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modulus of f(z)

 
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May24-12, 06:06 AM   #1
 

modulus of f(z)


For expressions like:

f(z) = (1+z2)/(1+z4) how does one write the modulus of that in terms of lzl?
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May24-12, 06:28 AM   #2
 
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hi aaaa202!
Quote by aaaa202 View Post
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 ??
May24-12, 06:48 AM   #3
 
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?
May24-12, 07:01 AM   #4
 
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modulus of f(z)


in that case, write z = re
May24-12, 08:36 AM   #5
 
hmm yes okay. So in that case i would for instance get:

(1+lzl2ei2θ)/(1+lzl4ei4θ). But still that doesnt really give me the modulus for the whole expression?
May24-12, 09:32 AM   #6
 
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Quote by aaaa202 View Post
Well I want to … show that lzllf(z)l goes to 0 as lzl goes to infinity.
Quote by aaaa202 View Post
(1+lzl2ei2θ)/(1+lzl4ei4θ).
ok, so multiply by |z| and then let |z| -> ∞ …
what difference do the θ terms make?
(btw, we normally write 2iθ rather that i2θ)
May24-12, 10:44 AM   #7
 
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..?
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