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

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4. 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):

(a) f (z) = 1/z4 (z &#8800; 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)

https://brainmass.com/math/trigonometry/function-indicated-domain-definition-33520

#### 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).

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