We will cover following topics

Analytic Functions
- Singular Points
Cauchy-Riemann Equations
- Harmonic Functions
- Milne-Thomson’s Method
Cauchy’s Theorem
Cauchy’s Integral Formulae
- Liouville’s Theorem
- Rouche’s Theorem
PYQs
- Analytic Functions
- Cauchy-Riemann Equations

Analytic Functions

A function $f (z)$ is said to be analysic at a point $z_{0}$ if there exists a ndghbothood $| z - z_{0} | < δ$ at all points of which $f^{'} (z)$ exists.

Singular Points

A point at which $f (z)$ fails to be analytic is called a singular point or singularity of $f (z)$ . Some types of singularities include:

i) Isolated Singularities ii) Poles iii) Removable Singularities

Cauchy-Riemann Equations

A necessary condition that $w = f (z) = u (x, y) + i c (x, y)$ be analytic in a region $R$ is that, in $R$ , $u$ and $v$ satisfy the Cauchy-Riemann equations

\frac{a}{\partial x} = \frac{\partial v}{\partial y}, \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x}

If the partial derivatives in above equations are continuous in $R$ , then the Cauchy-Ricmann equations are sufficient conditions that $f (z)$ be analytic in $R$ .

Harmonic Functions

A function $h (x, y)$ is called harmonic if it satisfies the Laplace’s equation $h_{x x} + h_{y y} = 0$ .

If $f (z) = u + i v$ is analytic and $u, v$ both satisfy Laplace’s equation, then $u$ and $v$ are called conjugate harmonic functions.

Milne-Thomson’s Method

This method is used to construct an analytic function when real or imaginary part of the function is given. Let $f (x, y) = u (x, y) + i v (x, y)$ be the required function.

Case 1: $u (x, y)$ is given. Then, $f (z)$ is obtained by integrating $f^{'} (z) = u_{x} (z, 0) - i u_{y} (z, 0)$ .

Case 2: $v (x, y)$ is given. Then, $f (z)$ is obtained by integrating $f^{'} (z) = v_{y} (z, 0) + i v_{x} (z, 0)$ .

Cauchy’s Theorem

According to Cauchy’s Theorem, if $f (z)$ is regular and $f^{'} (z)$ is continuous at each point within and on a closed contour $C$ , then $\int_{C} f (z) d z = 0$

Cauchy’s Integral Formulae

According to Cauchy’s Integral Formula, if $f (z)$ is analytic within and on any closed contour $C$ and if $a$ be a point within the contour $C$ , then

f (a) = \frac{1}{2 π i} \int \frac{f (z)}{z - a} d z

Liouville’s Theorem

Suppose that for all $z$ in the entire complex plane, i) $f (z)$ is analytic and
ii) $f (z)$ is bounded, i.e. $| f (z) | < M$ for some constant $M$ .

Then $f (z)$ must be a constant.

Rouche’s Theorem

Suppose $f (z)$ and $g (z)$ are analytic inside and on a simple closed curve $C$ and suppose $| g (z) | < | f (z) |$ on $C$ . Then $f (z) + g (z)$ and $f (z)$ have the same number of zeros inside $C$ .

PYQs

Analytic Functions

1) Use Cauchy integral formula to evaluate $\int_{c} \frac{e^{3 z}}{(z + 1)^{4}} d z$ , where $c$ is the circle $| z | = 2$ .

[2012, 15M]

From Cauchy Integral Formula, we have

f^{n} (a) = \frac{n!}{2 π i} \int_{C} \frac{f (z) d z}{(z - a)^{n + 1}}

Put $a = - 1$ , $n = 3$

f^{3} (- 1) = \frac{3!}{2 π i} \int_{C} \frac{f (z) d z}{(z + 1)^{4}} \dots (i)

Take $f (z) = e^{3 z}$ , then $f^{n} (z) = 3^{n} e^{3 z}$

∴ (3) (- 1) = 3^{3} e^{3} = \frac{27}{e^{3}}

$∴$ from $(i)$

\frac{27}{e^{3}} = \frac{3!}{2 π i} \int \frac{e^{3 z} d z}{(z + 1)^{4}}

\Rightarrow \int_{c} \frac{e^{3 z} d z}{(z + 1)^{4}} = \frac{9 π i}{e^{3}}

2) Prove that the function $f$ defined by $f (z) = {\begin{cases} \frac{z^{5}}{| z |^{4}}, & z \neq 0 \\ 0, & z = 0 \end{cases}$ is not differentiable at $z = 0$ .

