Complex analysis inequality proof

[shebbbbo]

Prove for all Z E C

|e^z-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

 

[Dick]

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

 

[shebbbbo]

great thanks...

quick question:

does |e^z| = e^|z| ?

 

[Dick]

great thanks...

quick question:

does |e^z| = e^|z| ?

You tell me. Use the series expansion to compare them.

 

[shebbbbo]

thanks!

 

[Dick]

thanks!

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

 

[shebbbbo]

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...

 

[Dick]

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 positive.

 

[shebbbbo]

yeah good point.

thanks for all your help!