Residue Calculation for Complex Analysis - Exercise Solution Discrepancy

[iflare]

Hello!

I am studing for my Complex Analysis exam and solving the exercises for Residues given by the professor.

The problem is that for some exercises I get to a solution different from the one of the professor , and I am not sure that the mistake is in my calculations.

I would greatly appreciate it, if somebody could solve it and tell me what a solution he/she came up with.

Here is the exercise:
Calculate the residue of the complex-valued function $$f(z)$$ at $$z=-\imath$$, as:

$$f(z) = \frac{\sin(z)}{(z^2 + 1)^2}$$​

My answer:

$$Res(f(z),\imath) = -\frac{1}{4e}$$​

The professor's answer:

$$Res(f(z),\imath) = \frac{\imath}{2}\cosh(1)$$​

Thanks a lot!

 

[cristo]

Well, what did you get? Post your attempt, and we'll be able to see if you went wrong.

 

[iflare]

Well, what did you get? Post your attempt, and we'll be able to see if you went wrong.

Thank you for getting involved , here is what I did, I used the formula:

$$Res(f(z), z_0) = \lim_{z \to z_0} \left( \frac{1}{(m-1)!}\, \dfrac{d^{m-1}}{dz^{m-1}} \left( (z-z_0)^m f(z) \right) \right)$$​

where $$m$$ is the order of the pole, and $$z_0$$ is the pole.

In the particular case of this exercise,

$$\begin{align*} m &= 2 \\ z_0 &= -i \end{align*}$$



$$\begin{align*} Res(f(z), -i) &= \lim_{z \to z_0} \left( \frac{1}{(m-1)!}\, \dfrac{d^{m-1}}{dz^{m-1}} \left( (z-z_0)^m f(z) \right) \right) \\ &= \lim_{z \to -i} \left( \frac{1}{1!} \, \dfrac{d}{dz} \left( (z+i)^2 f(z) \right) \right) \end{align*}$$
​

 

[iflare]

My Calculations

Here are the calculations:

$$\begin{align*} Res(f(z), -i) &= \lim_{z \to z_0} \left( \frac{1}{(m-1)!} \, \dfrac{d^{m-1}}{dz^{m-1}} \left( (z-z_0)^m f(z) \right) \right) \\ &= \lim_{z \to -i} \left( \frac{1}{1!} \, \dfrac{d}{dz} \left( (z+i)^2 f(z) \right) \right) \\ &= \lim_{z \to -i} \left( \dfrac{d}{dz} \left( (z+i)^2 \frac{\sin(z)}{(z+i)^2 (z-i)^2} \right) \right) \\ &= \lim_{z \to -i} \left( \dfrac{d}{dz} \left(\frac{\sin(z)}{(z-i)^2} \right) \right) \\ &= \lim_{z \to -i} \left( \dfrac{\cos(z) \cdot (z-i)^2 - \sin(z) \cdot 2(z-i)(z-i)'}{(z-i)^4} \right) \\ &= \dfrac{\cos(-i) \cdot (-i-i)^2 - \sin(-i) \cdot 2(-i-i)}{(-i-i)^4} = \dfrac{\cos(-i) \cdot 4(-1) +4i \sin(-i)}{(-2i)^4} \\ &= \dfrac{-4\cos(-i) +4i \sin(-i)}{16} = \dfrac{-\cos(-i) +i \sin(-i)}{4} = \dfrac{-\cos(i) -i \sin(i)}{4}\\ &= -\dfrac{\cos(i) + i \sin(i)}{4} = -\dfrac{e^{i \cdot i}}{4}= -\dfrac{e^{-1}}{4} = -\dfrac{1}{4e}\\ \end{align*}$$
​

 

[iflare]

The Professor's Solution

Here is the solution of the professor. This is a screenshot of the page of his lecture notes on which he solves the exercise. In his calculations:

$$j=\sqrt{-1}$$​

 

[benorin]

iflare, the fact of the matter is that your prof goofed the derivative and your work is correct.

 

[iflare]

iflare, the fact of the matter is that your prof goofed the derivative and your work is correct.

That is just great , thank you very much for the help!

 

[MathematicalPhysicist]

i don't think it's so great that the prof goofed here!

 

[mathwonk]

i haven't taught this in a long time, so this may be wrong, but i think the idea is to separate the pole from the holomorphic part at -i.

so we have sin(z)/(z-i)^2 as the holomorphic part and 1/(z+i)^2 as the polar part.

now we want to multiply these together and pick off the coefficient of
1/(z+i)

Now to get the coefficient of 1/(z+i) it seems we just need the derivative of the holo part at -i.

by the quotient rule that should be [cos(z)(z-i) - 2sin(z)]/(z-i)^3, all evaluated at z= -i. yipes!

i.e. -[cos(i)+i sin(i)]/4 = -1/4e,

using the fact that cos(z) = (1/2)[e^z + e^(-z)], etc...

so its much easier than it looks above, but still tedious for me.

 

[mathwonk]

did you understand my post? the point was that if you multiply a holomorphic power series by 1/(z-a)^2, the residue at a will be given by the coefficient of the linear term of the powers eries. i.e.all the exponents of (z-a) go down by 2, so the linear etrm becomes the -1 order term.