Let z = x + iy be a complex variable. Consider that contour integration is performed along the unit circle in anticlockwise direction. Which one of the following statements is Not True?

Complex Variable Analysis: Identifying Non-Analytic Functions

Let's analyze each statement concerning complex variables and contour integration along the unit circle, performed in the anticlockwise direction, to determine which one is Not True.

Residue Calculation for Poles

Consider the first statement about the residue of $\frac{\mathrm{z}}{{\mathrm{z}}^2-1}\mathrm{at}\mathrm{z}=1$.

- The function is given by $f(z)=\frac{z}{z^2-1}$.
- We can factor the denominator as $z^2-1=(z-1)(z+1)$.
- So, $f(z)=\frac{z}{(z-1)(z+1)}$.
- The point $z=1$ is a simple pole of the function $f(z)$.
- The residue at a simple pole $z=a$ for a function $f(z)$ is given by the formula $\mathrm{Res}(f,a)={\lim }_{z\to a}(z-a)f(z)$.
- Therefore, the residue of $f(z)$ at $z=1$ is: $\mathrm{Res}(f,1)={\lim }_{z\to 1}(z-1)\frac{z}{(z-1)(z+1)}$ $\mathrm{Res}(f,1)={\lim }_{z\to 1}\frac{z}{z+1}$ $\mathrm{Res}(f,1)=\frac{1}{1+1}=\frac{1}{2}$

Thus, the statement "The residue of $\frac{\mathrm{z}}{{\mathrm{z}}^2-1}\mathrm{at}\mathrm{z}=1\mathrm{is}1/2$ " is True.

Contour Integration of Analytic Functions

Let's examine the integral $\underset{\mathrm{c}}{\oint }{\mathrm{z}}^2\mathrm{dz}=0$.

- The function $f(z)=z^2$ is a polynomial function. Polynomial functions are entire functions, meaning they are analytic everywhere in the complex plane.
- The contour $\mathrm{C}$ is the unit circle, which is a simple closed contour.
- According to Cauchy's Integral Theorem, if a function $f(z)$ is analytic inside and on a simple closed contour $\mathrm{C}$, then the integral of $f(z)$ over $\mathrm{C}$ is zero.
- Since $z^2$ is analytic everywhere, it is certainly analytic inside and on the unit circle.

Therefore, the statement $\underset{\mathrm{c}}{\oint }{\mathrm{z}}^2\mathrm{dz}=0$ is True.

Cauchy's Integral Formula and Residue Theorem

Now let's evaluate the statement $\frac{1}{2\pi \mathrm{i}}\underset{\mathrm{c}}{\oint }\frac{1}{\mathrm{z}}\mathrm{dz}=1$.

- The function is $f(z)=\frac{1}{z}$.
- The singularity of $f(z)$ occurs at $z=0$.
- The contour $\mathrm{C}$ is the unit circle $|z|=1$, which encloses the singularity at $z=0$.
- We can use Cauchy's Integral Formula for a function $f(z)$ analytic inside and on $\mathrm{C}$, and for a point $a$ inside $\mathrm{C}$: $\underset{\mathrm{c}}{\oint }\frac{f(z)}{z-a}\mathrm{dz}=2\pi \mathrm{i}f(a)$
- In our case, we can write $\frac{1}{z}$ as $\frac{1}{z-0}$. So, $f(z)=1$ (which is analytic everywhere) and $a=0$.
- Applying the formula: $\underset{\mathrm{c}}{\oint }\frac{1}{\mathrm{z}}\mathrm{dz}=2\pi \mathrm{i}\cdot f(0)=2\pi \mathrm{i}\cdot 1=2\pi \mathrm{i}$.
- Substituting this result back into the given expression: $\frac{1}{2\pi \mathrm{i}}\underset{\mathrm{c}}{\oint }\frac{1}{\mathrm{z}}\mathrm{dz}=\frac{1}{2\pi \mathrm{i}}(2\pi \mathrm{i})=1$
- Alternatively, using the Residue Theorem: The residue of $\frac{1}{z}$ at its simple pole $z=0$ is ${\lim }_{z\to 0}z\cdot \frac{1}{z}=1$. The Residue Theorem states that $\underset{\mathrm{c}}{\oint }f(z)\mathrm{dz}=2\pi \mathrm{i}\times (\text{sum of residues of }f(z)\text{ inside C})$. So, $\underset{\mathrm{c}}{\oint }\frac{1}{\mathrm{z}}\mathrm{dz}=2\pi \mathrm{i}\times 1=2\pi \mathrm{i}$. Therefore, $\frac{1}{2\pi \mathrm{i}}\underset{\mathrm{c}}{\oint }\frac{1}{\mathrm{z}}\mathrm{dz}=\frac{1}{2\pi \mathrm{i}}(2\pi \mathrm{i})=1$.

Thus, the statement $\frac{1}{2\pi \mathrm{i}}\underset{\mathrm{c}}{\oint }\frac{1}{\mathrm{z}}\mathrm{dz}=1$ is True.

Analytic Function Definition: Cauchy-Riemann Equations

Finally, let's analyze the statement that $\bar{\mathrm{z}}$ (Complex conjugate of z) is an analytical function.

- Let $z=x+iy$. Then its complex conjugate is $\bar{z}=x-iy$.
- For a complex function $f(z)=u(x,y)+iv(x,y)$ to be analytic at a point, its real and imaginary parts must satisfy the Cauchy-Riemann (CR) equations: 1. $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$ 2. $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$
- For $f(z)=\bar{z}=x-iy$: - The real part is $u(x,y)=x$. - The imaginary part is $v(x,y)=-y$.
- Let's compute the partial derivatives of $u$ and $v$: - $\frac{\partial u}{\partial x}=\frac{\partial }{\partial x}(x)=1$ - $\frac{\partial u}{\partial y}=\frac{\partial }{\partial y}(x)=0$ - $\frac{\partial v}{\partial x}=\frac{\partial }{\partial x}(-y)=0$ - $\frac{\partial v}{\partial y}=\frac{\partial }{\partial y}(-y)=-1$
- Now, let's check if the Cauchy-Riemann equations are satisfied: - For the first equation, $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$: Is $1=-1$? This is False. - For the second equation, $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$: Is $0=-(0)$? This is True.
- Since the first Cauchy-Riemann equation ($\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$) is not satisfied, the function $f(z)=\bar{z}$ is not analytic at any point.

Therefore, the statement "$\bar{\mathrm{z}}$ (Complex conjugate of z) is an analytical function" is Not True.

Statements SummaryBased on the detailed analysis of each statement, the statement that is Not True is the one claiming $\bar{\mathrm{z}}$ is an analytical function.