# Complex Analysis

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Q1 Find the poles of $f\left(z\right)=\frac{{z}^{2}+4}{{z}^{3}+2{z}^{2}+2z}$$f\left(z\right)=\frac{{z}^{2}+4}{{z}^{3}+2{z}^{2}+2z}$f(z)=(z^(2)+4)/(z^(3)+2z^(2)+2z)f(z)=\frac{z^2+4}{z^3+2 z^2+2 z}$f\left(z\right)=\frac{{z}^{2}+4}{{z}^{3}+2{z}^{2}+2z}$ and determine the residues at the poles.
Answer:
To find the poles of the function $f\left(z\right)=\frac{{z}^{2}+4}{{z}^{3}+2{z}^{2}+2z}$$f\left(z\right)=\frac{{z}^{2}+4}{{z}^{3}+2{z}^{2}+2z}$f(z)=(z^(2)+4)/(z^(3)+2z^(2)+2z)f(z) = \frac{z^2 + 4}{z^3 + 2z^2 + 2z}$f\left(z\right)=\frac{{z}^{2}+4}{{z}^{3}+2{z}^{2}+2z}$, we need to determine the values of $z$$z$zz$z$ for which the denominator is zero, as these are the points where the function becomes undefined. The poles are the roots of the equation:
${z}^{3}+2{z}^{2}+2z=0$${z}^{3}+2{z}^{2}+2z=0$z^(3)+2z^(2)+2z=0z^3 + 2z^2 + 2z = 0${z}^{3}+2{z}^{2}+2z=0$
We can factor out a $z$$z$zz$z$ from the equation:
$z\left({z}^{2}+2z+2\right)=0$$z\left({z}^{2}+2z+2\right)=0$z(z^(2)+2z+2)=0z(z^2 + 2z + 2) = 0$z\left({z}^{2}+2z+2\right)=0$
This gives us one root at $z=0$$z=0$z=0z = 0$z=0$. The quadratic part can be solved using the quadratic formula:
${z}^{2}+2z+2=0$${z}^{2}+2z+2=0$z^(2)+2z+2=0z^2 + 2z + 2 = 0${z}^{2}+2z+2=0$
The roots of this quadratic equation are given by:
$z=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$$z=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$z=(-b+-sqrt(b^(2)-4ac))/(2a)z = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$z=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
where $a=1$$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$$z=-1+i$z=-1+iz = -1 + i$z=-1+i$. Therefore, the function $f\left(z\right)=\frac{{z}^{2}+4}{{z}^{3}+2{z}^{2}+2z}$$f\left(z\right)=\frac{{z}^{2}+4}{{z}^{3}+2{z}^{2}+2z}$f(z)=(z^(2)+4)/(z^(3)+2z^(2)+2z)f(z) = \frac{z^2 + 4}{z^3 + 2z^2 + 2z}$f\left(z\right)=\frac{{z}^{2}+4}{{z}^{3}+2{z}^{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\left(z\right)$$f\left(z\right)$f(z)f(z)$f\left(z\right)$ is given by $\underset{z\to a}{lim}\left(z-a\right)f\left(z\right)$$\underset{z\to a}{lim} \left(z-a\right)f\left(z\right)$lim_(z rarr a)(z-a)f(z)\lim_{z \to a} (z – a)f(z)$\underset{z\to a}{lim}\left(z-a\right)f\left(z\right)$.
1. Residue at $z=0$$z=0$z=0z = 0$z=0$: Since $z=0$$z=0$z=0z = 0$z=0$ is a simple pole, the residue is $\underset{z\to 0}{lim}z\cdot \frac{{z}^{2}+4}{{z}^{3}+2{z}^{2}+2z}$$\underset{z\to 0}{lim} z\cdot \frac{{z}^{2}+4}{{z}^{3}+2{z}^{2}+2z}$lim_(z rarr0)z*(z^(2)+4)/(z^(3)+2z^(2)+2z)\lim_{z \to 0} z \cdot \frac{z^2 + 4}{z^3 + 2z^2 + 2z}$\underset{z\to 0}{lim}z\cdot \frac{{z}^{2}+4}{{z}^{3}+2{z}^{2}+2z}$.
