M4.6 Some identities on Analytic and harmonic functions

01. If $u$ and $v$ are harmonic functions show that $\left(\frac{\partial u}{\partial y}-\frac{\partial v}{\partial x}\right)+i\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)$ is analytic.
Ans:
Let $P=\left(\frac{\partial u}{\partial y}-\frac{\partial v}{\partial x}\right)$ and $Q=\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)$.
Now we have to show that $P+iQ$ is analytic, i.e., we have to show that $P$ and $Q$ will satisfies CR-equations
$\frac{\partial P}{\partial x}-\frac{\partial Q}{\partial y}=\frac{\partial }{\partial x}\left(\frac{\partial u}{\partial y}-\frac{\partial v}{\partial x}\right)-\frac{\partial }{\partial y}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)$
$\Longrightarrow \frac{\partial P}{\partial x}-\frac{\partial Q}{\partial y}=\frac{\partial^2 u}{\partial x \partial y}-\frac{\partial^2 v}{\partial x^2}-\frac{\partial^2 u}{\partial y \partial x}-\frac{\partial^2 v}{\partial y^2}=-\left(\frac{\partial^2 v}{\partial x^2}+\frac{\partial^2 v}{\partial y^2}\right)=0$
[because $\frac{\partial^2 v}{\partial x \partial y}=\frac{\partial^2 v}{\partial y \partial x}$ and since $u$ and $v$ are harmonic, $u$ and $v$ satisfies Laplace equations]. Hence $\frac{\partial P}{\partial x}-\frac{\partial Q}{\partial y}=0$$\Longrightarrow \frac{\partial P}{\partial x}=\frac{\partial Q}{\partial y}$
Similarly, consider
$\frac{\partial P}{\partial y}+\frac{\partial Q}{\partial x}=\frac{\partial }{\partial y}\left(\frac{\partial u}{\partial y}-\frac{\partial v}{\partial x}\right)+\frac{\partial }{\partial x}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)$
$\Longrightarrow \frac{\partial P}{\partial y}+\frac{\partial Q}{\partial x}=\frac{\partial^2 u}{\partial y^2}-\frac{\partial^2 v}{\partial y \partial x}+\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 v}{\partial x \partial y}=\left(\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}\right)=0$
[because $\frac{\partial^2 u}{\partial x \partial y}=\frac{\partial^2 u}{\partial y \partial x}$ and since $u$ and $v$ are harmonic, $u$ and $v$ satisfies Laplace equations]. Hence $\frac{\partial Q}{\partial x}+\frac{\partial Q}{\partial y}=0$$\Longrightarrow \frac{\partial Q}{\partial x}=-\frac{\partial Q}{\partial y}$
Hence $P+iQ$ is analytic.

---

02. If $f(z)$ is analytic, show that $\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)|f(z)|^2=4|f'(z)|^2$ is analytic.
Ans:
Let $f(z)=u+iv$ be analytic. $\Longrightarrow |f(z)|=\sqrt{u^2+v^2}\Longrightarrow |f(z)|^2=u^2+v^2=\phi$<\br> Hence, we need to show that $\phi_{xx}+\phi_{yy}=4|f'(z)|^2$.<\br> $$\phi=u^2+v^2\Longrightarrow \phi_x=2uu_x+2vv_x \Longrightarrow \phi_{xx}=2[uu_{xx}+u_x^2+vv_{xx}+v_x^2]\rightarrow (1)$$ $$\phi=u^2+v^2\Longrightarrow \phi_y=2uu_y+2vv_y \Longrightarrow \phi_{yy}=2[uu_{yy}+u_y^2+vv_{yy}+v_y^2]\rightarrow (2)$$ Now add (1) and (2)
$$\phi_{xx}+\phi_{yy}=2[u(u_{xx}+u_{yy})+v(v_{xx}+v_{yy})+u_x^2+v_x^2+u_y^2+v_y^2]\rightarrow (3)$$ Since $f(z)$ is analytic, $u$ and $v$ are harmonic $\Longrightarrow u_{xx}+u_{yy}=0,v_{xx}+v_{yy}=0$, also using CR-equations in (3), we get.
$$\phi_{xx}+\phi_{yy}=2[2u_x^2+2v_x^2]=4[u_x^2+v_x^2]\rightarrow (4)$$ But $f'(z)=u_x+iv_x \Longrightarrow |f'(z)|=u_x^2+v_x^2$, hence we can conclude that $$\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)|f(z)|^2=4|f'(z)|^2$$

