Q1 Find the poles of $f(z)=\frac{z^2+4}{z^3+2z^2+2z}$$f(z)=\frac{z^2+4}{z^3+2z^2+2z}$f(z)=(z^(2)+4)/(z^(3)+2z^(2)+2z)f(z)=\frac{z^2+4}{z^3+2 z^2+2 z}$f(z)=\frac{z^2+4}{z^3+2z^2+2z}$ and determine the residues at the poles.

Answer:

To find the poles of the function $f(z)=\frac{z^2+4}{z^3+2z^2+2z}$$f(z)=\frac{z^2+4}{z^3+2z^2+2z}$f(z)=(z^(2)+4)/(z^(3)+2z^(2)+2z)f(z) = \frac{z^2 + 4}{z^3 + 2z^2 + 2z}$f(z)=\frac{z^2+4}{z^3+2z^2+2z}$, we need to determine the values of $z$$z$zz$z$ for which the denominator is zero, as these are the points where the function becomes undefined. The poles are the roots of the equation:

We can factor out a $z$$z$zz$z$ from the equation:

This gives us one root at $z=0$$z=0$z=0z = 0$z=0$. The quadratic part can be solved using the quadratic formula:

The roots of this quadratic equation are given by:

where $a=1$$a=1$a=1a = 1$a=1$, $b=2$$b=2$b=2b = 2$b=2$, and $c=2$$c=2$c=2c = 2$c=2$. Let’s calculate these roots.

The roots of the quadratic equation $z^2+2z+2=0$$z^2+2z+2=0$z^(2)+2z+2=0z^2 + 2z + 2 = 0$z^2+2z+2=0$ are $z=-1-i$$z=-1-i$z=-1-iz = -1 – i$z=-1-i$ and $z=-1+i$$z=-1+i$z=-1+iz = -1 + i$z=-1+i$. Therefore, the function $f(z)=\frac{z^2+4}{z^3+2z^2+2z}$$f(z)=\frac{z^2+4}{z^3+2z^2+2z}$f(z)=(z^(2)+4)/(z^(3)+2z^(2)+2z)f(z) = \frac{z^2 + 4}{z^3 + 2z^2 + 2z}$f(z)=\frac{z^2+4}{z^3+2z^2+2z}$ has poles at $z=0$$z=0$z=0z = 0$z=0$, $z=-1-i$$z=-1-i$z=-1-iz = -1 – i$z=-1-i$, and $z=-1+i$$z=-1+i$z=-1+iz = -1 + i$z=-1+i$.

Next, we need to determine the residues at these poles. The residue at a simple pole $z=a$$z=a$z=az = a$z=a$ of a function $f(z)$$f(z)$f(z)f(z)$f(z)$ is given by $\underset{z\to a}{lim}(z-a)f(z)$$\underset{z\to a}{lim} (z-a)f(z)$lim_(z rarr a)(z-a)f(z)\lim_{z \to a} (z – a)f(z)$\underset{z\to a}{lim}(z-a)f(z)$.

The residues at the poles of the function $f(z)=\frac{z^2+4}{z^3+2z^2+2z}$$f(z)=\frac{z^2+4}{z^3+2z^2+2z}$f(z)=(z^(2)+4)/(z^(3)+2z^(2)+2z)f(z) = \frac{z^2 + 4}{z^3 + 2z^2 + 2z}$f(z)=\frac{z^2+4}{z^3+2z^2+2z}$ are as follows:

Q2 Evaluate ${\int }_c\frac{1}{2z-3}dz$${\int }_c \frac{1}{2z-3}dz$int _(c)(1)/(2z-3)dz\int_c \frac{1}{2 z-3} d z${\int }_c\frac{1}{2z-3}dz$ by using residue theorem, where C is the circle $|z|\le 2.$$|z|\le 2.$|z| <= 2.quad|z| \leq 2 . \quad$|z|\le 2.$

Answer:

To evaluate the integral ${\int }_C\frac{1}{2z-3}dz$${\int }_C \frac{1}{2z-3}dz$int _(C)(1)/(2z-3)dz\int_C \frac{1}{2z – 3} dz${\int }_C\frac{1}{2z-3}dz$ using the residue theorem, where $C$$C$CC$C$ is the circle $|z|\le 2$$|z|\le 2$|z| <= 2|z| \leq 2$|z|\le 2$, we need to follow these steps:

Let’s calculate the residue at $z=\frac{3}{2}$$z=\frac{3}{2}$z=(3)/(2)z = \frac{3}{2}$z=\frac{3}{2}$.

