Mathematics II (BT-202) Unit 4 Notes - RGPV

How unit 4 is examined

Analytic functions, Cauchy-Riemann equations, harmonic functions, complex line integrals, Cauchy's integral formula, residues and real integrals; harmonic conjugate, Cauchy-Riemann and Cauchy's integral formula carry the most marks.

Functions of Complex Variables: Analytic Functions

Medium weight

Definition. A function $f(z)=u(x,y)+iv(x,y)$ is analytic at a point if it is differentiable at that point and at every point of some neighbourhood of it; analytic in a region means analytic at every point of it.

Key points.

The derivative is $f'(z)=\lim_{h\to0}\frac{f(z+h)-f(z)}{h}$, and the limit must be the same along every path of approach.
Analyticity needs differentiability in a whole neighbourhood, so it is stronger than differentiability at a single point.
Necessary condition: analytic implies the Cauchy-Riemann equations $u_x=v_y,\ u_y=-v_x$ hold.
Sufficient condition: if $u,v$ have continuous first partials satisfying Cauchy-Riemann, then $f$ is analytic.
Polynomials, $e^z$, $\sin z$, $\cos z$ are analytic everywhere (entire); $\bar z$, $|z|$, $\operatorname{Re}z$ are analytic nowhere.
Then $f'(z)=u_x+iv_x=v_y-iu_y$.

Example. $f=z\bar z=x^2+y^2$, so $u=x^2+y^2,\ v=0$.

Step	Working
At $0$ by definition	$f'(0)=\lim_{z\to0}\frac{z\bar z}{z}=\lim\bar z=0$, so differentiable at $0$
Partials	$u_x=2x,\ u_y=2y,\ v_x=v_y=0$
Cauchy-Riemann	$2x=0$ and $2y=0$ hold only at $(0,0)$

Not analytic at the origin, since Cauchy-Riemann fails in every neighbourhood of $0$.

Constant modulus. Let $f=u+iv$ be analytic with $u^2+v^2=c^2$. Differentiating in $x$ and $y$: $uu_x+vv_x=0$ and $uu_y+vv_y=0$. Put $u_y=-v_x,\ v_y=u_x$ in the second: $-uv_x+vu_x=0$. Multiply the first by $u$ and this by $v$ and add: $(u^2+v^2)u_x=0$. If $c\ne0$ then $u_x=0$, and similarly $v_x=0$; so all four partials vanish and $f$ is constant (if $c=0$, $f=0$).

A function is analytic only if it is differentiable in a whole neighbourhood, and differentiability at one point does not give it.

Answer frame. Open with the definition of analytic; for $z\bar z$ show the limit at $0$ first, then the failure of Cauchy-Riemann nearby; for constant modulus write $u^2+v^2=c^2$, differentiate, use Cauchy-Riemann; close with the conclusion in one line.

Pitfall: Showing Cauchy-Riemann holds at one point and calling the function analytic there loses the marks.

Asked: [7 marks] (Nov 2022) Show that $f(z)=z\bar z$ is differentiable but not analytic at origin. Asked: [7 marks] (Dec 2023) Determine $p$ so that $f(z)=\frac12\log(x^2+y^2)+i\tan^{-1}\left(\frac{px}{y}\right)$ is analytic (worked under Cauchy-Riemann Equations; $p=-1$). Asked: [7 marks] (Jun 2023) Prove that an analytic function with constant modulus is constant.

Harmonic Conjugate

Definition. A real function $\phi(x,y)$ with continuous second partials is harmonic if it satisfies Laplace's equation $\phi_{xx}+\phi_{yy}=0$; if $u$ is harmonic and $f=u+iv$ is analytic, then $v$ is called the harmonic conjugate of $u$.

Key points.

The real and imaginary parts of an analytic function are both harmonic, because Cauchy-Riemann gives $u_{xx}=v_{yx}=-u_{yy}$.
To test harmonic, find $u_{xx}$ and $u_{yy}$ and show their sum is zero.
$v$ is the harmonic conjugate of $u$ only when $u+iv$ is analytic; $u$ is then the conjugate of $-v$, not of $v$.
The conjugate is unique up to an added constant.
If $u$ is not harmonic, no analytic function has $u$ as its real part.
Conjugates are found by integrating the Cauchy-Riemann equations or by the Milne-Thomson method below.

Milne-Thomson method (given $u$).