---

03. If $f(z)$ is a regular analytic, show that $\left(\frac{\partial}{\partial x}|f(z)|\right)^2+\left(\frac{\partial}{\partial y}|f(z)|\right)^2=|f'(z)|^2$
Ans:
Let $f(z)=u+iv$ be analytic. $\Longrightarrow |f(z)|=\sqrt{u^2+v^2}=\phi\Longrightarrow |f(z)|^2=u^2+v^2=\phi^2$<\br> Consider $\phi^2=u^2+v^2$ $$\Longrightarrow 2\phi \phi_x=2uu_x+2vv_x\Longrightarrow \phi \phi_x=uu_x+vv_x \rightarrow (1)$$ $$\text{similarly} 2\phi \phi_y=2uu_y+2vv_y\Longrightarrow \phi \phi_y=uu_y+vv_y\rightarrow (2)$$ Squaring and adding both (1) and (2), we will get $$\phi^2 (\phi_x^2+\phi_y^2)=(uu_x+vv_x)^2+(uu_y+vv_y)^2$$ $$=(u^2 u_x^2+v^2 v_x^2+2uvu_x v_x)+(u^2u_y^2+v^2v_y^2+2uvu_y v_y)$$ Since $f(z)$ is analytic, $u$ and $v$ satisfies CR-equations. $$\Longrightarrow \phi^2 (\phi_x^2+\phi_y^2)=u^2(u_x^2+v_x^2)+v^2(u_x^2+v_x^2)=(u^2+v^2)(u_x^2+v_x^2)=\phi^2 (u_x^2+v_x^2)$$ Hence, we get $$\phi_x^2+\phi_y^2=u_x^2+v_x^2$$ Also we have $f(z)=u+iv\Longrightarrow f'(z)=u_x+iv_x \Longrightarrow |f'(z)|=\sqrt{u_x^2+v_x^2}$, hence above equation will become $$\phi_x^2+\phi_y^2=|f'(z)|^2$$

M4.5 Application on Analytic Functions

Let vector field $V$$=v_x\hat{i}+v_y\hat{j}$ represents velocity vector field. If the fluid is irrotational then $\nabla\times V=0$. Also the velocity of an irrotational flow can be represented by gradient of a scalar function $\phi$. i.e., $$V=\nabla \phi=\frac{\partial \phi}{\partial x}\hat{i}+\frac{\partial \phi}{\partial y}\hat{j}\,\,\,\,-(1).$$ $\phi$ is called as velocity potential and the family of curves $\phi(x,y)=c$ are called as equipotenttial lines

If the fluid is incompressible then $div V=0$, using (1), we will get $\nabla^2 \phi=0$, hence $\phi$ is harmonic. In other words an incompressible, irrotational flow is governed by Laplace's equation.

If $\phi$ is harmonic then there exists conjugate harmonic function $\psi(x,y)$ called as stream function such that $w(z)=\phi(x,y)+i\psi(x,y)$ is analytic. Where the family of curves $\psi(x,y)=c'$ are called as stream lines
From orthogonal property we can say that stream lines and equipotential lines are orthogonal as shown in the following figure.

Equipotential Lines Vs Stream Lines

Airflow streamlines behind tall building(PC:https://sesamenet.eu)

