Another calculus proof

1. Mar 14, 2006

stunner5000pt

Prove that $$\lim_{z \rightarrow -i} 1/z = i$$ using the definition of a limit

ok from teh defition of limit we know that
$$|z+i| < \delta$$

also we need to show that $$|\frac{1}{z} - i| < \epsilon$$
$$|\frac{1}{z} - i| = |\frac{1}{z} + z - z + i| \leq |\frac{1}{z} - z| + |z+i| < |\frac{1}{z} - z| + \delta$$

stuck here... do i just say the above is less than delta 1 and then pick an epsilon which si the min of delta 1 and delta?

2. Mar 14, 2006

StatusX

Well, this is a pretty silly problem. You obviously don't need to use limits to get the answer. But if you must, your approach won't work, since |z-1/z| isn't approaching zero.

You want to prove that, given any $\epsilon$>0, there exists a $\delta$>0 such that if:

$$|z+i|<\delta$$

Then:

$$|1/z-i|<\epsilon[/itex] I'm sure you know this, but what you've written doesn't make it very clear that this is what you're trying to prove. So $\epsilon$ is some number, and you want to find the corresponding $\delta$ that makes the above true. Use: [tex]|1/z-i|=|\frac{1-iz}{z}|=|\frac{i+z}{iz}|=\frac{|i+z|}{|iz|}$$

Do you see where to go from here?

3. Mar 14, 2006

stunner5000pt

$$\frac{|i+z|}{|iz|} < \frac{\delta}{iz} = -\frac{i\delta}{z}$$

now im not sure ...

4. Mar 14, 2006

StatusX

The magnitude of a complex number is always real. Once you fix that, you need to pick delta so that this expression alway evaluates to something less than epsilon. (by the way, you used what you're trying to prove in your second step there. Not that there's anything wrong with that, it just shows what a silly problem this is)

Last edited: Mar 14, 2006
5. Mar 14, 2006

stunner5000pt

so that denominator turns into x^2 + y^2??

6. Mar 14, 2006

StatusX

OK. Now, given z is in a circle of radius delta around -i, what is the maximum value of the expression?

7. Mar 14, 2006

stunner5000pt

so the magnitude of the numerator is
$$|i+z| = \sqrt{x^2 + (y+1)^2}$$ ??

the denominator is simply $$\sqrt{x^2 + y^2}$$

8. Mar 14, 2006

stunner5000pt

so then the max vcalue would be the distance that z is awawy from teh edge of the circumference?

z + delta - |-i-z| ??

9. Mar 14, 2006

StatusX

Like I've been saying, you want to calculate the maximum value that |1/z-i| could acheive given that |z+i|<delta. Once you know this, you know how to choose delta, given an epsilon, that guarantees |1/z-i|<epsilon. This is the essence of the epsilon-delta definition of a limit. The maximum value of the expression you found:

$$\frac{\delta}{|z|^2}$$

can be found geometrically when you consider that z must lie inside a circle of radius delta around the point i. To find the max of this expression, you want to minimize the denominator.

Last edited: Mar 14, 2006
10. Mar 14, 2006

stunner5000pt

minimize the denominator...
not sure what that means here.
to minimize te z we need to make z small by finding z's smallest value in that cirle u describe
but the smallest value would be when z = -i - delta?j The center loess the radius on the left?

11. Mar 15, 2006

StatusX

Well, you might have a sign error there, but what's important is the minimum value of |z|2. And do you understand why you need this?

12. Mar 15, 2006

greytomato

always try to simplify the expression:

1/z = (1*z')/(z*z') = z'/|z^2| = ...