# Complex analysis inequality proof

by shebbbbo
Tags: analysis, complex, inequality, proof
 P: 17 Prove for all Z E C |ez-1| $\leq$ e|z| - 1 $\leq$ |z|e|z| I think this has to be proven using the triangle inequality but not sure how. Please help. :) thanks
 Sci Advisor HW Helper Thanks P: 24,980 Use the triangle inequality along with the power series expansion e^z=1+z+z^2/2!+...
 P: 17 great thanks... quick question: does |ez| = e|z| ????
HW Helper
Thanks
P: 24,980

## Complex analysis inequality proof

 Quote by shebbbbo great thanks... quick question: does |ez| = e|z| ????
You tell me. Use the series expansion to compare them.
 P: 17 thanks!
HW Helper
Thanks
P: 24,980
 Quote by shebbbbo thanks!
Glad you got it. Just out of curiousity, what did you conclude about the truth of |e^z|=e^|z|?
 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...