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Complex-valued function - finding argument and magnitude

  • Thread starter lveenis
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Homework Statement


Let H(ω) be a complex-value function of the real variable ω. For each of the cases below, find |H(ω)| and argH(ω).

a: H(ω)= 1/(1+iω)^10

b: H(ω)=(-2-iω)/(3+iω)^2

Homework Equations





The Attempt at a Solution



Our prof has not taught how to do these types of questions in terms of functions. I understand how to find the magnitude and principal argument of a complex number, but I'm completely lost how to approach this question.

for complex numbers I'd convert it to polar form using r = [itex]\sqrt{a2+b2}[/itex] for some complex number z = a+ib and then using tan-1(b/a) to get the principal argument (adjusting the result as necessary).

If someone could please give me a hint as to how to approach this problem I would be extremely grateful since it's due tomorrow!
 

Answers and Replies

  • #2
Simon Bridge
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Have you tried rationalizing the denominator as a first step?
 
  • #3
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I'm sorry if this is a stupid question, but I don't understand what you mean exactly?
 
  • #4
Simon Bridge
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you want [itex]z=x+iy[/itex] but you have something like [tex]z=\frac{1}{a+ib}[/tex]... you can make this look like the other one by "rationalizing the denominator". (Good search term to try.) You use the result [itex](a+ib)(a-ib)=a^2+b^2[/itex] like this:[tex]\frac{1}{a+ib} = \frac{1}{a+ib}\frac{a-ib}{a-ib}=\frac{a-ib}{a^2+b^2}[/tex]... now you can find the modulus and argument easily.

You may need to multiply out the powers in your problems first though.
 
  • #5
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That's what I was wondering about, the exponent part. Normally you would just put it into polar form to make the power easy to do. I was curious if there was a way to put it into polar form, or some other trick that I'm unaware of. Thank you for your help, I will try expanding it out and doing it using your suggestion
 
  • #6
Simon Bridge
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That is the trick for putting it into polar form.
Usually you'd have had some experience working through the binomial coefficients for complex powers by this stage... of course, this could be the lesson ;)
 
  • #7
rude man
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That's what I was wondering about, the exponent part. Normally you would just put it into polar form to make the power easy to do. I was curious if there was a way to put it into polar form, or some other trick that I'm unaware of. Thank you for your help, I will try expanding it out and doing it using your suggestion
You certainly can work this with polar math and without "rationalizing the denominator". General example:

f =(a + jb)/(c + jd)

= √{(a2 + b2)/(c2 + d2}exp{j(θ1 - θ2} where θ1 = tan-1(b/a) and θ2 = tan-1(d/c).

In fact, that's how I always do it.

Careful with the arc tangents. For example, the angle arc tan (-1/2) is not the same as arc tan 1/(-2). Preserve the signs of the numerator and denominator!
 
  • #8
Simon Bridge
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Expressing H as a ratio of exponentials may help deal with the powers too.
 

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