Problems
01. Find the complex potential of a fluid flow if the stream function is given by $\psi=\frac{-y}{x^2+y^2}$. Also find the velocity potential $\phi$.
Ans:
Let complex potential of a fluid is given by $f(z)=\phi(x,y)+i\psi(x,y)\Longrightarrow f'(z)=\phi_x+i\psi_x$
Since $f(z)$ is analytic, $\phi,\psi$ satisfies CR-equations$\Longrightarrow f'(z)=\psi_y+i\psi_x$
Given $\psi=\frac{-y}{x^2+y^2}\Longrightarrow \psi_x=\frac{2xy}{(x^2+y^2)^2}\,\,\,\, \&\,\,\, \psi_y=\frac{y^2-x^2}{(x^2+y^2)^2}$
$\Longrightarrow f'(z)=\frac{y^2-x^2}{(x^2+y^2)^2}+i\frac{2xy}{(x^2+y^2)^2}$
Now put $x=z,y=0 \Longrightarrow f'(z)=\frac{-z^2}{(z^2)^2}+i\cdot 0 \Longrightarrow f'(z)=\frac{-1}{z^2}$$\Longrightarrow f(z)=\frac{1}{z}$
Now put $z=x+i y\Longrightarrow f(z)=\frac{1}{x+i y}=\frac{x-iy}{x^2+y^2}=\frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2}$
Hence $\phi = \frac{x}{x^2+y^2}$
Mathematica Notebook Sourcecode which shows the above problem Graphically.
02. If $\phi+i\psi$ represents the complex potential of an electric field if $\psi=(x^2-y^2)+\frac{x}{x^2+y^2}$ find the complex potential as a function of $z$ and also find $\phi$.
Ans:
Let complex potential of a fluid is given by $f(z)=\phi(x,y)+i\psi(x,y)\Longrightarrow f'(z)=\phi_x+i\psi_x$
Since $f(z)$ is analytic, $\phi,\psi$ satisfies CR-equations$\Longrightarrow f'(z)=\psi_y+i\psi_x$
Given $\psi=(x^2-y^2)+\frac{x}{x^2+y^2}\Longrightarrow \psi_x=2x+\frac{y^2-x^2}{(x^2+y^2)^2}\,\,\,\, \&\,\,\, \psi_y=-2y-\frac{2xy}{(x^2+y^2)^2}$
$\Longrightarrow f'(z)=-2y-\frac{2xy}{(x^2+y^2)^2}+i\left(2x+\frac{y^2-x^2}{(x^2+y^2)^2}\right)$
Now put $x=z,y=0 \Longrightarrow f'(z)=i\left(2z-\frac{1}{z^2}\right) \Longrightarrow$$f(z)=i\left(z^2+\frac{1}{z}\right)$
Now put $z=x+i y\Longrightarrow f(z)=i\left((x+iy)^2+\frac{1}{x+i y}\right)=i\left((x^2-y^2+i2xy)+\frac{x-iy}{(x+iy)(x-iy)}\right)=\left(-2xy+\frac{y}{x^2+y^2}\right)+i\left((x^2-y^2)+\frac{x}{x^2+y^2}\right)$
Hence $\phi = -2xy+\frac{y}{x^2+y^2}$

M2.1 Vector Calculus Introduction

Scalars: The physical quantities having only magnitude are called as scalars. Ex: length, mass, time, speed, $a=4, c=\pi$
Vectors: The physical quantities having both magnitude and direction are called as vectors. Ex: velocity, force, $\vec{v}=a \hat{i}+b \hat{j}$ A vector in three dimension is of the form $\vec{v}=a\hat{i}+b \hat{j}+c \hat{k}$, where $\hat{i},\hat{j}, \hat{k}$ are the unit vectors in the direction of $X, Y$ and $Z$ respectively.
Addition of two vectors
Let $v_1=a\hat{i}+b \hat{j}, v_2=c\hat{i}+d \hat{j}$ then $v_1+v_2=(a+c)\hat{i}+(b+d) \hat{j}$ and $v_1-v_2=(a-c)\hat{i}+(b-d) \hat{j}$
Following is the example for vector addition, created using geogebra.

u(Green)=v1(Blue)+v2(Red)

M4.4 Problems on Analytic Functions

---

01. Show that $w=z+e^z$ is analytic and find $\frac{dw}{dz}$.
Ans: $$w=f(z)= z+e^z\Longrightarrow u+iv=(x+iy)+e^{(x+iy)} =(x+iy)+e^x\cdot e^{iy}=(x+iy)+e^x (\cos y+i\sin y)$$ $$\Longrightarrow u=x+e^x \cos y \,\,\,\,\,;\,\,\,\,\, v=y+e^x \sin y$$ $$\Longrightarrow u_x=1+e^x \cos y \,\,\,\,\,;\,\,\,\,\, v_x=e^x \sin y$$ $$\Longrightarrow u_y=-e^x \sin y \,\,\,\,\,;\,\,\,\,\, v_y=1+e^x \cos y$$ From the above set of equations we have $u_x=v_y; v_x=-u_y$, hence $u,v$ satisfies the CR-equations, Hence $f$ is analytic
To find $\frac{dw}{dz}$: Differentiate $w=f(z)=u+iv$ with respect to $x$ (derivative horizontally)
$$\Longrightarrow \frac{dw}{dz}=f'(z)=u_x+iv_x=1+e^x \cos y+i e^x \sin y=1+e^x (\cos y+i \sin y)=1+e^x e^{iy}=1+e^{x+iy}=1+e^z$$

