# Complex analysis inequality proof

1. ### shebbbbo

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.

thanks

2. ### Dick

25,849
Use the triangle inequality along with the power series expansion e^z=1+z+z^2/2!+...

3. ### shebbbbo

17
great thanks...

quick question:

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

4. ### Dick

25,849
You tell me. Use the series expansion to compare them.

17
thanks!

6. ### Dick

25,849
Glad you got it. Just out of curiousity, what did you conclude about the truth of |e^z|=e^|z|?

7. ### shebbbbo

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. ### Dick

25,849
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.

9. ### shebbbbo

17
yeah good point.