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Complex analysis inequality proof 
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#1
Sep1811, 10:56 PM

P: 17

Prove for all Z E C
e^{z}1 [itex]\leq[/itex] e^{z}  1 [itex]\leq[/itex] ze^{z} I think this has to be proven using the triangle inequality but not sure how. Please help. :) thanks 


#2
Sep1811, 11:11 PM

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Use the triangle inequality along with the power series expansion e^z=1+z+z^2/2!+...



#3
Sep1911, 02:58 AM

P: 17

great thanks...
quick question: does e^{z} = e^{z} ???? 


#4
Sep1911, 09:05 AM

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Complex analysis inequality proof



#5
Sep1911, 09:11 PM

P: 17

thanks!



#6
Sep1911, 09:23 PM

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#7
Sep1911, 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... 


#8
Sep1911, 11:27 PM

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#9
Sep1911, 11:47 PM

P: 17

yeah good point.
thanks for all your help! 


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