---

02. Show that $w=\sin z$ is analytic and hence find $\frac{dw}{dz}$.
Ans: $$w=f(z)= \sin z\Longrightarrow u+iv=\sin(x+iy)=\sin x\, \cos iy +\cos x\, \sin iy$$ (we know that $\sin ix=i\sin hx, \cos ix=cos hx$) $$\Longrightarrow u+iv=\sin x\, \cosh y +i\cos x\, \sin hy$$ $$\Longrightarrow u=\sin x\,\cosh y \,\,\,\,\,;\,\,\,\,\, v=\cos x\,\sinh y$$ $$\Longrightarrow u_x=\cos x\,\cosh y \,\,\,\,\,;\,\,\,\,\, v_x=-\sin x\,\sinh y$$ $$\Longrightarrow u_y=\sin x\,\sinh y \,\,\,\,\,;\,\,\,\,\, v_y=\cos x\,\cosh y$$ From the above set of equations we have $u_x=v_y; v_x=-u_y$, hence $u,v$ satisfies the CR-equations, Hence $f$ is analytic

---

03. Construct the analytic function $f(z)=u+iv$ given that $u=\log{\sqrt{(x^2+y^2)}}$
Ans:
Let $f(z)=u+iv$, since $f$ is analytic, $u,v$ satisfies CR-equations and we have $u$ only, hence we have $$f'(z)=u_x+i v_x=u_x-iu_y$$ Now $u=\log{\sqrt{(x^2+y^2)}}=\log{(x^2+y^2)}^{1/2}=\frac{1}{2}\log{(x^2+y^2)}$ $$\Longrightarrow u_x=\frac{1}{2}\frac{2x}{x^2+y^2}\,\,\;\,\,\,\,\,\,u_y=\frac{1}{2}\frac{2y}{x^2+y^2}$$ $$\Longrightarrow f'(z)=\frac{1}{2}\frac{2x}{x^2+y^2}-i\frac{1}{2}\frac{2y}{x^2+y^2}$$ Now substitute $x=z$ and $y=0$ in the above equation we get $$f'(z)=1/z\Longrightarrow f(z)=\log{(z)}$$ Verified using Maple

---

04. Construct the analytic function $f(z)=u+iv$ given that $u=e^{2x}(x\cos{2y}-y\sin{2y})$
Ans:
Let $f(z)=u+iv$, since $f$ is analytic, $u,v$ satisfies CR-equations and we have $u$ only, hence we have $$f'(z)=u_x+i v_x=u_x-iu_y$$ $$\Longrightarrow u_x=2e^{2x}(x\cos{2y}-y\sin{2y})+e^{2x}(\cos{2y})\,\,\;\,\,\,\,\,\,u_y=e^{2x}(-2x\sin{2y}-\sin{2y}-2y\cos{2y})$$ $$\Longrightarrow f'(z)=(2e^{2x}(x\cos{2y}-y\sin{2y})+e^{2x}(\cos{2y}))+i(e^{2x}(-2x\sin{2y}-\sin{2y}-2y\cos{2y}))$$ Now substitute $x=z$ and $y=0$ in the above equation we get $$f'(z)=2e^{2z}z+e^{2z}+i(0)=e^{2z}(1+2z)\Longrightarrow f(z)=(1+2z)\frac{e^{2z}}{2}-\frac{e^{2z}}{2}+c=z e^{2z}+c$$

---

05. Construct the analytic function $f(z)=u+iv$ given that $v=e^{x}(x\sin{y}+y\cos{y})$
Ans:
Let $f(z)=u+iv$, since $f$ is analytic, $u,v$ satisfies CR-equations and we have $v$ only, hence we have $$f'(z)=u_x+i v_x=v_y+iv_x$$ $$\Longrightarrow v_x=e^x(x\sin{y}+y\cos{y})+e^x(\sin(y))\,\,\;\,\,\,\,\,\,v_y=e^x(x\cos{y}+\cos{y}-y\sin{y})$$ $$\Longrightarrow f'(z)=(e^x(x\cos{y}+\cos{y}-y\sin{y}))+i(e^x(x\sin{y}+y\cos{y})+e^x(\sin(y)))$$ Now substitute $x=z$ and $y=0$ in the above equation we get $$f'(z)=e^z(z+1)\Longrightarrow f(z)=(z+1)e^{z}-e^{z}+c=z e^{z}+c$$

