Originally posted by: RaynorWolfcastle
I do, but I really don't feel like looking it up right now; I'll check it out tomorrow if you like.
Originally posted by: hypn0tik
Originally posted by: RaynorWolfcastle
I do, but I really don't feel like looking it up right now; I'll check it out tomorrow if you like.
Hahaha. Well, I have a quiz tomorrow hence the late night post.
Don't worry about it though, I think I'll be ok.
Originally posted by: shimsham
Originally posted by: hypn0tik
Originally posted by: RaynorWolfcastle
I do, but I really don't feel like looking it up right now; I'll check it out tomorrow if you like.
Hahaha. Well, I have a quiz tomorrow hence the late night post.
Don't worry about it though, I think I'll be ok.
then tell us.
Originally posted by: hypn0tik
Originally posted by: shimsham
Originally posted by: hypn0tik
Originally posted by: RaynorWolfcastle
I do, but I really don't feel like looking it up right now; I'll check it out tomorrow if you like.
Hahaha. Well, I have a quiz tomorrow hence the late night post.
Don't worry about it though, I think I'll be ok.
then tell us.
State the location of all the poles for the following function and give the order of each ple. Use the principal branch of the given functions if there is any ambiguity.
1/[cosh(z) -ae^z] where a is real.
The poles occur when:
cosh(z) -ae^z = 0
1/2 e^z +1/2 e^-z -ae^z = 0
(1-2a)e^z +e^-z = 0
(1-2a)e^2z + 1 = 0
e^2z = 1/(2a-1)
z = 1/2 log (1/(2a-1))
Right?
There are no answers in the back so I can't verify my result.
Originally posted by: shimsham
Originally posted by: hypn0tik
Originally posted by: shimsham
Originally posted by: hypn0tik
Originally posted by: RaynorWolfcastle
I do, but I really don't feel like looking it up right now; I'll check it out tomorrow if you like.
Hahaha. Well, I have a quiz tomorrow hence the late night post.
Don't worry about it though, I think I'll be ok.
then tell us.
State the location of all the poles for the following function and give the order of each ple. Use the principal branch of the given functions if there is any ambiguity.
1/[cosh(z) -ae^z] where a is real.
The poles occur when:
cosh(z) -ae^z = 0
1/2 e^z +1/2 e^-z -ae^z = 0
(1-2a)e^z +e^-z = 0
(1-2a)e^2z + 1 = 0
e^2z = 1/(2a-1)
z = 1/2 log (1/(2a-1))
Right?
There are no answers in the back so I can't verify my result.
uhh...er....
drpizza? einstein?

 
				
		