Step 1: Find u_x and u_y; check u_xx + u_yy = 0.
Step 2: f'(z) = u_x - i u_y; replace x by z and y by 0.
Step 3: Integrate to get f(z) = ∫ [u_x(z,0) - i u_y(z,0)] dz + C.
Step 4: Separate the imaginary part of f to read off v.

(Given $v$: $f'(z)=v_y+iv_x$ with $x=z,\ y=0$.)

Example 1. $u=e^x\cos y$: $u_x=e^x\cos y,\ u_y=-e^x\sin y$; $u_{xx}+u_{yy}=e^x\cos y-e^x\cos y=0$, so harmonic. At $(z,0)$: $f'(z)=e^z-i\cdot0=e^z$, hence $f(z)=e^z+C$ and $v=e^x\sin y+c$.

Example 2. $u=e^{-2x}\sin2y$: $u_{xx}=4e^{-2x}\sin2y,\ u_{yy}=-4e^{-2x}\sin2y$, sum $0$, so harmonic. $u_x(z,0)=0,\ u_y(z,0)=2e^{-2z}$, so $f'(z)=-2ie^{-2z}$ and $f=ie^{-2z}+C$; conjugate $v=e^{-2x}\cos2y+c$.

Example 3. $u=e^{-x}(x\sin y-y\cos y)$: $u_x=e^{-x}(\sin y-x\sin y+y\cos y)$, $u_{xx}=e^{-x}(x\sin y-y\cos y-2\sin y)$; $u_y=e^{-x}(x\cos y-\cos y+y\sin y)$, $u_{yy}=e^{-x}(2\sin y-x\sin y+y\cos y)$. Adding, $u_{xx}+u_{yy}=0$, so $u$ is harmonic.

==If $u$ satisfies $u_{xx}+u_{yy}=0$ it is harmonic, and its conjugate $v$ makes $u+iv$ analytic.==

Answer frame. Open with the definition of harmonic; first prove $u_{xx}+u_{yy}=0$ line by line; then state Milne-Thomson, substitute $x=z,y=0$, integrate; close with $f(z)$ and the conjugate $v$.

Pitfall: Forgetting to test Laplace's equation first, or leaving out the constant $C$.

Asked: [7 marks] (Jun 2022, Nov 2022, Dec 2023) Construct the analytic function $f(z)$ whose real part is $e^x\cos y$; show $u=e^{-2x}\sin2y$ is harmonic and find its conjugate; show $u=e^x\cos y$ is harmonic and find its conjugate. Asked: [7 marks] (Dec 2024) Show that $u=e^{-x}(x\sin y-y\cos y)$ is harmonic.

Cauchy-Riemann Equations (without proof)

Definition. If $f(z)=u+iv$ is differentiable at $z=x+iy$, then the first partials exist and satisfy the Cauchy-Riemann equations $$u_x=v_y,\qquad u_y=-v_x.$$

Key points.

They are a necessary condition for differentiability; with continuous partials they are also sufficient.
In polar form ($z=re^{i\theta}$): $u_r=\frac1r v_\theta,\ \ v_r=-\frac1r u_\theta$.
The derivative is $f'(z)=u_x+iv_x=v_y-iu_y$.
Consequently $u,v$ are harmonic, and the curves $u=c_1,\ v=c_2$ cut orthogonally.
To find an unknown constant, apply both equations and solve.
For $|f|^2=u^2+v^2$: $\nabla^2|f|^2=4|f'|^2$.

Example (find $p$). $u=\frac12\log(x^2+y^2),\ v=\tan^{-1}(px/y)$. Then $u_x=\frac{x}{x^2+y^2}$, $u_y=\frac{y}{x^2+y^2}$, $v_x=\frac{py}{y^2+p^2x^2}$, $v_y=\frac{-px}{y^2+p^2x^2}$. From $u_x=v_y$: $\frac{1}{x^2+y^2}=\frac{-p}{y^2+p^2x^2}$, needing $p^2=1$ and $-p=1$. $p=-1$ (then $u_y=-v_x$ also holds). For $f=e^x(\cos ky+i\sin ky)$: $u_x=e^x\cos ky$, $v_y=ke^x\cos ky$, so $k=1$.

Proof of $\nabla^2|f|^2=4|f'|^2$. $\frac{\partial^2}{\partial x^2}(u^2+v^2)=2(u_x^2+v_x^2)+2(uu_{xx}+vv_{xx})$ and the same in $y$ with $u_y,v_y$. Adding, and using $u_{xx}+u_{yy}=0=v_{xx}+v_{yy}$: $2(u_x^2+u_y^2+v_x^2+v_y^2)$. By Cauchy-Riemann $u_y^2=v_x^2$ and $v_y^2=u_x^2$, giving $4(u_x^2+v_x^2)=\mathbf{4|f'(z)|^2}$.