---

06. Construct the analytic function $f(z)=u+iv$ given that $u=\frac{x^4-y^4-2x}{x^2+y^2}\,\,\,$-(OQP,MQP)
Ans:
Let $f(z)=u+iv$, since $f$ is analytic, $u,v$ satisfies CR-equations and we have $u$ only, hence we have $$f'(z)=u_x+i v_x=u_x-iu_y$$ $$\Longrightarrow u_x=\frac{(x^2+y^2)(4x^3-2)-(x^4-y^4-2x)2x}{(x^2+y^2)^2}\,\,\;\,\,\,\,\,\,u_y=\frac{(x^2+y^2)(-4y^3)-(x^4-y^4-2x)2y}{(x^2+y^2)^2}$$ $$\Longrightarrow f'(z)=\left(\frac{(x^2+y^2)(4x^3-2)-(x^4-y^4-2x)2x}{(x^2+y^2)^2}\right)-i\left(\frac{(x^2+y^2)(-4y^3)-(x^4-y^4-2x)2y}{(x^2+y^2)^2}\right)$$ Now substitute $x=z$ and $y=0$ in the above equation we get $$f'(z)=\frac{2z^5+2z^2}{z^4}-i(0)=2z+\frac{1}{z^2}\Longrightarrow f(z)=z^2-\frac{2}{z}+c$$

---

07. Construct the analytic function $f(z)=u+iv$ given that $v=\left(r-\frac{k^2}{r}\right)\sin \theta\,\,\,$-(MQP)
Ans:
Let $f(z)=u+iv$, since $f$ is analytic, $u,v$ satisfies CR-equations and we have $u$ only, hence we have $$f'(z)=f'\left(re^{i\theta}\right)e^{-i\theta}=(u_r+iv_r)\Longrightarrow f'(z)=e^{-i\theta} (u_r+iv_r)=e^{-i\theta} \left(\frac{v_\theta}{r}+iv_r\right)$$ $$\Longrightarrow v_r=\left(1+\frac{k^2}{r^2}\right)\sin\theta\,\,\;\,\,\,\,\,\,v_\theta=\left(r-\frac{k^2}{r}\right)\cos\theta$$ $$\Longrightarrow f'(z)=e^{-i\theta} \left[\frac{1}{r}\left(r-\frac{k^2}{r}\right)\cos\theta+i\left(1+\frac{k^2}{r^2}\right)\sin\theta\right]$$ Now substitute $r=z$ and $\theta=0$ in the above equation we get $$f'(z)=1-\frac{k^2}{z}\Longrightarrow f(z)=z+\frac{k^2}{z}+c$$

---

08. Construct the analytic function $f(z)=u+iv$ given that $u-v=e^x (\cos y-\sin y)$
Ans:
$$u-v=e^x (\cos y-\sin y),$$ Differentiate with respect to $x$ $$\Longrightarrow u_x-v_x=e^x (\cos y-\sin y)\,\,\,\,\,\,\,-(1)$$ Differentiate with respect to $y$ $$\Longrightarrow u_y-v_y=e^x (-\sin y-\cos y)$$ Using CR-equations, we can write the same as $$\Longrightarrow -v_x-u_x=e^x (-\sin y-\cos y) \,\,\,\,\,\,\,-(2)$$ $$(1)+(2) \Longrightarrow -2v_x=-2e^x\sin y,\,\,\,\,\Longrightarrow v_x=e^x\sin y$$ $$(1)-(2) \Longrightarrow 2u_x=2e^x\cos y,\,\,\,\,\Longrightarrow u_x=e^x\cos y$$

---

Do it yourself
01. Construct the analytic function $f(z)=u+iv$ given that $u-v=(x-y)(x^2+4xy+y^2)$.
02. Construct the analytic function $f(z)=u+iv$ given that $u+v=\frac{1}{r^2}(\cos 2\theta -\sin 2\theta)$.
03. Construct the analytic function $f(z)=u+iv$ given that sum of its real and imaginary parts is $x^3-y^3+3xy(x-y)$.

M4.3 Harmonioc Property & Orthogonal Property of a Complex Function

Harmonic Function

A function $\phi$ is said to be harmonic if it satisfies the Laplace equation $\nabla^2 \phi=0.$ i.e., $$\frac{\partial^2 \phi}{\partial x^2}+\frac{\partial^2 \phi}{\partial y^2}=0 \longrightarrow Cartesian Form$$ $$\frac{\partial^2 \phi}{\partial r^2}+\frac{1}{r}\frac{\partial \phi}{\partial r}+\frac{1}{r^2}\frac{\partial \phi^2}{\partial \theta^2}=0 \longrightarrow Polar Form$$

