Use the Cauchy Integral formula to evaluate foC (closed curve) f(z) dz,
where C is oriented anticlockwise, for f(z) = (z + i)^2 / (z + 3 - 2i)^3
and C is the circle |z + 2| = 5.
Thanks.
To evaluate the integral using the Cauchy Integral formula, we need to first check if the given function f(z) is analytic inside and on the closed curve C.
The Cauchy Integral formula states that if f(z) is analytic inside and on a simple closed curve C, and a is a point inside C, then the integral of f(z) dz over the closed curve C is equal to 2πi times the sum of the residues of f(z) at all the poles inside C.
Let's start by finding the poles of the function f(z).
The function f(z) has a pole at z = -3 + 2i, as it makes the denominator zero.
Now, let's check if the pole z = -3 + 2i is inside the closed curve C.
The equation of the circle C is |z + 2| = 5. We can rewrite it as:
|z + 2|^2 = 5^2.
Expanding the modulus squared, we get:
(z + 2)(z + 2*) = 25,
where * denotes complex conjugate.
Rearranging the equation, we get:
z + 2 = ±√25.
z = -2 ± √25.
We have two values for z, -2 + 5 = 3 and -2 - 5 = -7.
Now, let's check if the pole z = -3 + 2i lies between these two values or not.
We can see that -2 + √25 = 3 lies between -3 + 2i and -2 - √25 = -7.
Therefore, the pole z = -3 + 2i is inside the closed curve C.
Now, using the Cauchy Integral formula, we have:
∮C f(z) dz = 2πi times the residue of f(z) at the pole z = -3 + 2i.
To find the residue, we need to calculate the limit of (z - (-3 + 2i))^3 times f(z) as z approaches -3 + 2i.
The function f(z) can be rewritten as:
f(z) = (z + i)^2 / [(z - (-3 + 2i))^3].
Taking the limit as z approaches -3 + 2i, we substitute z = -3 + 2i in f(z):
f(-3 + 2i) = (-3 + 2i + i)^2 / [(-3 + 2i - (-3 + 2i))^3]
= (-3 + 3i)^2 / (0^3)
= 9 / 0.
We can observe that the limit of f(z) as z approaches -3 + 2i is undefined. Hence, we need to apply the Residue theorem to calculate the residue using differentiation techniques.
To find the residue, we differentiate f(z) with respect to z, and evaluate the result at the pole z = -3 + 2i.
Differentiating f(z) with respect to z, we get:
f'(z) = 2(z + i) / [(z + 3 - 2i)^3].
Substituting z = -3 + 2i in f'(z), we get:
f'(-3 + 2i) = 2(-3 + 2i + i) / [(-3 + 2i + 3 - 2i)^3]
= 2(-2 + 3i) / [0^3]
= 2(-2 + 3i) / 0.
Again, we observe that the limit of f'(z) as z approaches -3 + 2i is undefined.
This indicates that the function f(z) has a pole of order 2 at z = -3 + 2i, and we need to find its second-order residue.
To find the second-order residue, we can differentiate f'(z) with respect to z and evaluate the result at the pole z = -3 + 2i.
Differentiating f'(z) with respect to z, we get:
f''(z) = 2 / [(z + 3 - 2i)^3].
Substituting z = -3 + 2i in f''(z), we get:
f''(-3 + 2i) = 2 / [(-3 + 2i + 3 - 2i)^3]
= 2 / 0.
Again, we observe that the limit of f''(z) as z approaches -3 + 2i is undefined.
This indicates that the second-order residue at z = -3 + 2i is also undefined.
Therefore, the Cauchy Integral formula cannot be applied directly in this case to evaluate the integral over the closed curve C.
If you have any other questions or need further assistance, feel free to ask.