1. Residue at $z=-1-i$$z=-1-i$z=-1-iz = -1 – i$z=-1-i$: This is also a simple pole. The residue is $\underset{z\to -1-i}{lim}\left(z+1+i\right)\cdot \frac{{z}^{2}+4}{{z}^{3}+2{z}^{2}+2z}$$\underset{z\to -1-i}{lim} \left(z+1+i\right)\cdot \frac{{z}^{2}+4}{{z}^{3}+2{z}^{2}+2z}$lim_(z rarr-1-i)(z+1+i)*(z^(2)+4)/(z^(3)+2z^(2)+2z)\lim_{z \to -1 – i} (z + 1 + i) \cdot \frac{z^2 + 4}{z^3 + 2z^2 + 2z}$\underset{z\to -1-i}{lim}\left(z+1+i\right)\cdot \frac{{z}^{2}+4}{{z}^{3}+2{z}^{2}+2z}$.
2. Residue at $z=-1+i$$z=-1+i$z=-1+iz = -1 + i$z=-1+i$: Similarly, the residue is $\underset{z\to -1+i}{lim}\left(z+1-i\right)\cdot \frac{{z}^{2}+4}{{z}^{3}+2{z}^{2}+2z}$$\underset{z\to -1+i}{lim} \left(z+1-i\right)\cdot \frac{{z}^{2}+4}{{z}^{3}+2{z}^{2}+2z}$lim_(z rarr-1+i)(z+1-i)*(z^(2)+4)/(z^(3)+2z^(2)+2z)\lim_{z \to -1 + i} (z + 1 – i) \cdot \frac{z^2 + 4}{z^3 + 2z^2 + 2z}$\underset{z\to -1+i}{lim}\left(z+1-i\right)\cdot \frac{{z}^{2}+4}{{z}^{3}+2{z}^{2}+2z}$.
The residues at the poles of the function $f\left(z\right)=\frac{{z}^{2}+4}{{z}^{3}+2{z}^{2}+2z}$$f\left(z\right)=\frac{{z}^{2}+4}{{z}^{3}+2{z}^{2}+2z}$f(z)=(z^(2)+4)/(z^(3)+2z^(2)+2z)f(z) = \frac{z^2 + 4}{z^3 + 2z^2 + 2z}$f\left(z\right)=\frac{{z}^{2}+4}{{z}^{3}+2{z}^{2}+2z}$ are as follows:
1. At $z=0$$z=0$z=0z = 0$z=0$: The residue is $2$$2$22$2$.
2. At $z=-1-i$$z=-1-i$z=-1-iz = -1 – i$z=-1-i$: The residue is $-\frac{1}{2}-\frac{3i}{2}$$-\frac{1}{2}-\frac{3i}{2}$-(1)/(2)-(3i)/(2)-\frac{1}{2} – \frac{3i}{2}$-\frac{1}{2}-\frac{3i}{2}$.
3. At $z=-1+i$$z=-1+i$z=-1+iz = -1 + i$z=-1+i$: The residue is $-\frac{1}{2}+\frac{3i}{2}$$-\frac{1}{2}+\frac{3i}{2}$-(1)/(2)+(3i)/(2)-\frac{1}{2} + \frac{3i}{2}$-\frac{1}{2}+\frac{3i}{2}$.
Q2 Evaluate ${\int }_{c}\frac{1}{2z-3}dz$${\int }_{c} \frac{1}{2z-3}dz$int _(c)(1)/(2z-3)dz\int_c \frac{1}{2 z-3} d z${\int }_{c}\frac{1}{2z-3}dz$ by using residue theorem, where C is the circle $|z|\le 2.\phantom{\rule{1em}{0ex}}$$|z|\le 2.\phantom{\rule{1em}{0ex}}$|z| <= 2.quad|z| \leq 2 . \quad$|z|\le 2.\phantom{\rule{1em}{0ex}}$
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$$|z|\le 2$|z| <= 2|z| \leq 2$|z|\le 2$, we need to follow these steps:
1. Identify Poles Inside C: First, we identify the poles of the function $f\left(z\right)=\frac{1}{2z-3}$$f\left(z\right)=\frac{1}{2z-3}$f(z)=(1)/(2z-3)f(z) = \frac{1}{2z – 3}$f\left(z\right)=\frac{1}{2z-3}$ that lie inside the circle $|z|\le 2$$|z|\le 2$|z| <= 2|z| \leq 2$|z|\le 2$. The function has a single pole at $z=\frac{3}{2}$$z=\frac{3}{2}$z=(3)/(2)z = \frac{3}{2}$z=\frac{3}{2}$, which is inside the circle $|z|\le 2$$|z|\le 2$|z| <= 2|z| \leq 2$|z|\le 2$.