Harmonic Property

The real and imaginary parts of an analytic function are harmonic
Proof: Cartesian Form
Let $f(z)=u(x,y)+iv(x,y)$ be analytic. We have to show that $u$ and $v$ satisfies the Laplace equations. i.e., $\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0$ & $\frac{\partial^2 v}{\partial x^2}+\frac{\partial^2 v}{\partial y^2}=0$
Since $f(z)$ is analytic, $u$ and $v$ should satisfies CR-equations $$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial x}\,\,\,\,\,\,\,(1)$$ $$\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}\,\,\,\,\,\,\,(2)$$ Differentiate (1) with respect to $x$ and differentiate (2) with respect to $y$ partially, we get $$\frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 v}{\partial x \partial y};\,\,\,\, \frac{\partial^2 v}{\partial y \partial x}=-\frac{\partial^2 u}{\partial y^2}$$ But we know that $\frac{\partial^2 v}{\partial x \partial y}=\frac{\partial^2 v}{\partial y \partial x}$ Hence $$\frac{\partial^2 u}{\partial x^2}=-\frac{\partial^2 u}{\partial y^2}\Longrightarrow \frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0$$ Similarly, by differentiating (1) with respect to $y$ and differentiate (2) with respect to $x$ partially, we get $$\frac{\partial^2 u}{\partial y \partial x}=\frac{\partial^2 v}{\partial y^2};\,\,\,\, \frac{\partial^2 v}{\partial x^2}=-\frac{\partial^2 u}{\partial x \partial y}$$ Since $\frac{\partial^2 v}{\partial x \partial y}=\frac{\partial^2 v}{\partial y \partial x}$, $$\frac{\partial^2 v}{\partial y^2}=-\frac{\partial^2 v}{\partial x^2}\Longrightarrow \frac{\partial^2 v}{\partial x^2}+\frac{\partial^2 v}{\partial y^2}=0$$ Hence the real and imaginary parts of an analytic function are harmonic.
Polar Form
Let $f(z)=u(r,\theta)+iv(r,\theta)$ be analytic. We have to show that $u$ and $v$ satisfies the Laplace equations. i.e., $\frac{\partial^2 u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^2}\frac{\partial u^2}{\partial \theta^2}=0$ and $\frac{\partial^2 v}{\partial r^2}+\frac{1}{r}\frac{\partial v}{\partial r}+\frac{1}{r^2}\frac{\partial v^2}{\partial \theta^2}=0$. Since $f(z)$ is analytic $u$ and $v$ should satisfies CR-equations $$\frac{\partial u}{\partial r}=\frac{1}{r}\frac{\partial v}{\partial \theta}$$ $$\frac{\partial v}{\partial r}=-\frac{1}{r}\frac{\partial u}{\partial \theta}$$ Differentiate (1) with respect to $r$ and differentiate (2) with respect to $\theta$ partially, we get $$r\frac{\partial^2 u}{\partial r^2}+\frac{\partial u}{\partial r}=\frac{\partial^2 v}{\partial r \partial\theta};\,\,\,\, r\frac{\partial^2 v}{\partial \theta \partial r}= -\frac{\partial^2 u}{\partial \theta^2}$$ It is true that $\frac{\partial^2 v}{\partial r \partial\theta}=\frac{\partial^2 v}{\partial \theta \partial r}$, hence by substituting one in another we get $$r\frac{\partial^2 u}{\partial r^2}+\frac{\partial u}{\partial r}=-\frac{1}{r}\frac{\partial^2 u}{\partial \theta^2}$$ Divide throughout by $r$ we get $$\frac{\partial^2 u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^2}\frac{\partial u^2}{\partial \theta^2}=0.$$ Hence $u$ is Harmonic. Similarly we can obtain $\frac{\partial^2 v}{\partial r^2}+\frac{1}{r}\frac{\partial v}{\partial r}+\frac{1}{r^2}\frac{\partial v^2}{\partial \theta^2}=0$ by differentiating (1) with respect to $\theta$ and differentiate (2) with respect to $r$ partially.

Orthogonal Property

If $f(z)=u+iv$ is analytic then the family of curves $u(x,y)=c_1, v(x,y)=c_2, c_1$ and $c_2$ being constants, intersect each other orthogonally.
Proof: Cartesian Form
We say that two curves cuts orthogonally at some point if the product of the slope of the tangents ($\frac{dy}{dx}$) drawn to those curve at that point is equal to $-1$.
Let $m_1$ be slope of the tangent drawn to the curve $u(x,y)=c_1$. Differentiate $u(x,y)=c_1$ with respect to $x$ by treating $y$ as function of $x$, we get $$\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y} \cdot \frac{dy}{dx}=0 \Longrightarrow m_1=\frac{dy}{dx}=-\frac{\partial u}{\partial x}/\frac{\partial u}{\partial y}$$ Similarly let $m_2$ be slope of the tangent drawn to the curve $v(x,y)=c_2$. Differentiate $v(x,y)=c_2$ with respect to $x$ by treating $y$ as function of $x$, we get $$m_2=\frac{dy}{dx}=-\frac{\partial v}{\partial x}/\frac{\partial v}{\partial y}$$ $$\Longrightarrow m_1\cdot m_2=\frac{\partial u}{\partial x}/\frac{\partial u}{\partial y} \cdot \frac{\partial v}{\partial x}/\frac{\partial v}{\partial y}$$ Using CR-equation in the above equation we get $$\Longrightarrow m_1\cdot m_2=\frac{\partial v}{\partial y}/\frac{\partial u}{\partial y} \cdot -\frac{\partial u}{\partial y}/\frac{\partial v}{\partial y}=-1$$ Hence the curves intersect orthogonally.

