Complex Analysis

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Q1 Find the poles of f ( z ) = z 2 + 4 z 3 + 2 z 2 + 2 z f ( z ) = z 2 + 4 z 3 + 2 z 2 + 2 z f(z)=(z^(2)+4)/(z^(3)+2z^(2)+2z)f(z)=\frac{z^2+4}{z^3+2 z^2+2 z}f(z)=z2+4z3+2z2+2z and determine the residues at the poles.
Answer:
To find the poles of the function f ( z ) = z 2 + 4 z 3 + 2 z 2 + 2 z f ( z ) = z 2 + 4 z 3 + 2 z 2 + 2 z f(z)=(z^(2)+4)/(z^(3)+2z^(2)+2z)f(z) = \frac{z^2 + 4}{z^3 + 2z^2 + 2z}f(z)=z2+4z3+2z2+2z, we need to determine the values of z z zzz for which the denominator is zero, as these are the points where the function becomes undefined. The poles are the roots of the equation:
z 3 + 2 z 2 + 2 z = 0 z 3 + 2 z 2 + 2 z = 0 z^(3)+2z^(2)+2z=0z^3 + 2z^2 + 2z = 0z3+2z2+2z=0
We can factor out a z z zzz from the equation:
z ( z 2 + 2 z + 2 ) = 0 z ( z 2 + 2 z + 2 ) = 0 z(z^(2)+2z+2)=0z(z^2 + 2z + 2) = 0z(z2+2z+2)=0
This gives us one root at z = 0 z = 0 z=0z = 0z=0. The quadratic part can be solved using the quadratic formula:
z 2 + 2 z + 2 = 0 z 2 + 2 z + 2 = 0 z^(2)+2z+2=0z^2 + 2z + 2 = 0z2+2z+2=0
The roots of this quadratic equation are given by:
z = b ± b 2 4 a c 2 a z = b ± b 2 4 a c 2 a z=(-b+-sqrt(b^(2)-4ac))/(2a)z = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}z=b±b24ac2a
where a = 1 a = 1 a=1a = 1a=1, b = 2 b = 2 b=2b = 2b=2, and c = 2 c = 2 c=2c = 2c=2. Let’s calculate these roots.
The roots of the quadratic equation z 2 + 2 z + 2 = 0 z 2 + 2 z + 2 = 0 z^(2)+2z+2=0z^2 + 2z + 2 = 0z2+2z+2=0 are z = 1 i z = 1 i z=-1-iz = -1 – iz=1i and z = 1 + i z = 1 + i z=-1+iz = -1 + iz=1+i. Therefore, the function f ( z ) = z 2 + 4 z 3 + 2 z 2 + 2 z f ( z ) = z 2 + 4 z 3 + 2 z 2 + 2 z f(z)=(z^(2)+4)/(z^(3)+2z^(2)+2z)f(z) = \frac{z^2 + 4}{z^3 + 2z^2 + 2z}f(z)=z2+4z3+2z2+2z has poles at z = 0 z = 0 z=0z = 0z=0, z = 1 i z = 1 i z=-1-iz = -1 – iz=1i, and z = 1 + i z = 1 + i z=-1+iz = -1 + iz=1+i.
Next, we need to determine the residues at these poles. The residue at a simple pole z = a z = a z=az = az=a of a function f ( z ) f ( z ) f(z)f(z)f(z) is given by lim z a ( z a ) f ( z ) lim z a ( z a ) f ( z ) lim_(z rarr a)(z-a)f(z)\lim_{z \to a} (z – a)f(z)limza(za)f(z).
  1. Residue at z = 0 z = 0 z=0z = 0z=0: Since z = 0 z = 0 z=0z = 0z=0 is a simple pole, the residue is lim z 0 z z 2 + 4 z 3 + 2 z 2 + 2 z lim z 0 z z 2 + 4 z 3 + 2 z 2 + 2 z lim_(z rarr0)z*(z^(2)+4)/(z^(3)+2z^(2)+2z)\lim_{z \to 0} z \cdot \frac{z^2 + 4}{z^3 + 2z^2 + 2z}limz0zz2+4z3+2z2+2z.