[2007, 12M]

Try yourself

3) If all zeros of a polynomial $P (z)$ lies in a half plane then show that zeros of the derivatives $P^{'} (z)$ also lie in the same half plane.

[2004, 15M]

We can assume without loss of generality that the zeros of $P (z)$ lie in the half plane Re $z < 0 .$ Let $P (z) = \prod_{j = 1}^{n} (z - α_{j})$ where $α_{j} = x_{j} + i y_{j}, x_{j} < 0$

If $Re z \geq 0,$ then $P (z) \neq 0$ and

\begin{aligned} \frac{P^{'} (z)}{P (z)} & = \sum_{j = 1}^{n} \frac{1}{z - α_{j}} \\ = \sum_{j = 1}^{n} \frac{1}{x - x_{j} + i (y - y_{j})} \\ = \sum_{j = 1}^{n} \frac{x - x_{j} - i (y - y_{j})}{{(x - x_{j})}^{2} + {(y - y_{j})}^{2}} \end{aligned}

Since $x_{j} < 0, 1 \leq j \leq n,$ it follows that

Re (\frac{P^{'} (z)}{P (z)}) = \sum_{j = 1}^{n} \frac{x - x_{j}}{{(x - x_{j})}^{2} + {(y - y_{j})}^{2}} > 0

whenever Re $z = x \geq 0 .$ Thus $\frac{x^{'} γ}{P (z)}$ and therefore $P^{'} (z)$ has no zeros in the right half plane Re $z \geq 0$ . Hence all zeros of $P^{'} (z)$ lie in the same half plane in which the zeros of $P (z)$ lie.

4) Suppose that $f$ and $g$ are two analytic functions on the set $ϕ$ of all complex numbers with $f (\frac{1}{n}) = g (\frac{1}{n})$ for $n = 1$ , 2, 3, $\dots$ . Then show that $f (z) = g (z)$ for each $z$ in $ϕ$ .

[2002, 12M]

Let $G (z) = f (z) - g (z),$ then $G (\frac{1}{n}) = 0$ for $n = 1, 2, \dots .$ We shall show that $G (z) \equiv 0$ for $z \in C$ which would prove the result.

Let $G (z) = \sum_{n = 0}^{\infty} a_{n} z^{n}$ be the power series of $G (z)$ with center 0 and radius of convergence $R,$ clearly $R > 0 .$ We shall now prove that $a_{n} = 0$ for every $n$ If $a_{n} \neq 0$ for some $n,$ let $a_{k}$ be the first non-zero coefficient. Then $G (z) = z^{k} (a_{k} + a_{k + 1} z + \dots) = z^{k} H (z)$ Clearly $H (z)$ is analytic in $| z | < R,$ and $H (0) \neq 0 .$ We now claim that $H (z) \neq 0$ in a neighborhood $| z | < δ$ of $0 .$ Let $ϵ = \frac{| H (0) |}{2},$ then continuity of $H (z)$ at $z = 0$ implies that there exists a $δ > 0$ such that $| z | < δ \Rightarrow | H (z) - H (0) | < ϵ$ or $| H (0) | - ϵ < | H (z) | < | H (0) | + ϵ$ for $| z | < δ .$

Thus $| H (z) | > \frac{| H (0) |}{2} > 0$ for $| z | < δ .$ Consequently, $G (z) \neq 0$ for any $z$ in $0 < | z | < δ .$ But this is not possible, as $| z | < δ$ contains all but finitely many $\frac{1}{n},$ at which $G (z)$ vanishes. Thus our assumption that $a_{n} \neq 0$ for some $n$ is false, thus $G (z) \equiv 0$ in $| z | < R$

Let $z^{'}$ be any point in $C,$ and let $r (t), a \leq t \leq b$ be a continuous curve joining 0 and $z^{'}$ . Using uniform continuity of $r (t),$ we get a partition $a = t_{0} < t_{1} < \dots < t_{n} = b$ of $[a, b]$ such that $r (t_{0}) = 0, r (t_{1}) = z_{1}, \dots, r (t_{n}) = r (b) = z^{'},$ and $| z_{j} - z_{j - 1} | < R$ .

