Can someone explain to me how this is not 1/4? Complex Analy

1. Aug 6, 2015

RJLiberator

1. The problem statement, all variables and given/known data
Hi all, I posted the image of the solution here.

My questions concerns the evaluation of the Residue at z=i.

2. Relevant equations
None needed, all giving in the question -- simple algebraic mistake made is likely to be the problem here.

3. The attempt at a solution

We see that phi function is clearly equal to 1/(z+i)^2 as the show.
Since it is a pole of order two, we must differentiate this function first before evaluation at singular point i.

The derivative of this function should be -2/(z+i)^(3)

Now we can evaluate this at point z=i.
-2/(2i)^3 = -2/8i^3 = -1/4i^3. When I calculate this, this becomes i/4

Where am I going wrong in my complex algebra?

Last edited: Aug 6, 2015
2. Aug 6, 2015

QIsReluctant

Looks like a simple algebra mistake. When you simplify for the second time you eliminate the negative sign, effectively dividing by -1. That's why your answer differs by a factor of -1.

3. Aug 6, 2015

RJLiberator

Wopps. good catch, that was a mistake in my post, but not the mistake i wanted to clarify. (I've seen edited my original post for future readers).

The problem is, when I evaluate this, it comes to -1/4i^3 and I believe this is equal to i/4 as per wolfram alpha, etc.
However, in the image, it shows that this is not the case. Instead, the image shows the solution to be 1/4i.

What's going on here? My only other line of thinking is this: They split up -1/4i^3 into -1/(4i^2*i)

Oh...

That works...

-1/(4i^2*i) = 1/4i
But, why would they not simplify this further so that it is i/4? The final answer would then be the negative of what the solution offers.

4. Aug 6, 2015

QIsReluctant

But, why would they not simplify this further so that it is i/4?
Because that would not be a valid simplification. 1/i = -i, and so we would need -i/4 and not i/4.

I'd be interested in seeing this Wolfram Alpha output you speak of ...

5. Aug 6, 2015

RJLiberator

Ah, I see. The that is indeed the mistake I was performing. Did not have parentheses around the denominator in wolfram alpha.

So it does come out to be -i/4. Which with that negative factor, makes them equivalent.

You have successfully helped me solve this problem and furthered my understanding.
Thank you.

6. Aug 6, 2015

QIsReluctant

Aha! That's the reason.