Writing the Modulus of f(z) in Terms of lzl

[aaaa202]

For expressions like:

f(z) = (1+z²)/(1+z⁴) how does one write the modulus of that in terms of lzl?

 

[tiny-tim]

hi aaaa202!

For expressions like:

f(z) = (1+z²)/(1+z⁴) how does one write the modulus of that in terms of lzl?

i don't really understand the question …

how does one write eg the modulus of 1+z² on its own in terms of lzl only ??

 

[aaaa202]

Well I want to find lf(z)l and show that lzllf(z)l goes to 0 as lzl goes to infinity. So wouldn't I need to write the function above in terms of lzl, multiply by lzl, and show that it goes to zero?

 

[tiny-tim]

in that case, write z = re^iθ

 

[aaaa202]

hmm yes okay. So in that case i would for instance get:

(1+lzl²e^i2θ)/(1+lzl⁴e^i4θ). But still that doesn't really give me the modulus for the whole expression?

 

[tiny-tim]

Well I want to … show that lzllf(z)l goes to 0 as lzl goes to infinity.

(1+lzl²e^i2θ)/(1+lzl⁴e^i4θ).

ok, so multiply by |z| and then let |z| -> ∞ …

what difference do the θ terms make? ​

(btw, we normally write 2iθ rather that i2θ)

 

[aaaa202]

oh well but the expression above is f(z) not lf(z)l = mod(f(z)), and I wanted to show that lzllf(z)l -> 0 as lzl->∞..

Edit: Oh well if f(z) goes to zero then surely lf(z)l does too.. So nevermind, unless you can have lf(z)l->0 even though f(z) doesn't? Nah that wouldn't make sense right..?