Complex-valued function - finding argument and magnitude

Click For Summary
SUMMARY

The discussion focuses on finding the magnitude |H(ω)| and argument argH(ω) of complex-valued functions H(ω) defined as H(ω) = 1/(1+iω)^10 and H(ω) = (-2-iω)/(3+iω)^2. Participants suggest rationalizing the denominator and converting to polar form using the modulus and argument formulas. Key techniques include using the binomial theorem for complex powers and ensuring correct sign preservation in arc tangent calculations. The conversation emphasizes the importance of understanding polar representation for simplifying complex functions.

PREREQUISITES
  • Understanding of complex numbers and their polar form
  • Familiarity with the modulus and argument of complex numbers
  • Knowledge of rationalizing denominators in complex fractions
  • Experience with binomial expansion for complex powers
NEXT STEPS
  • Learn how to rationalize complex denominators effectively
  • Study the binomial theorem for complex numbers
  • Explore polar form representation of complex functions
  • Investigate the properties of arc tangent in complex analysis
USEFUL FOR

Students studying complex analysis, mathematicians dealing with complex functions, and anyone looking to enhance their understanding of complex number operations.

lveenis
Messages
12
Reaction score
0

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 = \sqrt{a<sup>2</sup>+b<sup>2</sup>} 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!
 
Physics news on Phys.org
Have you tried rationalizing the denominator as a first step?
 
I'm sorry if this is a stupid question, but I don't understand what you mean exactly?
 
you want z=x+iy but you have something like z=\frac{1}{a+ib}... you can make this look like the other one by "rationalizing the denominator". (Good search term to try.) You use the result (a+ib)(a-ib)=a^2+b^2 like this:\frac{1}{a+ib} = \frac{1}{a+ib}\frac{a-ib}{a-ib}=\frac{a-ib}{a^2+b^2}... now you can find the modulus and argument easily.

You may need to multiply out the powers in your problems first though.
 
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
 
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 ;)
 
lveenis said:
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!
 
Expressing H as a ratio of exponentials may help deal with the powers too.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 4 ·
Replies
4
Views
82K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
8K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
2K