What is the Limit of This Complex Function as z Approaches i?

[TheCanadian]

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?

 

[Charles Link]

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.

 

[fresh_42]

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.

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

 

[Charles Link]

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

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.

 

[fresh_42]

Can you supply a little more detail and/or try computing it again? z^4-1 is in the denominator.

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.

 

[TheCanadian]

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.

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.

 

[Charles Link]

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.

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

 

[TheCanadian]

It needs to be $z^2+1$.

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

 

[fresh_42]

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

 

[Charles Link]

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

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.)

 

[PeroK]

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?

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.

 

[Charles Link]

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.

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.

 

[PeroK]

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.