Stokes' theorem (for the short note). $\oint_C\vec F\cdot d\vec r=\iint_S(\nabla\times\vec F)\cdot\hat n\,dS$: the circulation of $\vec F$ around a closed curve $C$ equals the flux of its curl through any surface $S$ bounded by $C$, with $C$ oriented by the right-hand rule.

==The Cauchy-Riemann equations $u_x=v_y$ and $u_y=-v_x$ are necessary for $f=u+iv$ to be analytic.==

Answer frame. Short note: state the equations in Cartesian and polar form, say necessary (and sufficient with continuity), give $f'(z)$ and an example, then Stokes' statement. Find-the-constant: name $u,v$, differentiate, equate, solve. Proof: expand, add, apply harmonicity.

Pitfall: The examiner's key lists $P=1$ once, but the correct value is $p=-1$; check by substituting back.

Asked: [7 marks] (Dec 2024, Jun 2025) Determine $P$ so that $f(z)=\frac12\log(x^2+y^2)+i\tan^{-1}(Px/y)$ is analytic; find all $K$ such that $f(z)=e^x(\cos ky+i\sin ky)$ is analytic. Asked: [7 marks] (Jun 2022) Prove that $\left[\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right]|f(z)|^2=4|f'(z)|^2$. Asked: [7 marks] (Dec 2023) Write short note on (i) Cauchy-Riemann equations (ii) Stokes theorem.

Line Integral

Definition. The complex line integral of $f(z)=u+iv$ along a curve $C$ is $\int_C f(z)\,dz=\int_C(u+iv)(dx+i\,dy)$.

Key points.

Put $dz=dx+i\,dy$ and reduce to real integrals along the given path.
On a path $y=g(x)$ use $dy=g'(x)dx$; on a parametrised path $z=z(t)$ use $dz=z'(t)dt$.
On a horizontal segment $dy=0,\ dz=dx$; on a vertical segment $dx=0,\ dz=i\,dy$.
Reversing the path changes the sign, and integrals over consecutive pieces add.
For a non-analytic integrand the value depends on the path, so use the path given.

Example 1 (two-piece path). $\int_0^{1+i}(x-y+ix^2)dz$: along $y=0,\ x:0\to1$, $\int_0^1(x+ix^2)dx=\frac12+\frac i3$. Along $x=1,\ dz=i\,dy,\ y:0\to1$, $i\int_0^1(1-y+i)dy=i\cdot\frac12-1=-1+\frac i2$. Total $=-\frac12+\frac{5i}{6}$.

Example 2. $\int_{(0,0)}^{(1,1)}(3x^2+4xy+ix^2)dz$ along $y=x^2$: $dz=(1+2ix)dx$, integrand $3x^2+4x^3+ix^2$. Product real part $3x^2+2x^3$, imaginary part $x^2+6x^3+8x^4$. Integrating $0$ to $1$: $\frac32+\frac{103}{30}i$.

==Reduce the complex line integral to real integrals by substituting $dz=dx+i\,dy$ and the equation of the path.==

Answer frame. Open with $\int f\,dz=\int(u+iv)(dx+i\,dy)$; draw the path if it has pieces; substitute per piece, integrate, add; close with the boxed complex number.

Asked: [7 marks] (Jun 2023, Jun 2025) Solve $\int_0^{1+i}(x-y+ix^2)dz$ along the real axis from $0$ to $1$ then parallel to the imaginary axis from $1$ to $1+i$; evaluate $\int_{(0,0)}^{(1,1)}(3x^2+4xy+ix^2)dz$ along $y=x^2$.

Cauchy-Goursat theorem (without proof)

Definition. If $f(z)$ is analytic inside and on a simple closed curve $C$, then $\oint_C f(z)\,dz=0$. Goursat's contribution is that continuity of $f'$ is not needed.

Key points.

The curve must be simple and closed, and $f$ analytic everywhere inside and on it.
A singularity inside $C$ breaks the theorem, and this is why residues are needed.
Consequence: the integral between two points is independent of the path within the region.
For a multiply connected region, the integral over the outer boundary equals the sum over the inner boundaries.

