be careful with such assumptions:

(1) Is it true tan (Arg(z)) = Im(z)/Re(z)? YES ALWAYS.

(2) Is it true Arg(z) = arctan (Im(z)/Re(z))? NOT ALWAYS.

in fact, statement (2) is only true when Re(z) > 0 **

What about when Re(z) < 0 ? Then you'll find that:

a) when Im(z) >= 0,
Arg(z) = Pi + arctan(Im(z)/Re(z))

b) when Im(z) < 0
Arg(z) = Pi - arctan(Im(z)/Re(z))

** (letting Re(z) = 0 is a bit of a grey area though, some might object to dividing by zero and then taking the arctan of that infinite value. Of course when Re(z) = 0, then Arg z = +- Pi/2 depending on the sign of Im(z), and hoping that Im(z) =/= 0 !!)

e.g. take z = - 1 + i
Arg(z) = 3/4 * Pi

Re(z) = -1
Im(z) = 1
therefore Re(z)/Im(z) = -1
therefore arctan (Re(z)/Im(z)) = arctan(-1)
= -Pi/4
=/= Arg(z)

 

[freaking_out]

yeah thats why my teacher tells me to draw a diagram to make sure when findings args of complex no.

 

[underthesun]

Originally posted by wogboy

(2) Is it true Arg(z) = arctan (Im(z)/Re(z))? NOT ALWAYS.

What about this? I'll do a sample question:

The locus of z, given by the equation of arg(z-1) - arg(z+1) = pi/4

Find the centre of the locus (which is a circle).

Now,

arg(z-1) = pi/4 + arg(z+1)

using arg(z) = tan^-1(^im(z)<i>/</i>_re(z)),

arctan(y / x-1) = pi/4 + arctan(y/ x+1)
we tan both sides.

y/(x-1) = tan(pi/4 + arctan(y/x+1))

by using compound angle formula, you'll actually get the relation between x and y, which happens to be the equation of the circle. From then you can get the radius. Which happens to be the right answer.

It's nifty, but i'm just curious whether you're allowed to use the above steps in exams.

 

[freaking_out]

Originally posted by underthesun
What about this? I'll do a sample question:

The locus of z, given by the equation of arg(z-1) - arg(z+1) = pi/4

Find the centre of the locus (which is a circle).

Now,

arg(z-1) = pi/4 + arg(z+1)

using arg(z) = tan^-1(^im(z)<i>/</i>_re(z)),

arctan(y / x-1) = pi/4 + arctan(y/ x+1)
we tan both sides.

y/(x-1) = tan(pi/4 + arctan(y/x+1))

by using compound angle formula, you'll actually get the relation between x and y, which happens to be the equation of the circle. From then you can get the radius. Which happens to be the right answer.

It's nifty, but i'm just curious whether you're allowed to use the above steps in exams.

hmm... looks phoney to me....but for the above question, u can just do it geometrically.

another way, is since the arg((z-1)/(z+1))=pi/4, therefore the imaginary part must equal to real part...then work from there.

 

[underthesun]

Originally posted by freaking_out
hmm... looks phoney to me....but for the above question, u can just do it geometrically.

another way, is since the arg((z-1)/(z+1))=pi/4, therefore the imaginary part must equal to real part...then work from there.

Yeah, the excel book says to work it out geometrically. However, if the complex numbers don't lie along an axis it would be very hard using the geometric method. Using the arctan method can get you the answer very very nicely, so im just curious..

 

[wogboy]

You can use the arctan method if you want, but if you do you just need to be a little careful with the quadrants. In which quadrants is it possible that arg(z-1) be Pi/4 radians greater than arg(z+1)? You need to draw up a quick Argand diagram to find out.

Try it out in the 1st quadrant (where x>0 & y>0). Pick a spot at random here, and plot a point & call it z. Then plot z+1 and z-1 (which are 1 unit to the left & right respectively of z). You'll see that arg(z-1) is always > arg(z+1) so this allows for the locus to exist here. Try the same thing in the 2nd quadrant and you'll see something similar, that arg(z-1) is always > than arg(z+1).

Now try it in the 3rd & 4th quadrants (where y<0), and you will see that arg(z-1) is always < than arg(z+1) hence making it impossible for the locus to exist here.

This locus does not extend into the region y<0. What about when y=0? Well for this circle locus we get (centre at (0,1) & radius sqrt(2)), at y=0, x = 1 or x= -1. So z=1 or z=-1. Therefore either z-1=0 or z+1=0. This will mean arg(z-1) + arg(z+1) will be undefined in both cases, since arg(0) is undefined.

So the answer to this locus question is the circle of radius sqrt(2) and centre (0,1), and ONLY the part of it above the x-axis (i.e. where y>0)

The moral of the story: If you want to assume that arg(z) = arctan Im(z)/Re(z), that's fine. However if you make this assumption, then you must check each quadrant at the end, since this assumption can give some extra unwanted extensions to the locus (e.g. the part of the circle where y<0). If you check and verify each quadrant at the end, then it should be fine.

 

[freaking_out]

faaar, looking at the size of wogboys posts...

the moral of the story: stick to the normal methods.

 

[underthesun]

Thanks a lot, that makes it much clearer

 

[OLDMAN]

Just to say that I've always appreciated wogboy's thorough commentary on various topics, especially inverse trig.