Expert: Paul Klarreich - 1/19/2007

Question
Hello! The may be trivial:

This is for z = x + iy  a complex number

I am to find all z such that the principal argument of the compliment of z, z-bar = x - iy ,  is equal to the negative principal argument of z .

Or Arg(z-bar) = - Arg(z)  for which z?

Anywho, after some basics I've come to

Arccos(x) = arcsin(-y) = - arcsin (y)

and, by symmetry , arcsin (-y) = -arcsin (y)

So I have the the following relation which must hold:

arccos(x) = - arcsin(y)

How do I solve this from here?

Thanks.

Answer
Questioner:   Andrew Pearson
Category:  Calculus
 
Subject:  Arg(z)
Question:  Hello! The may be trivial:

>> Everything is trivial AFTER you do it.

This is for z = x + iy  a complex number

I am to find all z such that the principal argument of the compliment of z, z-bar = x - iy ,  is equal to the negative principal argument of z .

Or Arg(z-bar) = - Arg(z)  for which z?

Anywho, after some basics I've come to

Arccos(x) = arcsin(-y) = - arcsin (y)

and, by symmetry , arcsin (-y) = -arcsin (y)

So I have the the following relation which must hold:

arccos(x) = - arcsin(y)

How do I solve this from here?

Thanks.

....................................
Hi, Andrew,

I think you are referring to the COMPLEX CONJUGATE of z, not complement.  That's what's usually called 'z-bar'

Suppose we call arg(z) = theta (I'll write just t for theta.)

Then I think the following is true:

z     = r (cos t + i sin t)

z-bar = r (cos t - i sin t)  << complex conjugate.
     = r (cos(-t) + i sin(-t)) << using cos(-t) = cos t,  sin(-t) = - sin t

So arg(zbar) = -t = - arg(zbar)

BUT what exactly is meant by the 'principal argument' of z?  That might vary from one author to another.  

If the P.A.(z) means it must be 0 to 2pi, then this can never happen, since a number from 0 to 2pi cannot be equal to a negative number.

But I think P.A. must mean  -pi to pi.  In that case,

Arg(zbar) = - arg(z) for all z.