answer:To find the order and type (pole or zero) of the function f(z) = frac{z^3 + 2z^2 + z}{(z-1)(z-2)^2} at z = 0, 1, and 2, we will analyze the numerator and denominator of the function at each of these points. 1. At z = 0: The numerator is 0^3 + 2(0)^2 + 0 = 0. The denominator is (0-1)(0-2)^2 = 1 cdot 4 = 4. Since the numerator is zero and the denominator is non-zero, there is a zero at z = 0. To find the order of the zero, we can look at the lowest power of z in the numerator, which is z. Therefore, the order of the zero at z = 0 is 1. 2. At z = 1: The numerator is 1^3 + 2(1)^2 + 1 = 4. The denominator is (1-1)(1-2)^2 = 0 cdot 1 = 0. Since the numerator is non-zero and the denominator is zero, there is a pole at z = 1. To find the order of the pole, we can look at the power of (z-1) in the denominator, which is 1. Therefore, the order of the pole at z = 1 is 1. 3. At z = 2: The numerator is 2^3 + 2(2)^2 + 2 = 16. The denominator is (2-1)(2-2)^2 = 1 cdot 0 = 0. Since the numerator is non-zero and the denominator is zero, there is a pole at z = 2. To find the order of the pole, we can look at the power of (z-2) in the denominator, which is 2. Therefore, the order of the pole at z = 2 is 2. In summary: - There is a zero of order 1 at z = 0. - There is a pole of order 1 at z = 1. - There is a pole of order 2 at z = 2.

answer:Let's first recall the definition of the residue of a function at a pole. If f(z) has a pole of order m at z=a, then the residue of f(z) at z=a is the coefficient of the frac{1}{(z-a)^{m-1}} term in the Laurent series expansion of f(z) around z=a. Since f(z) has a zero of order 3 at z=2+3i and a pole of order 2 at z=1-i, we can write f(z) as: f(z) = (z - (2+3i))^3 g(z) where g(z) is an analytic function with a pole of order 2 at z=1-i. Now, we can rewrite g(z) as: g(z) = frac{h(z)}{(z - (1-i))^2} where h(z) is an analytic function with no zeros or poles on or inside the circle |z|=5. Combining the expressions for f(z) and g(z), we have: f(z) = frac{(z - (2+3i))^3 h(z)}{(z - (1-i))^2} Now, we want to find the residue of f(z) at z=1-i. To do this, we can find the coefficient of the frac{1}{(z-(1-i))} term in the Laurent series expansion of f(z) around z=1-i. Let's first expand (z - (2+3i))^3 using the binomial theorem: (z - (2+3i))^3 = (z-(1-i) + (1+4i))^3 = (1+4i)^3 + 3(1+4i)^2(z-(1-i)) + 3(1+4i)(z-(1-i))^2 + (z-(1-i))^3 Now, we can substitute this expression into the expression for f(z): f(z) = frac{((1+4i)^3 + 3(1+4i)^2(z-(1-i)) + 3(1+4i)(z-(1-i))^2 + (z-(1-i))^3) h(z)}{(z - (1-i))^2} The residue of f(z) at z=1-i is the coefficient of the frac{1}{(z-(1-i))} term in this expression. We can see that the only term that contributes to this coefficient is the 3(1+4i)^2(z-(1-i)) h(z) term. Therefore, the residue of f(z) at z=1-i is: 3(1+4i)^2 h(1-i) Since h(z) is analytic at z=1-i, we can simply evaluate h(1-i) to find the residue. However, we are not given any information about h(z), so we cannot determine the exact value of the residue. The final answer is: text{Residue of } f(z) text{ at } z=1-i = 3(1+4i)^2 h(1-i)

answer:To find the order of the pole of the given function f(z) = (z-1)/(z^2 + 2z + 5) at the point z = 2, we first need to check if z = 2 is indeed a pole of the function. A pole of a function is a point where the function becomes infinite due to the denominator being zero. In this case, we need to check if the denominator becomes zero at z = 2: z^2 + 2z + 5 = (2)^2 + 2(2) + 5 = 4 + 4 + 5 = 13 Since the denominator does not become zero at z = 2, there is no pole at z = 2. Therefore, the order of the pole at z = 2 is not applicable, as there is no pole at this point.

answer:To determine the singularities of the function f(z), we first need to find the points where the denominator is equal to zero. The denominator is given by: z^2 + 2z + 1 This is a quadratic equation, which can be factored as: (z + 1)^2 Setting this equal to zero, we find that the only singularity is at z = -1. Now we need to determine the type and order of the singularity. Since the denominator has a factor of (z + 1)^2, the singularity at z = -1 is a pole. To find the order of the pole, we look at the exponent of the factor (z + 1) in the denominator. In this case, the exponent is 2, so the order of the pole is 2. In summary, the function f(z) has a pole of order 2 at z = -1.