  1. Residue at z = 1 i z = 1 i z=-1-iz = -1 – iz=1i: This is also a simple pole. The residue is lim z 1 i ( z + 1 + i ) z 2 + 4 z 3 + 2 z 2 + 2 z lim z 1 i ( z + 1 + i ) z 2 + 4 z 3 + 2 z 2 + 2 z lim_(z rarr-1-i)(z+1+i)*(z^(2)+4)/(z^(3)+2z^(2)+2z)\lim_{z \to -1 – i} (z + 1 + i) \cdot \frac{z^2 + 4}{z^3 + 2z^2 + 2z}limz1i(z+1+i)z2+4z3+2z2+2z.
  2. Residue at z = 1 + i z = 1 + i z=-1+iz = -1 + iz=1+i: Similarly, the residue is lim z 1 + i ( z + 1 i ) z 2 + 4 z 3 + 2 z 2 + 2 z lim z 1 + i ( z + 1 i ) z 2 + 4 z 3 + 2 z 2 + 2 z lim_(z rarr-1+i)(z+1-i)*(z^(2)+4)/(z^(3)+2z^(2)+2z)\lim_{z \to -1 + i} (z + 1 – i) \cdot \frac{z^2 + 4}{z^3 + 2z^2 + 2z}limz1+i(z+1i)z2+4z3+2z2+2z.
The residues at the poles of the function f ( z ) = z 2 + 4 z 3 + 2 z 2 + 2 z f ( z ) = z 2 + 4 z 3 + 2 z 2 + 2 z f(z)=(z^(2)+4)/(z^(3)+2z^(2)+2z)f(z) = \frac{z^2 + 4}{z^3 + 2z^2 + 2z}f(z)=z2+4z3+2z2+2z are as follows:
  1. At z = 0 z = 0 z=0z = 0z=0: The residue is 2 2 222.
  2. At z = 1 i z = 1 i z=-1-iz = -1 – iz=1i: The residue is 1 2 3 i 2 1 2 3 i 2 -(1)/(2)-(3i)/(2)-\frac{1}{2} – \frac{3i}{2}123i2.
  3. At z = 1 + i z = 1 + i z=-1+iz = -1 + iz=1+i: The residue is 1 2 + 3 i 2 1 2 + 3 i 2 -(1)/(2)+(3i)/(2)-\frac{1}{2} + \frac{3i}{2}12+3i2.
Q2 Evaluate c 1 2 z 3 d z c 1 2 z 3 d z int _(c)(1)/(2z-3)dz\int_c \frac{1}{2 z-3} d zc12z3dz by using residue theorem, where C is the circle | z | 2 . | z | 2 . |z| <= 2.quad|z| \leq 2 . \quad|z|2.
Answer:
To evaluate the integral C 1 2 z 3 d z C 1 2 z 3 d z int _(C)(1)/(2z-3)dz\int_C \frac{1}{2z – 3} dzC12z3dz using the residue theorem, where C C CCC is the circle | z | 2 | z | 2 |z| <= 2|z| \leq 2|z|2, we need to follow these steps:
  1. Identify Poles Inside C: First, we identify the poles of the function f ( z ) = 1 2 z 3 f ( z ) = 1 2 z 3 f(z)=(1)/(2z-3)f(z) = \frac{1}{2z – 3}f(z)=12z3 that lie inside the circle | z | 2 | z | 2 |z| <= 2|z| \leq 2|z|2. The function has a single pole at z = 3 2 z = 3 2 z=(3)/(2)z = \frac{3}{2}z=32, which is inside the circle | z | 2 | z | 2 |z| <= 2|z| \leq 2|z|2.
  2. Calculate Residue at the Pole: Next, we calculate the residue of f ( z ) f ( z ) f(z)f(z)f(z) at the pole z = 3 2 z = 3 2 z=(3)/(2)z = \frac{3}{2}z=32. The residue of a function at a simple pole z = a z = a z=az = az=a is given by lim z a ( z a ) f ( z ) lim z a ( z a ) f ( z ) lim_(z rarr a)(z-a)f(z)\lim_{z \to a} (z – a)f(z)limza(za)f(z). In this case, the residue at z = 3 2 z = 3 2 z=(3)/(2)z = \frac{3}{2}z=32 is lim z 3 2 ( z 3 2 ) 1 2 z 3 lim z 3 2 z 3 2 1 2 z 3 lim_(z rarr(3)/(2))(z-(3)/(2))(1)/(2z-3)\lim_{z \to \frac{3}{2}} \left(z – \frac{3}{2}\right) \frac{1}{2z – 3}limz32(z32)12z3.
