# Complex Analysis - Residues

## Main Question or Discussion Point

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!

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## Answers and Replies

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

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*}

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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*}

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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:
$$j=\sqrt{-1}$$​

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

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

mathwonk
Science Advisor
Homework Helper
i havent 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.

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mathwonk
Science Advisor
Homework Helper
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.