2. Calculate Residue at the Pole: Next, we calculate the residue of $f\left(z\right)$$f\left(z\right)$f(z)f(z)$f\left(z\right)$ at the pole $z=\frac{3}{2}$$z=\frac{3}{2}$z=(3)/(2)z = \frac{3}{2}$z=\frac{3}{2}$. The residue of a function at a simple pole $z=a$$z=a$z=az = a$z=a$ is given by $\underset{z\to a}{lim}\left(z-a\right)f\left(z\right)$$\underset{z\to a}{lim} \left(z-a\right)f\left(z\right)$lim_(z rarr a)(z-a)f(z)\lim_{z \to a} (z – a)f(z)$\underset{z\to a}{lim}\left(z-a\right)f\left(z\right)$. In this case, the residue at $z=\frac{3}{2}$$z=\frac{3}{2}$z=(3)/(2)z = \frac{3}{2}$z=\frac{3}{2}$ is $\underset{z\to \frac{3}{2}}{lim}\left(z-\frac{3}{2}\right)\frac{1}{2z-3}$$\underset{z\to \frac{3}{2}}{lim} \left(z-\frac{3}{2}\right)\frac{1}{2z-3}$lim_(z rarr(3)/(2))(z-(3)/(2))(1)/(2z-3)\lim_{z \to \frac{3}{2}} \left(z – \frac{3}{2}\right) \frac{1}{2z – 3}$\underset{z\to \frac{3}{2}}{lim}\left(z-\frac{3}{2}\right)\frac{1}{2z-3}$.
3. Apply the Residue Theorem: The residue theorem states that if a function is analytic inside and on a simple closed contour $C$$C$CC$C$, except for a finite number of poles inside $C$$C$CC$C$, then the integral of the function around $C$$C$CC$C$ 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$.
Let’s calculate the residue at $z=\frac{3}{2}$$z=\frac{3}{2}$z=(3)/(2)z = \frac{3}{2}$z=\frac{3}{2}$.
The residue of the function $f\left(z\right)=\frac{1}{2z-3}$$f\left(z\right)=\frac{1}{2z-3}$f(z)=(1)/(2z-3)f(z) = \frac{1}{2z – 3}$f\left(z\right)=\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\left(z\right)$$f\left(z\right)$f(z)f(z)$f\left(z\right)$ around the circle $C$$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:
${\int }_{C}\frac{1}{2z-3}dz=2\pi i×\frac{1}{2}=\pi i$${\int }_{C} \frac{1}{2z-3}dz=2\pi i×\frac{1}{2}=\pi i$int _(C)(1)/(2z-3)dz=2pi i xx(1)/(2)=pi i\int_C \frac{1}{2z – 3} dz = 2\pi i \times \frac{1}{2} = \pi i${\int }_{C}\frac{1}{2z-3}dz=2\pi i×\frac{1}{2}=\pi i$
Therefore, the value of the integral ${\int }_{C}\frac{1}{2z-3}dz$${\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\left(z\right)=\frac{1}{z-5}$$f\left(z\right)=\frac{1}{z-5}$f(z)=(1)/(z-5)f(z) = \frac{1}{z – 5}$f\left(z\right)=\frac{1}{z-5}$.
The function $f\left(z\right)$$f\left(z\right)$f(z)f(z)$f\left(z\right)$ 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\left(z\right)$$f\left(z\right)$f(z)f(z)$f\left(z\right)$ inside the circle $|z|\le 1$$|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.
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