The Cauchy-Goursat theorem states that if a function is analytic (i.e., complex differentiable) at all points within and on a simple closed contour C, then the integral of the function around C is zero.

Let's apply this theorem to each of the given functions:

(a) f(z) = z^2/(z-4)
This function is analytic everywhere in the complex plane except at z = 4. Since 4 is not within or on the unit circle |z| = 1, the function is analytic at all points within and on the contour. Therefore, by the Cauchy-Goursat theorem, ∫C f(z) dz = 0.

(b) f(z) = sin(z)/(z^2+4)
This function is analytic everywhere in the complex plane. Therefore, by the Cauchy-Goursat theorem, ∫C f(z) dz = 0.

(c) f(z) = tan(z)
The function tan(z) has singularities at z = (n + 1/2)π, where n is an integer. However, none of these singularities lie within or on the unit circle |z| = 1. Therefore, by the Cauchy-Goursat theorem, ∫C f(z) dz = 0.

(d) f(z) = Log(z + 3)
The function Log(z + 3) has a branch cut along the negative real axis from -3 to -∞. As long as the contour |z| = 1 does not cross this branch cut, the function is analytic at all points within and on the contour. Therefore, by the Cauchy-Goursat theorem, ∫C f(z) dz = 0.

Question

The Cauchy-Goursat theorem states that if a function is analytic (i.e., complex differentiable) at all points within and on a simple closed contour C, then the integral of the function around C is zero.

Let's apply this theorem to each of the given functions:

(a) f(z) = z^2/(z-4)
This function is analytic everywhere in the complex plane except at z = 4. Since 4 is not within or on the unit circle |z| = 1, the function is analytic at all points within and on the contour. Therefore, by the Cauchy-Goursat theorem, ∫C f(z) dz = 0.

(b) f(z) = sin(z)/(z^2+4)
This function is analytic everywhere in the complex plane. Therefore, by the Cauchy-Goursat theorem, ∫C f(z) dz = 0.

(c) f(z) = tan(z)
The function tan(z) has singularities at z = (n + 1/2)π, where n is an integer. However, none of these singularities lie within or on the unit circle |z| = 1. Therefore, by the Cauchy-Goursat theorem, ∫C f(z) dz = 0.

(d) f(z) = Log(z + 3)
The function Log(z + 3) has a branch cut along the negative real axis from -3 to -∞. As long as the contour |z| = 1 does not cross this branch cut, the function is analytic at all points within and on the contour. Therefore, by the Cauchy-Goursat theorem, ∫C f(z) dz = 0.

Knowee AI · Accepted Answer

The Cauchy-Goursat theorem states that if a function is analytic (i.e., complex differentiable) at all points within and on a simple closed contour C, then the integral of the function around C is zero.

Let's apply this theorem to each of the given functions:

(a) f(z) = z^2/(z-4)
This function is analytic everywhere in the complex plane except at z = 4. Since 4 is not within or on the unit circle |z| = 1, the function is analytic at all points within and on the contour. Therefore, by the Cauchy-Goursat theorem, ∫C f(z) dz = 0.

(b) f(z) = sin(z)/(z^2+4)
This function is analytic everywhere in the complex plane. Therefore, by the Cauchy-Goursat theorem, ∫C f(z) dz = 0.

(c) f(z) = tan(z)
The function tan(z) has singularities at z = (n + 1/2)π, where n is an integer. However, none of these singularities lie within or on the unit circle |z| = 1. Therefore, by the Cauchy-Goursat theorem, ∫C f(z) dz = 0.

(d) f(z) = Log(z + 3)
The function Log(z + 3) has a branch cut along the negative real axis from -3 to -∞. As long as the contour |z| = 1 does not cross this branch cut, the function is analytic at all points within and on the contour. Therefore, by the Cauchy-Goursat theorem, ∫C f(z) dz = 0.

Apply the Cauchy-Gaursat theorem to show that RC f (z) dz = 0 when the contourC is the unit circle |z| = 1, in either direction, and when(a) f (z) = z2z−4 (b) f (z) = sin zz2+4 c f (z) = tan z (d) f (z) = Log(z + 3)

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