# Thread: Limit of complex function

1. ## Limit of complex function

Hi,

Can someone explain how to evaluate:

$\lim_{\Re (z)\to-\infty} \left|\frac{1+\frac{1}{2}z}{1-\frac{1}{2}z}\right|$

Would I need L'Hopital's rule?

2. Originally Posted by scorpion007
Hi,

Can someone explain how to evaluate:

$\lim_{\Re (z)\to-\infty} \left|\frac{1+\frac{1}{2}z}{1-\frac{1}{2}z}\right|$

Would I need L'Hopital's rule?
I assume the imaginary part of z is remaining finite .....? $\lim_{x \rightarrow - \infty} \left| \frac{1 + \frac{1}{2} x + i \frac{1}{2} y}{1 - \frac{1}{2} x - i \frac{1}{2} y} \right| = \lim_{x \rightarrow - \infty} \left| \frac{\frac{1}{x} + \frac{1}{2} + i \frac{y}{2x}}{\frac{1}{x} - \frac{1}{2} - i \frac{y}{2x}}\right|$

3. Yes, I believe the Im(z) remains finite. I'm basically trying to understand how Wikipedia found the limit on this page: Stiff equation - Wikipedia, the free encyclopedia , section "Example: The Euler and trapezoidal methods".

So, about what you did: I assume $y$ is the imaginary part of z? Where did the $i$ vanish to?

But from what you did, it looks like the limit becomes: (since the other terms vanish).

$\lim_{x\to -\infty} \left|\frac{ +\frac{1}{2} }{- \frac{1}{2} }\right| = |-1| = 1$.

Which is what wikipedia got, but is my working right?

4. Originally Posted by scorpion007
Yes, I believe the Im(z) remains finite. I'm basically trying to understand how Wikipedia found the limit on this page: Stiff equation - Wikipedia, the free encyclopedia , section "Example: The Euler and trapezoidal methods".

So, about what you did: I assume $y$ is the imaginary part of z? Where did the $i$ vanish to?

But from what you did, it looks like the limit becomes: (since the other terms vanish).

$\lim_{x\to -\infty} \left|\frac{ +\frac{1}{2} }{- \frac{1}{2} }\right| = |-1| = 1$.

Which is what wikipedia got, but is my working right?
Well you know what they say, it's all fun amd laughter until someone loses an i ..... I made a typo which I've since fxed. What you have looks fne.

5. Originally Posted by scorpion007
Hi,

Can someone explain how to evaluate:

$\lim_{\Re (z)\to-\infty} \left|\frac{1+\frac{1}{2}z}{1-\frac{1}{2}z}\right|$

Would I need L'Hopital's rule?
I think it's much easier: put z = x + iy, so [1 +(1/2)z]/[1 - (1/2)z] = [2 + x + iy]/[2 - x - iy] = 1 + 2[x +iy]/[2 - x - iy] --> 1 + 2*(-1) = -1 when x = Re(z) --> -oo

By the way, the same result we get when Re(z) --> oo

Tonio

6. tonio, thanks for the comment, but I don't quite understand how you got from:

$\frac{2 + x + iy}{2 - x - iy} = 1 + \frac{2(x +iy)}{2 - x - iy}$

(even though they are indeed equal)

and then got 1 + 2*(-1) = -1 when x = Re(z) --> -oo

7. Originally Posted by scorpion007
tonio, thanks for the comment, but I don't quite understand how you got from:

$\frac{2 + x + iy}{2 - x - iy} = 1 + \frac{2(x +iy)}{2 - x - iy}$

(even though they are indeed equal)

and then got 1 + 2*(-1) = -1 when x = Re(z) --> -oo
$\frac{2 + x + iy}{2 - x - iy} = \frac{(2 - x - iy) + 2x + 2iy}{2 - x - iy} = \frac{(2 - x - iy)}{2 - x - iy} + \frac{2x + 2iy}{2 - x - iy}$

8. Originally Posted by scorpion007
tonio, thanks for the comment, but I don't quite understand how you got from:

$\frac{2 + x + iy}{2 - x - iy} = 1 + \frac{2(x +iy)}{2 - x - iy}$

(even though they are indeed equal)

and then got 1 + 2*(-1) = -1 when x = Re(z) --> -oo

it's a basic algebra thing: if we have a quotient Q/P, where Q, P are polynomials on some unknow t of the same degree (on t), then we can always write Q/P = 1 - Q'/P, for some pol. in t of degree at most deg(Q)

After that all is simple limits: as it is x --> -oo all the rest is taken as constants, and thus (x + constant)/(-x + constant) --> -1

Tonio