Residues, describing contours: Find res[exp(1/z)/z^2-16,4], res[exp(1/z)/z^2-16,-4],

tuaisceart

New member
Joined
Nov 6, 2017
Messages
2
Any help would be great.

Find;

i) res[exp(1/z)/z^2-16,4]

ii) res[exp(1/z)/z^2-16,-4]

iii) res[exp(1/z)/z^2-16,0]

And then describe all contours of the integral

iv) ~ exp(1/z)/z^2+16 dz = 0
 

tkhunny

Moderator
Staff member
Joined
Apr 12, 2005
Messages
10,214
Well, there is one thing that can help. If you're in Complex Analysis, you should know the difference between

1/z^2 - 16 = \(\displaystyle \dfrac{1}{z^{2}}-16\)

and

1/(z^2 - 16) = \(\displaystyle \dfrac{1}{z^{2}-16}\)

After that, have you considered identifying the Order of the Pole and simply computing the appropriate limit?

Please follow the forum guidelines - this includes showing YOUR work.
 

tuaisceart

New member
Joined
Nov 6, 2017
Messages
2
Apologies, it was an error on my part, I forgot to include the brackets.

So far I have exp(1/z)/(z+4)(z-4) so z = +/- 4

For z = 4 ; exp(1/4)/(4+4)/(4-4) = 0.161

For z = -4 ; exp(1/-4)/(-4-4)/(-4+4) = -0.097

For z = 0; exp(1)/-1 = -2.72

For the second part, the contour exp(1/z)/(z+4i)(z-4i);

z = +/- 4i

2(pi)i * [exp(1/4i)/8i + exp(1/-4i)/-8i]

Not sure if I am correct so any bit of help would be great. Thanks
 

tkhunny

Moderator
Staff member
Joined
Apr 12, 2005
Messages
10,214
Let's get a little better with notation, please. Good notation will save you. It matters.

How can you write "/(4-4)" with a straight face? That's no good.

Are you sure about z = 0?
 
Top