• Converse is not true, for example the curves $u=\frac{x^2}{y}$ and $v=x^2+2y^2$ intersects orthogonally but it does not satisfies CR-equations, hence the function $f(z)=u+iv$ is not analytic.

M4.2 Cauchy Riemannian Equations in Cartesian Form/Polar Form

Cauchy Riemannian Equations in Cartesian Form - (VTU 2020,2019,2018,2017)

The necessary condition for a complex function $f(z)=u(x,y)+i v(x,y)$ is said to be analytic is that, there exists four first order continuous partial derivatives $\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \frac{\partial v}{\partial x}, \frac{\partial v}{\partial y}$ such that it satisfies the conditions $$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial x}$$ $$\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}$$ Proof: Let $f(z)=u(x,y)+iv(x,y)$ be a complex function and is analytic at $z=x+iy$. According to definition of Analytic function we have $f'(z)=lim_{\delta z\to z}{\frac{f(z+\delta z)-f(z)}{\delta z}}$. Let $\delta z= \delta x+i \delta y$ then $f(z+\delta z)= u(x+\delta x,y+\delta y)+i v(x+\delta x,y+\delta y)$ $$\Longrightarrow f’(z)=\frac{[u(x+\delta x,y+\delta y)-u(x,y)]+i[v(x+\delta x,y+\delta y)-v(x,y)]}{\delta x+ i \delta y}$$ $$\Longrightarrow f’(z)= lim_{\delta z\to 0}\frac{[u(x+\delta x,y+\delta y)-u(x,y)]}{ \delta x+ i \delta y }+i lim_{\delta z\to 0} \frac{[v(x+\delta x,y+\delta y)-v(x,y)]}{\delta x+ i \delta y}$$ Case 01: If $z$ approaches horizontally, i.e., $\delta y=0$ then $\delta z = \delta x$ $$\Longrightarrow f’(z)= lim_{\delta x\to 0}\frac{[u(x+\delta x,y)-u(x,y)]}{\delta x}+i lim_{\delta x\to 0} \frac{[v(x+\delta x,y)-v(x,y)]}{\delta x}$$ $$\Longrightarrow f’(z)= \frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x} \longrightarrow (1)$$
Case 02: If $z$ approaches vertically, i.e., $\delta x=0$ then $\delta z = i\delta y$ $$\Longrightarrow f’(z)= lim_{i\delta y\to 0}\frac{[u(x,y+\delta y)-u(x,y)]}{i\delta y}+ lim_{i\delta y\to 0} \frac{[v(x,y+\delta y)-v(x,y)]}{\delta y}$$ $$\Longrightarrow f’(z)= -i\frac{\partial u}{\partial y}+\frac{\partial v}{\partial y}\longrightarrow (2)$$ Comparing (1) and (2) we will get $$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial x}, \frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}$$

Cauchy Riemannian Equations in Polar Form - (VTU 2020,2019,2018,2017)