Now the disc $K_{0} = | z - 0 | < R$ contains $z_{1}$ , the center of disc $K_{1} = | z - z_{1} | < R$ . since $G (z_{1}) = 0$ as $z_{1} \in K_{0} \cap K_{1},$ and $K_{0} \cap K_{1}$ contains a sequence of points $y_{n}$ such that $y_{n} \to z_{1}$ and $G (y_{n}) = 0,$ we can prove as before that $G (z) \equiv 0$ in $K_{1}$ . Proceeding in this way, in $n$ steps we get $G (z) \equiv 0$ in $K_{n},$ or $G (z^{'}) = 0 .$ since $z^{'}$ is an arbitrary point of $C,$ we get $G (z) \equiv 0$ in $C$ .

5) Prove that the Riemann zeta function $ζ$ defined by $ζ (z) = \sum_{n = 1}^{\infty} n^{- z}$ converges for $Re z > 1$ and converges uniformly for $Re z \geq 1 + ε$ where $ε > 0$ is arbitrary small.

[2001, 12M]

$| \frac{1}{n^{z}} | = | \frac{1}{n^{x} \cdot n^{i y}} | = | \frac{1}{n^{x}} | ∵ | \frac{1}{n^{i y}} | = | \frac{1}{e^{i y \log n}} | = 1$ since $\sum_{n = 1}^{\infty} \frac{1}{n^{x}}$ converges for $x > 1,$ it follows that $\sum_{n = 1}^{\infty} n^{- z}$ converges absolutely for $Re z > 1$ . If Re $z \geq 1 + ϵ,$ then $\frac{1}{n^{x}} \leq \frac{1}{n^{1 + ϵ}}$ and $\sum_{n = 1}^{\infty} | n^{- z} | \leq \sum_{n = 1}^{\infty} \frac{1}{n^{1 + ϵ}}$ for Re $z \geq 1 + ϵ$ . Weierstrass’ M-test gives that the given series converges uniformly and absolutely for $Re z \geq 1 + ϵ$

Cauchy-Riemann Equations

1) Suppose $f (z)$ is analytic function on a domain $D \in C$ and satisfies the equation $I m f (z) = (R e f (z))^{2}$ , $Z \in D$ . Show that $f (z)$ is constant in $D$ .

[2019, 10M]

Hint: Take $f (z) = u + i u^{2}$ and apply CR equations.

In the final step, we get,

$u_{x} = 0$ , $u_{y} = 0$ , $v_{x} = 0$ , $v_{y} = 0$

Hence, $u = c_{1}$ and $v = c_{2}$

$⟹$ $f (z) = c_{1} + i c_{2}$ is constant.

2) Prove that the function: $u (z, y) = (x - 1)^{3} - 3 x y^{2} + 3 y^{2}$ is harmonic and find its harmonic conjugate and the corresponding analytic function $f (z)$ in terms of $z$ .

[2018, 10M]

Let, $f (z) = u + i v$ $u = (x - 1)^{3} - 3 x y^{2} + 3 y^{2}$ $u_{x} = 3 (x - 1)^{2} - 3 y^{2}$ $u_{x x} = 6 (x - 1) \dots (1)$ $u y = - 6 x y + 6 y$ $u_{y y} = - 6 x + 6 = - 6 (x - 1) \dots (2)$

$∴ u_{x x} + u_{y y} = 0$ Hence, function $u$ is harmonic.

We can use Milne’s method to find harmonic conjugate of $u$ .

$ϕ_{1} (z, 0) = {(\frac{\partial u}{\partial x})}_{x = z, y = 0} = 3 (z - 1)^{2}$ $ϕ_{2} (z, 0) = {(\frac{\partial u}{\partial y})}_{x = z, y = 0} = 0$ $f (z) = \int [ϕ_{1} (z, 0) - i ϕ_{2} (z, 0)] d z + c$ $= \int 3 (z - 1)^{2} d z + c$ $= (z - 1)^{3} + c$

∴ f (z) = (z - 1)^{3} + c

\begin{aligned} f (z) & = [(x - 1) + i y]^{3} + c \\ = (x - 1)^{3} - i y^{3} + 3 (x - 1) i y [(x - 1) + i y] \\ = (x - 1)^{3} - 3 (x - 1) y^{2} - i y^{3} + 3 i (x - 1)^{2} y \\ = (x - 1)^{3} - 3 (x - 1) y^{2} + i [3 (x - 1)^{2} y - y^{3}] \\ ∴ v (x, y) = 3 (x - 1)^{2} y - y^{3} \end{aligned}

3) Let $f = u + i v$ be analytic function on the unit disc $D = {z \in C : | z | < 1}$ . Show that $\frac{\partial^{2} u}{\partial x^{2}} + \frac{\partial^{2} u}{\partial y^{2}}$ =0= $\frac{\partial^{2} v}{\partial x^{2}} + \frac{\partial^{2} v}{\partial y^{2}}$ at all points of $D$ .

