Complex analysis inequality proof


by shebbbbo
Tags: analysis, complex, inequality, proof
shebbbbo
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#1
Sep18-11, 10:56 PM
P: 17
Prove for all Z E C

|ez-1| [itex]\leq[/itex] e|z| - 1 [itex]\leq[/itex] |z|e|z|

I think this has to be proven using the triangle inequality but not sure how.

Please help. :)

thanks
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Dick
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#2
Sep18-11, 11:11 PM
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Use the triangle inequality along with the power series expansion e^z=1+z+z^2/2!+...
shebbbbo
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#3
Sep19-11, 02:58 AM
P: 17
great thanks...

quick question:

does |ez| = e|z| ????

Dick
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#4
Sep19-11, 09:05 AM
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Complex analysis inequality proof


Quote Quote by shebbbbo View Post
great thanks...

quick question:

does |ez| = e|z| ????
You tell me. Use the series expansion to compare them.
shebbbbo
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#5
Sep19-11, 09:11 PM
P: 17
thanks!
Dick
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#6
Sep19-11, 09:23 PM
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Quote Quote by shebbbbo View Post
thanks!
Glad you got it. Just out of curiousity, what did you conclude about the truth of |e^z|=e^|z|?
shebbbbo
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#7
Sep19-11, 11:18 PM
P: 17
Once i expanded them i realized they looked exactly like the triangle inequality where the modulus of the summation of terms was less than or equal to the modulus of each term summed.

i didnt try to conclude |e^z| = e^|z|

from what i was reading i think they are equal when z is real and the inequality holds when z has some imaginary part. but not too sure...
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#8
Sep19-11, 11:27 PM
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Quote Quote by shebbbbo View Post
Once i expanded them i realized they looked exactly like the triangle inequality where the modulus of the summation of terms was less than or equal to the modulus of each term summed.

i didnt try to conclude |e^z| = e^|z|

from what i was reading i think they are equal when z is real and the inequality holds when z has some imaginary part. but not too sure...
Good. That's about it. Except that they aren't necessarily equal when z is real and negative either, yes? It's only clearly true if z is real and postive.
shebbbbo
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#9
Sep19-11, 11:47 PM
P: 17
yeah good point.

thanks for all your help!


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