how do I identify the regions on which the functions below and analytic?

$\displaystyle \sqrt{e^z+ + 1}$

$\displaystyle \frac{1}{e^z -1}$

Do I just differentiate them and then take the cauchy-reimann equations?

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- Nov 11th 2012, 04:11 PMflametag3Differentiating complex variables
how do I identify the regions on which the functions below and analytic?

$\displaystyle \sqrt{e^z+ + 1}$

$\displaystyle \frac{1}{e^z -1}$

Do I just differentiate them and then take the cauchy-reimann equations? - Nov 11th 2012, 11:49 PMchiroRe: Differentiating complex variables
Hey flametag3.

If these functions are analytic then they will be finite everywhere where they are analytic which means the modulus will be < infinity.

So recall that the modulus square can be calculated by doing z * z_bar for a complex number z so from this hint, one way is calculate the modulus of the function mapping (which is just a complex variable and subsequently just a complex number).

Another way is by examination showing that for example e^z is defined everywhere for all complex z and so is e^z + 1 and square root of this is defined everywhere, while 1/(e^z - 1) is not defined at z = 0.

You can show this with the modulus where the modulus is always less than infinity as long as z is not infinity but for the second one if z = 0 then the modulus blows up to infinity. - Nov 13th 2012, 03:09 PMflametag3Re: Differentiating complex variables
Thank you for the information. I do understand what you are saying and reading the book now makes a little more sense. is it possible that you could show me this for one of the parts? it would be greatly appreciated.

- Nov 15th 2012, 12:32 AMchiroRe: Differentiating complex variables
Well first we consider f(z) = e^z + 1.

e^z + 1 can be shown to be analytic in a number of ways including showing that its taylor expansion is the usual expansion (z^n/n!) for n = 1 to infinity + 2 so its residue is 0 meaning analytic everywhere.

You can also write z = x+y and then show this through Cauchy-Riemann.

Now consider f(z) = SQRT(z) = SQRT(r)*e^(i*z/2). We have shown e^z is analytic for all z in C and SQRT(r) is analytic for all r > 0 by real variables so SQRT(z) is analytic.

Now consider compositions of analytic functions and show that if both are entire, then the composition is as well.

If you want to consider just the function in terms of its bounded-ness then we know that |e^z| < infinity for all z in C (not included the infinity point) since |e^z| = e^(x) * |e^(iy)| = e^x < infinity and |e^z + 1| <= |e^z| + 1 < infinity. So that is bounded.

Now for SQRT(z) we have |SQRT(z)| = |r^(1/2)*e^(iy)| = r^(1/2) < infinity for all r >= 0 for all z in C.

So with the chain of maps e^(z) + 1 -> SQRT(z) -> f(z) you have all of these are bounded at every composition.

In terms of compositions, this is the kind of thing that you will see if you look in pure mathematics particularly for things like analysis and topology. I don't know what is covered in your book though.

Also to formulize you should show that e^(z) + 1 maps to the complex plane in the range and since SQRT(z) maps back to the complex numbers and e^(z) + 1 maps to z in the range then the final map will be bounded based on the arguments above.

I'm not exactly sure how your professor expects it in terms of specifics, but those should help you fill in the gaps.