[2017, 15M]

If $f (z)$ is analytic in $R,$ then the Cauchy-Riemann equations

\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \dots (i)

and

\frac{\partial v}{\partial x} = - \frac{\partial u}{\partial y} \dots (i i)

are satisfied in $R$ and g $u$ and $v$ have continuous second partial derivatives. We can differentiate both sides of $(i)$ with respect to $x$ and $(i i)$ with respect to $y$ to obtain

\frac{\partial^{2} u}{\partial x^{2}} = \frac{\partial^{2} v}{\partial x \partial y}

and

\frac{\partial^{2} v}{\partial y \partial x} = - \frac{\partial^{2} u}{\partial y^{2}}

from which

\frac{\partial^{2} u}{\partial x^{2}} = - \frac{\partial^{2} u}{\partial y^{2}} or \frac{\partial^{2} u}{\partial x^{2}} + \frac{\partial^{2} u}{\partial y^{2}} = 0

Similarly, by differentiating both sides of $(i)$ with respect to $y$ and $(i i)$ with respect to $x$ , we get $\frac{\partial^{2} v}{\partial x^{2}} + \frac{\partial^{2} v}{\partial y^{2}} = 0$

4) Is $v (x, y) = x^{3} - 3 {x y}^{2} + 2 y$ a harmonic function? Prove your claim. If yes, find its conjugate harmonic function $u (x, y)$ and hence, obtain the analytic function whose real and imaginary parts are $u$ and $v$ respectively.

[2016, 10M]

Let $f$ be the function $f = u + i v$ Given: $v = x^{3} - 3 x y^{2} + 2 y$ Therefore, $\begin{array}{l} v_{x} = 3 x^{2} - 3 y^{2}, v_{x x} = 6 x \dots (i) \\ v_{y} = - 6 x y + 2, v_{y y} = - 6 x \dots (i i) \end{array}$ Now, $v_{x x} + v_{y y} = + 6 x - 6 x = 0$ $\Rightarrow v$ is a harmonic function. Let $u$ be harmonic conjugate of $v$ . By Cauchy-Riemann equations, $\begin{aligned} u_{x} & = v_{y} \Rightarrow u_{x} = - 6 x y + 2 \\ \frac{\partial u}{\partial x} & = - 6 x y + 2 \end{aligned}$ $\Rightarrow$ $\int d u = \int (- 6 x y + 2) d x$ $\Rightarrow u = - 3 x^{2} y + 2 x + g (y)$ , where $g$ is an arbitrary function of $y$ $\dots (i i i)$ Also, $u_{y} = - v_{x} \Rightarrow u_{y} = - (3 x^{2} - 3 y^{2})$ $\Rightarrow$ $\frac{\partial u}{\partial y} = - 3 x^{2} + 3 y^{2}$ $\Rightarrow$ $\int d u = \int (- 3 x^{2} + 3 y^{2}) d y$ $\Rightarrow u = - 3 x^{2} y + y^{3} + n (x),$ where $n (x)$ is an arbitrary fucntion of $x \dots$ (iv) From $(i i i)$ and $(i v)$ $u = - 3 x^{2} y + 2 x + y^{3} + c, where c is a constant$ So, $f = (- 3 x^{2} y + 2 x + y^{3} + c) + i (x^{3} - 3 x y^{2} + 2 y)$

5) Show that the function $v (x, y) = \ln (x^{2} + y^{2}) + x + y$ is harmonic. Find its conjugate harmonic function $u (x, y)$ . Also, find the corresponding analytic function $f (z) = u + i v$ in terms of $z$ .

[2015, 10M]

6) Prove that the function $f (z) = u + i v$ , where $f (z) = \frac{x^{3} (1 + i) - y^{3} (1 - i)}{x^{2} + y^{2}}$ , $z \neq 0$ ; $f (0) = 0$ satisfies Cauchy-Riemann equations at the origin, but the derivative of $f$ at $z = 0$ does not exist.

[2014, 10M]

7) Prove that if $b e^{a + 1} < 1$ , where $a$ and $b$ are positive and real, then the function $z^{n} e^{- a} - b e^{z}$ has $n$ zeros in the unit circle.