The necessary condition for a complex function $f(z)=u(r,\theta)+i v(r,\theta)$ is said to be analytic is that, there exists four first order continuous partial derivatives $\frac{\partial u}{\partial r}, \frac{\partial u}{\partial \theta}, \frac{\partial v}{\partial r}, \frac{\partial v}{\partial \theta}$ such that it satisfies the conditions $$\frac{\partial u}{\partial r}=\frac{1}{r}\frac{\partial v}{\partial \theta}$$ $$\frac{\partial v}{\partial r}=-\frac{1}{r}\frac{\partial u}{\partial \theta}$$ Proof:Let $f(z)=u(r,\theta)+iv(r,\theta)$ be a complex function and is analytic at $z=r e^{i\theta}$. $$\Longrightarrow f(z)=f(r e^{i\theta})= u(r,\theta)+iv(r,\theta)=u+iv$$ Since $f$ is an Analytic function, $f’(z)$ is exists and unique. Differentiate $f$ with respect to $r$ $$\Longrightarrow f’(z)=f’(r e^{i\theta}) e^{i\theta}= \frac{\partial u}{\partial r}+i\frac{\partial v}{\partial r} \longrightarrow (1)$$ Differentiate $f$ with respect to $\theta$ $$\Longrightarrow f’(z)=f’(r e^{i\theta})\cdot r\cdot e^{i\theta}\cdot i= \frac{\partial u}{\partial \theta}+i\frac{\partial v}{\partial \theta} \longrightarrow (2)$$ Put (1) in (2), we will get $$\Longrightarrow r\cdot i\left(\frac{\partial u}{\partial r}+i\frac{\partial v}{\partial r}\right)= \frac{\partial u}{\partial \theta}+i\frac{\partial v}{\partial \theta}$$ $$\Longrightarrow ir\frac{\partial u}{\partial r}-r\frac{\partial v}{\partial r}= \frac{\partial u}{\partial \theta}+i\frac{\partial v}{\partial \theta}$$ Comparing real and imaginary part of the above equation we get $$\frac{\partial u}{\partial r}=\frac{1}{r}\frac{\partial v}{\partial \theta}$$ $$\frac{\partial v}{\partial r}=-\frac{1}{r}\frac{\partial u}{\partial \theta}$$

M4.1 Complex Analysis-Introdution

Review of basic definitions & Key points

• The numbers are of the form $x+iy$ are called as complex numbers, where $x$ & $y$ are Real numbers and $i=\sqrt{-1}$.
• $i=\sqrt{-1}$ is obtained from the equation $x^2=-1$.
• $1,i,i^2,i^3$ are the fourth roots of the unity and $i^2=-1, i^3=-i, i^4=1$.
• Geometrically a complex number $x+iy$ is a point in Cartesian plane as shown in the following figure
• From the figure we can say that $r=\sqrt{x^2+y^2}$ and $\theta=tan^{-1}{y/x}$ are called as modulus and amplitude of the complex number $z$ denoted by $|z|$ and $amp(z)$ or $arg(z)$ respectively
• $z\bar=x-iy$ is the complex conjugate of the complex number $z=x+iy$

Euler's Formula

We know the following series 

• $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...$
• $\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-+...$
• $\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-+...$

If we multiply $i$ to the above equation we get

• $i \sin(x)=ix-\frac{ix^3}{3!}+\frac{ix^5}{5!}-+...$ Now
• $\cos(x)+i \sin(x)=1+ix-\frac{x^2}{2!}-\frac{ix^3}{3!}+\frac{x^4}{4!}+\frac{ix^5}{5!}-+... \,\,\,\,\,\,\,\,\,\, (1)$
• Now 
• $e^{ix}=1+ix-\frac{x^2}{2!}-\frac{ix^3}{3!}+\frac{x^4}{4!}+\frac{ix^5}{5!}-+... \,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)$

If we compare (1) and (2) we can conclude that

$cos(x) + i sin(x) = e^{ix}$

• $e^{-ix} = \cos(x)-i \sin(x)$ Hence,
• $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$  & $\sin(x)=\frac{e^{ix}-e^{-ix}}{2}$
• De-Moivre Theorem: $(\cos(\theta)+i \sin(\theta))^n=\cos(n\theta)+i \sin(n\theta)$

Complex Functions

Complex functions are mapping from complex numbers to complex numbers. Ex: $f(z)=\frac{1}{z}=\frac{1}{x+iy}=\frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2}=u(x,y)+iv(x,y)$

Limit of a Complex Function

Let $f(z)$ be a complex valued function, limit $l$ as $z$ tends to some point $z_0$, there exists $\epsilon, \delta$, when $|z\to z_0|<\epsilon, |f(z)−l|<\delta$. i.e., $lim_{z\to z_0}f(z)=l$

Continuity of a complex function

If $lim_{z\to z_0} f(z)=f(z_0)$ then $f(z)$ is continuous at $z=z_0$
Example

Differentiablility of a complex function

Let $\delta z=z−z_0$, as $z\to z_0$ or $\delta z\to 0, lim_{z\to z_0}\frac{f(z+\delta z)−f(z)}{\delta z}$ is the derivative of $f$ at $z_0$

Analytic Functions

A complex function $w=f(z)$ is said to be analytic at some point $z_0$ if $\frac{dw}{dz}=f′(z)=lim_{\delta z \to z_0} \frac{f(z+\delta z)−f(z)}{\delta z}$ exists and unique.