# I Limit of complex function

1. Sep 22, 2016

I am trying to find the limit of $\frac {z^2 + i}{z^4 - 1}$ as $z$ approaches $i$.

I've broken the solution down to: $\frac {(z + \sqrt{i})(z - \sqrt{i})}{(z+1)(z-1)(z+i)(z-i)}$ but this does not seem to get me anywhere. The solution says $-0.5$ but I don't quite understand how they arrived at that answer. I may be missing something very obvious, but any thoughts on how I can change my approach?

2. Sep 22, 2016

I wonder if the book made a mistake or you copied the problem incorrectly. The approach I would recommend is a Taylor expansion of the denominator about z=i and similarly for the numerator. I could have miscalculated, but I don't get a finite limit. I think the numerator should say z^2+1. ..editing... Doing that does give the book's answer.

Last edited: Sep 22, 2016
3. Sep 22, 2016

### Staff: Mentor

And I made a polynomial division of the inverse quotient and got $0$.

4. Sep 22, 2016

editing =I missed the term "inverse quotient". But make the numerator $z^2+1$ and you should get the book's answer. I think it's a typo somewhere i.e. either the OP copied it wrong or the book has a typo.

5. Sep 22, 2016

### Staff: Mentor

Yes. I divided $(z^4-1) : (z^2+i) = z^2 - i - \frac{2}{z^2+i}$ which is $0$ for $z=i$. That is to say you were right, the limit in question doesn't exists.

6. Sep 22, 2016

That is possible. My e-copies are slightly blurry, but I've attached the problem and solution. I will admit the limit of $z$ isn't very clear in the image, but that seems to be an i based on previous questions.

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7. Sep 22, 2016

It needs to be $z^2+1$ in the numerator.

8. Sep 22, 2016

Then the problem is resolved. Likely just a typo in the book—thanks for pointing it out.

9. Sep 22, 2016

### Staff: Mentor

However, I've typed it into Wolfram and got -1 as the result. They made a Taylor expansion at i.
I'm confused.

10. Sep 22, 2016

Take a few minutes and recompute it... The denominator gives (using $f(z)=z^4-1$) $\$ $f(z)=f(i)+f'(i)(z-i)+...=0+4z^3 (z-i) +... =4(i^3)(z-i)$ Meanwhile $z^2+1$ factors as $(z+i)(z-i)$. (Could do Taylor series in numerator also, but this is quicker.)

11. Sep 23, 2016

### PeroK

Limits are only tricky if they result in something of the form $0/0$ or $\pm \infty/\pm \infty$. In this case, you simply set $z = i$ to get something of the form $(-1 + i)/0$ and you know the limit does not exist without the need for further analysis.

12. Sep 23, 2016

The problem on this one is the OP had the answer the book gave, and it did not agree with what we were computing. Now that we believe it should read $(z^2+1)/(z^4-1)$ , a simple factoring of the denominator $(z^2+1)/((z^2-1)(z^2+1))$ is probably the quickest approach. The Taylor series also worked well for this one.
Yes, but with the question as originally quoted there was not a need for any analysis. Nor was any analysis going to change the simple answer that could be obtained by setting $z = i$ in the original expression.