[2013, 10M]

8) Show that the function defined by $f (z) = {\begin{matrix} \frac{x^{3} y^{5} (x + i y)}{x^{6} + y^{10}}, z \neq 0 \\ 0, z = 0 \end{matrix}$ is not analytic at the origin though it satisfies Cauchy Riemann-Equations at the origin.

[2012, 12M]

9) If $f (z) = u + i v$ is an analytic function of $z = x + i y$ and $u - v = \frac{e^{y} - \cos x + \sin x}{\cosh y - \cos x}$ , find $f (z)$ subject to the condition, $f (\frac{π}{2}) = (3 - i) / 2$ .

[2011, 12M]

10) Show that $u (x, y) = 2 x - x^{3} + 3 x y^{2}$ is a harmonic function. Find a harmonic conjugate of $u (x, y)$ . Hence, find the analytic function $f$ for which $u (x, y)$ is the real part.

[2010, 12M]

11) Prove that all the roots of $z^{7} - 5 z^{3} + 12 = 0$ lie between the circles $| z | = 1$ and $| z | = 2$ .

[2006, 15M]

Let $g (z) = z^{7} - 5 z^{3}, f (z) = 12,$ then i) On $| z | = 1, | g (z) | \leq | z^{7} | + 5 | z^{3} | = 6 < 12 = | f (z) |$ . ii) Both $g (z)$ and $f (z)$ are analytic on and within $| z | < 1$ iii) Both $g (z)$ and $f (z)$ have no zeros on $| z | = 1$

By Rouche’s theorem, $f (z) + g (z) = z^{7} - 5 z^{3} + 12$ and $f (z)$ have the same number of zeros inside $| z | = 1 .$ But $f (z) = 12$ has no zeros anywhere and in particular in the region $| z | < 1$ therefore $z^{7} - 5 z^{3} + 12$ has no zeros inside the unit circle. Now we take $g (z) = 12 - 5 z^{3}, f (z) = z^{7}$

i) On $| z | = 2, | g (z) | \leq 12 + 5 | z^{3} | = 52 < 2^{7} = | f (z) |$ . ii) Both $g (z)$ and $f (z)$ are analytic on and within $| z | < 2$

Therefore by Rouche’s theorem, $g (z) + f (z) = z^{7} - 5 z^{3} + 12$ and $f (z)$ have the same number of zeros inside $| z | = 2 .$ since $f (z) = z^{7}$ has 7 zeros $(z = 0$ is a zero of order 7 of $f (z))$ inside $| z | = 2,$ the given polynomial has seven zeros inside $| z | = 2$ i.e. all its zeros lie inside $| z | = 2$ since $z^{7} - 5 z^{3} + 12$ has no zeros inside and on $| z | = 1,$ therefore all zeros lie in the ring $1 < | z | < 2$ .

12) If $f (z) = u + i v$ is an analytic function of the complex variable $z$ and $u - v = e^{x} (\cos y - \sin y)$ , determine $f (z)$ in terms of $z$ .

[2005, 12M]

Let $F (z) = (1 + i) f (z) = (1 + i) (u + i v) = (u - v) + i (u + v) .$ Now $\frac{\partial^{2} (u - v)}{\partial x^{2}} + \frac{\partial^{2} (u - v)}{\partial y^{2}} = e^{x} (\cos y - \sin y) + e^{x} (- \cos y + \sin y) = 0$ Let $F (z) = U + i V,$ where $U = u - v$ is harmonic. If $f (z)$ is analytic, then so is $F (z)$ and $\frac{d F}{d z} = \frac{\partial U}{\partial x} + i \frac{\partial V}{\partial x} = \frac{\partial U}{\partial x} - i \frac{\partial U}{\partial y}$ as $\frac{\partial V}{\partial x} = - \frac{\partial U}{\partial y}$ by the Cauchy-Riemann equations. Thus $\begin{aligned} F^{'} (z) & = e^{x} (\cos y - \sin y) - i e^{x} (- \sin y - \cos y) \\ = e^{x} (\cos y + i \sin y) + i e^{x} (\cos y + i \sin y) \\ = (1 + i) e^{x} \cdot e^{i y} = (1 + i) e^{z} \end{aligned}$ Thus $F (z) = (1 + i) e^{z}$ and $f (z) = e^{z},$ which is the required function.

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