Inverse of arg(z)Date: 10/10/2003 at 14:40:08 From: Debbie Subject: the function arg(z) I would like to know if there is an inverse to the function arg(z). In other words, if I know, for example, that arg(z) < 0, what can I say about z? Should it be positive, negative, real, imaginary? Another example: if arg(y) > -pi/2, what can I say about y in this case? I find the concept of arg(z) a little confusing, I would like to see a graphic representation of it or something to help me extract information about its argument. If z = -i, then arg(z) = -pi/2, thus the case above of arg(y) > -pi/2 implies that y should be greater than -i ? I am a bit confused. Date: 10/11/2003 at 15:20:35 From: Doctor Luis Subject: Re: the function arg(z) Hi Debbie, Note that arg(z) is simply the angle in the polar representation of the complex number z = x + iy. That is, z = r * e^(i * theta) The radius r is the modulus, or absolute value of z, denoted by r = sqrt(x^2 + y^2) = mod(z) = |z| The angle theta is the argument of z, denoted by theta = arctan(y/x) = arg(z) Now, because z lives in a two dimensional plane, the inverse of arg(z) is not really defined. There are infinitely many complex numbers with the same argument. To see why, draw any ray leading away from the origin of the complex plane. Every point on this ray is a complex number z, and they all have the same value of arg(z). Since functions can only be defined when a given input gives exactly one output, we conclude that the function arg(z) has no inverse. I hope this helped! Let us know if you have any more questions. - Doctor Luis, The Math Forum http://mathforum.org/dr.math/ Date: 10/11/2003 at 15:32:59 From: Debbie Subject: Thank you (the function arg(z)) Thank you very much for your answer. It helped me clarify a lot about what arg(z) is. My problem is that I am implementing a computer program where I need to input a parameter, alpha, for which the formula prescribes: -pi/2 < arg(alpha) < 0 I don't know what value to put in for alpha. From what you say, it seems that there are infinite many values of alpha that would satisfy this inequality. Am I right? Thank you very much again! Debbie Date: 10/11/2003 at 16:34:22 From: Doctor Luis Subject: Re: Thank you (the function arg(z)) Hi Debbie, In the following figure, the set of complex numbers alpha such that -pi/2 < arg(alpha) < 0 is denoted in red. The axes and the origin aren't included of course. So, you're essentially picking a number in the fourth quadrant. Any complex number alpha = x+iy, with x>0 and y<0, will do. Does this make sense? - Doctor Luis, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/