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modulus of f(z) |
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| May24-12, 06:06 AM | #1 |
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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? |
| May24-12, 06:28 AM | #2 |
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hi aaaa202!
![]() …how does one write eg the modulus of 1+z2 on its own in terms of lzl only ??
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| May24-12, 06:48 AM | #3 |
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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?
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| May24-12, 07:01 AM | #4 |
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modulus of f(z)
in that case, write z = reiθ
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| May24-12, 08:36 AM | #5 |
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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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what difference do the θ terms make?(btw, we normally write 2iθ rather that i2θ) |
| May24-12, 10:44 AM | #7 |
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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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