Share
Explore BrainMass

Use the given theorem to show that each of these functions is differentiable in the indicated domain of definition, and then find f ΄(z):

The problems are from complex variable class. Please specify the terms that you use if necessary and explain each step of your solution. If there is anything unclear in the problem, please tell me. Thank you very much.

4. Use the given theorem to show that each of these functions is differentiable in the indicated domain of definition, and then find f ΄(z):

(a) f (z) = 1/z4 (z ≠ 0)
(b) f (z) = sqrt(r)?ei&#952;/2 (r > 0, &#945; < &#952; < &#945; + 2&#960;) (sqrt means square root)
(c) f (z) = e-&#952; cos (ln r) + i e-&#952; sin (ln r) (r > 0, 0 < &#952; <2&#960;)

Theorem. Let the function
f (z) = u (r, &#952;) + i v (r, &#952;)
be defined throughout some &#949; neighborhood of a nonzero point z0 = r0 ei&#952;0, and suppose that the first-order partial derivatives of the functions u and v with respect to r and &#952; exist everywhere in the neighborhood. If those partial derivatives are continuous at (r0, &#952;0) and satisfy the polar form
r ur = v&#952;, u&#952; = -r vr
of the Cauchy-Riemann equations at (r0, &#952;0), then f &#900;(z0) exists.

The derivative f &#900;(z0) here can be written
f &#900;(z0) = e-i&#952;(ur + i vr),
where the right-hand side is to be evaluated at (r0, &#952;0)

Attachments

Solution Preview

a)
1/z^4= 1/r^4*e^(-4itehta)=1/r^4*cos(4theta)-i/r^4*sin(4theta)

so:
u(r,theta)=1/r^4*cos(4theta)
v(r,theta)=-1/r^4*sin(4theta)

Now we have:

ur=-4cos(4theta)/r^5
utheta=-4sin(4theta)/r^4
vr=4sin(4theta)/r^5
vtheta=-4cos(4theta)/r^4

Now we can verify that
r*ur=vtheta=-4cos(4theta)/r^4
and
utheta=-r*vr=-4sin(4theta)/r^4

Therefore the function is differentiable in the region.

f'= ...

Solution Summary

This solution is comprised of a detailed explanation to use the given theorem to show that each of these functions is differentiable in the indicated domain of definition, and then find f &#900;(z).

$2.19