  3. Apply the Residue Theorem: The residue theorem states that if a function is analytic inside and on a simple closed contour C C CCC, except for a finite number of poles inside C C CCC, then the integral of the function around C C CCC is 2 π i 2 π i 2pi i2\pi i2πi times the sum of the residues at the poles inside C C CCC.
Let’s calculate the residue at z = 3 2 z = 3 2 z=(3)/(2)z = \frac{3}{2}z=32.
The residue of the function f ( z ) = 1 2 z 3 f ( z ) = 1 2 z 3 f(z)=(1)/(2z-3)f(z) = \frac{1}{2z – 3}f(z)=12z3 at the pole z = 3 2 z = 3 2 z=(3)/(2)z = \frac{3}{2}z=32 is 1 2 1 2 (1)/(2)\frac{1}{2}12.
Now, applying the residue theorem, the integral of f ( z ) f ( z ) f(z)f(z)f(z) around the circle C C CCC (where | z | 2 | z | 2 |z| <= 2|z| \leq 2|z|2) is 2 π i 2 π i 2pi i2\pi i2πi times the sum of the residues at the poles inside C C CCC. Since there is only one pole inside C C CCC at z = 3 2 z = 3 2 z=(3)/(2)z = \frac{3}{2}z=32 with a residue of 1 2 1 2 (1)/(2)\frac{1}{2}12, the integral is:
C 1 2 z 3 d z = 2 π i × 1 2 = π i C 1 2 z 3 d z = 2 π i × 1 2 = π i int _(C)(1)/(2z-3)dz=2pi i xx(1)/(2)=pi i\int_C \frac{1}{2z – 3} dz = 2\pi i \times \frac{1}{2} = \pi iC12z3dz=2πi×12=πi
Therefore, the value of the integral C 1 2 z 3 d z C 1 2 z 3 d z int _(C)(1)/(2z-3)dz\int_C \frac{1}{2z – 3} dzC12z3dz is π i π i pi i\pi iπi.
Q3 Obtain the complex integral: c 1 z 5 d z c 1 z 5 d z int _(c)(1)/(z-5)dz\int_c \frac{1}{z-5} d zc1z5dz where C : | z | 1 C : | z | 1 C:|z| <= 1C:|z| \leq 1C:|z|1 ?
Answer:
To evaluate the integral C 1 z 5 d z C 1 z 5 d z int _(C)(1)/(z-5)dz\int_C \frac{1}{z – 5} dzC1z5dz, where C C CCC is the circle | z | 1 | z | 1 |z| <= 1|z| \leq 1|z|1, we can use the residue theorem. However, in this case, it’s important to note the location of the pole of the function f ( z ) = 1 z 5 f ( z ) = 1 z 5 f(z)=(1)/(z-5)f(z) = \frac{1}{z – 5}f(z)=1z5.
The function f ( z ) f ( z ) f(z)f(z)f(z) has a single pole at z = 5 z = 5 z=5z = 5z=5. This pole is not inside the circle | z | 1 | z | 1 |z| <= 1|z| \leq 1|z|1, as it lies outside the region enclosed by C C CCC.
According to the residue theorem, the integral of a function around a closed contour is 2 π i 2 π i 2pi i2\pi i2πi times the sum of the residues of the function at the poles inside the contour. Since there are no poles of f ( z ) f ( z ) f(z)f(z)f(z) inside the circle | z | 1 | z | 1 |z| <= 1|z| \leq 1|z|1, the sum of the residues inside C C CCC is zero.
Therefore, the integral C 1 z 5 d z C 1 z 5 d z int _(C)(1)/(z-5)dz\int_C \frac{1}{z – 5} dzC1z5dz over the circle | z | 1 | z | 1 |z| <= 1|z| \leq 1|z|1 is equal to 0 0 000. This result is consistent with Cauchy’s theorem, which states that the integral of a holomorphic function over a closed contour is zero if the function has no poles inside the contour.
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