Now you can take the limit as ##z## approaches ##-1##.

[jjr]

Homework Statement

Calculate the following limit if it exists

$\lim_{z\to -1}\frac{\sqrt{z}-i+\sqrt{z+1}}{\sqrt{z^2-1}}$

the branch of root is chosen so that $\sqrt{-1}=i$

Homework Equations

The Attempt at a Solution

I tried most of the same things that I tried earlier today ( https://www.physicsforums.com/threads/complex-limit-help.813800/ ).

1. No obvious way to simplify the expression and get rid of the zero in the denominator.

2. Tried to multiply the numerator and denominator with the complex conjugate $\sqrt{\bar{z}^2-1}$, which did give me a $\sqrt{2}$ in the denominator. The problem with this is that if $z$ goes to $-1$, then I suppose $\bar{z}$ goes to $-1$ as well. This means, in effect, that I multiplied with 0, and I don't think it's legit. (Yields a wrong answer (0) anyway).

3. Tried using l'Hopitals rule, but it doesn't remove any of the radicals which will still go to 0 no matter how many times I apply it.

4. Tried the transformation $z = re^{i\theta}$, but can't see how it would get me any further.

Any ideas?

Thanks,
J

 

[Dazed&Confused]

edit: mistake

 

[jjr]

I will take your word, because it makes perfect sense and it's the correct answer (suppose I should've mentioned that). Thanks.

 

[Dazed&Confused]

No there was a mistake there. The first term should be multiplied by $\sqrt{z} + i$

You get $$\frac{(\sqrt{z}-i)(\sqrt{z}+i)}{\sqrt{(z-1)(z+1)}(\sqrt{z} + i)} + \frac{\sqrt{z+1}}{\sqrt{(z-1)(z+1)}} = \frac{\sqrt{z+1}}{(\sqrt{z-1})(\sqrt{z} + i )} + \frac{1}{\sqrt{z-1}}$$