The residue of the function $f(z)=\frac{1}{2z-3}$$f(z)=\frac{1}{2z-3}$f(z)=(1)/(2z-3)f(z) = \frac{1}{2z – 3}$f(z)=\frac{1}{2z-3}$ at the pole $z=\frac{3}{2}$$z=\frac{3}{2}$z=(3)/(2)z = \frac{3}{2}$z=\frac{3}{2}$ is $\frac{1}{2}$$\frac{1}{2}$(1)/(2)\frac{1}{2}$\frac{1}{2}$.

Now, applying the residue theorem, the integral of $f(z)$$f(z)$f(z)f(z)$f(z)$ around the circle $C$$C$CC$C$ (where $|z|\le 2$$|z|\le 2$|z| <= 2|z| \leq 2$|z|\le 2$) is $2\pi i$$2\pi i$2pi i2\pi i$2\pi i$ times the sum of the residues at the poles inside $C$$C$CC$C$. Since there is only one pole inside $C$$C$CC$C$ at $z=\frac{3}{2}$$z=\frac{3}{2}$z=(3)/(2)z = \frac{3}{2}$z=\frac{3}{2}$ with a residue of $\frac{1}{2}$$\frac{1}{2}$(1)/(2)\frac{1}{2}$\frac{1}{2}$, the integral is:

Therefore, the value of the integral ${\int }_C\frac{1}{2z-3}dz$${\int }_C \frac{1}{2z-3}dz$int _(C)(1)/(2z-3)dz\int_C \frac{1}{2z – 3} dz${\int }_C\frac{1}{2z-3}dz$ is $\pi i$$\pi i$pi i\pi i$\pi i$.

Q3 Obtain the complex integral: ${\int }_c\frac{1}{z-5}dz$${\int }_c \frac{1}{z-5}dz$int _(c)(1)/(z-5)dz\int_c \frac{1}{z-5} d z${\int }_c\frac{1}{z-5}dz$ where $C:|z|\le 1$$C:|z|\le 1$C:|z| <= 1C:|z| \leq 1$C:|z|\le 1$ ?

Answer:

To evaluate the integral ${\int }_C\frac{1}{z-5}dz$${\int }_C \frac{1}{z-5}dz$int _(C)(1)/(z-5)dz\int_C \frac{1}{z – 5} dz${\int }_C\frac{1}{z-5}dz$, where $C$$C$CC$C$ is the circle $|z|\le 1$$|z|\le 1$|z| <= 1|z| \leq 1$|z|\le 1$, we can use the residue theorem. However, in this case, it’s important to note the location of the pole of the function $f(z)=\frac{1}{z-5}$$f(z)=\frac{1}{z-5}$f(z)=(1)/(z-5)f(z) = \frac{1}{z – 5}$f(z)=\frac{1}{z-5}$.

The function $f(z)$$f(z)$f(z)f(z)$f(z)$ has a single pole at $z=5$$z=5$z=5z = 5$z=5$. This pole is not inside the circle $|z|\le 1$$|z|\le 1$|z| <= 1|z| \leq 1$|z|\le 1$, as it lies outside the region enclosed by $C$$C$CC$C$.

According to the residue theorem, the integral of a function around a closed contour is $2\pi i$$2\pi i$2pi i2\pi i$2\pi i$ times the sum of the residues of the function at the poles inside the contour. Since there are no poles of $f(z)$$f(z)$f(z)f(z)$f(z)$ inside the circle $|z|\le 1$$|z|\le 1$|z| <= 1|z| \leq 1$|z|\le 1$, the sum of the residues inside $C$$C$CC$C$ is zero.

Therefore, the integral ${\int }_C\frac{1}{z-5}dz$${\int }_C \frac{1}{z-5}dz$int _(C)(1)/(z-5)dz\int_C \frac{1}{z – 5} dz${\int }_C\frac{1}{z-5}dz$ over the circle $|z|\le 1$$|z|\le 1$|z| <= 1|z| \leq 1$|z|\le 1$ is equal to $0$$0$00$0$. 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.