# Thread: Limit of a complex variable

1. ## Limit of a complex variable

lim z to i (z^2+i)/(z^4-1)=?

2. Take into account that is $\displaystyle z^{4}-1= (z^{2}+1) (z^{2}-1)= (z-i) (z+i) (z-1) (z+1)$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. Yes, I did the same and I've found:

lim z->i (z^2+i)/(z^4-1) = lim z->i (1/(z^2-i)) = 1/(i^2-i) = 1/(-1-i) = (-1/((1+i)) * (1-i/(1-i)) = (-1+i)/(1-i^2) = 1+i/(1+1) = (1+i)/2

but in the solution manual it says the answer is: -1/2

4. Originally Posted by chisigma
Take into account that is $\displaystyle z^{4}-1= (z^{2}-i) (z^{2}+i)$...

I'm afraid it is not: $\displaystyle (z^2+i)(z^2-i)=z^4-i^2=z^4+1$ ...A little blunder, but

what is true is $\displaystyle z^4-1=(z^2-1)(z^2+1)=(z-1)(z+1)(z-i)(z+i)$

Tonio

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
.

5. Originally Posted by essedra
Yes, I did the same and I've found:

lim z->i (z^2+i)/(z^4-1) = lim z->i (1/(z^2-i)) = 1/(i^2-i) = 1/(-1-i) = (-1/((1+i)) * (1-i/(1-i)) = (-1+i)/(1-i^2) = 1+i/(1+1) = (1+i)/2

but in the solution manual it says the answer is: -1/2

The solution manual is wrong unless you miswrote the numerator and it actually is $\displaystyle z^2+1$. As you

wrote it, the limit doesn't exist.

Tonio

6. I've also checked out the cramster.com solution, in the nominator they've written z^2+1, I and found -1/2. So, you're right that there's an error whether in the question or the solution manual. You helped me to correct my misunderstanding, anyway. Many thanks for your help.

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