Cauchy Integral formula (without proof)

Definition. If $f(z)$ is analytic inside and on a simple closed curve $C$ and $a$ lies inside $C$, then $$f(a)=\frac1{2\pi i}\oint_C\frac{f(z)}{z-a}\,dz,\qquad f^{(n)}(a)=\frac{n!}{2\pi i}\oint_C\frac{f(z)}{(z-a)^{n+1}}\,dz.$$

Key points.

The formula gives the value of an analytic function inside $C$ from its values on $C$.
So $\oint_C\frac{f(z)}{z-a}dz=2\pi i\,f(a)$ and $\oint_C\frac{f(z)}{(z-a)^{n+1}}dz=\frac{2\pi i}{n!}f^{(n)}(a)$.
First locate the singularities and check which lie inside $C$; those outside contribute nothing.
Write the integrand as $\frac{f(z)}{(z-a)^{n+1}}$ with $f$ analytic in and on $C$.
If there are several points inside, split by partial fractions and apply the formula to each part.
An analytic function has derivatives of every order.

Example 1. $\oint_{|z|=2}\frac{e^{2z}}{(z+1)^3}dz$: $z=-1$ is inside, $n+1=3$, $f=e^{2z},\ f''=4e^{2z}$. Value $=\frac{2\pi i}{2!}\cdot4e^{-2}=$ $4\pi ie^{-2}$.

Example 2. $\oint_{|z|=3}\frac{\sin\pi z^2+\cos\pi z^2}{(z-1)(z-2)}dz$: both poles inside; $\frac1{(z-1)(z-2)}=\frac1{z-2}-\frac1{z-1}$. With $f=\sin\pi z^2+\cos\pi z^2$, $f(2)=1,\ f(1)=-1$. Value $=2\pi i[f(2)-f(1)]=$ $4\pi i$.

Example 3. $\oint_{|z|=2}\frac{e^{2z}}{(z-1)^2(z-3)}dz$: $z=3$ is outside, $z=1$ inside. $f=\frac{e^{2z}}{z-3},\ f'(1)=\frac{2e^2(-2)-e^2}{4}=-\frac{5e^2}4$. Value $=2\pi i f'(1)=$ $-\frac{5\pi i e^2}{2}$.

Example 4. $\oint_{|z-1|=1/2}\frac{z^3e^{-z}}{(z-1)^3}dz$: $f=z^3e^{-z},\ f''=e^{-z}(z^3-6z^2+6z),\ f''(1)=e^{-1}$. Value $=\frac{2\pi i}{2}e^{-1}=$ $\dfrac{\pi i}{e}$.

==$\oint_C\frac{f(z)}{(z-a)^{n+1}}dz=\frac{2\pi i}{n!}f^{(n)}(a)$ when $f$ is analytic in and on $C$ and $a$ is inside.==

Answer frame. Open with the formula and its conditions; find singularities and mark which lie inside $|z|=r$; identify $f$ and $n$; differentiate, substitute, multiply by $2\pi i/n!$; close with the value.

Pitfall: Including a singularity that lies outside $C$, or forgetting the $n!$.

Asked: [7 marks] (Jun 2022, Nov 2022, Dec 2024) Find $\int_C\frac{e^{2z}}{(z+1)^3}dz$, $|z|=2$; evaluate $\int_C\frac{e^{2z}}{(z-1)^2(z-3)}dz$, $|z|=2$; evaluate $\int_c\frac{z^3e^{-z}}{(z-1)^3}dz$, $|z-1|=\frac12$. Asked: [7 marks] (Jun 2023) Use Cauchy integral formula to solve $\oint_C\frac{\sin\pi z^2+\cos\pi z^2}{(z-1)(z-2)}dz$, $C$: $|z|=3$.

Singular Points

Definition. A point $z_0$ where $f(z)$ is not analytic is a singular point; it is isolated if $f$ is analytic in some punctured neighbourhood of it.

Key points.

Removable singularity: the limit exists, as $\frac{\sin z}{z}$ at $0$.
Pole of order $m$: $f=\frac{g(z)}{(z-z_0)^m}$ with $g(z_0)\ne0$, as $\frac1{(z-1)^3}$ at $1$.
Essential singularity: the Laurent series has infinitely many negative powers, as $e^{1/z}$ at $0$.
The type is read from the principal part of the Laurent series about $z_0$.

Poles & Residues

Definition. The residue of $f$ at an isolated singularity $z_0$ is the coefficient $a_{-1}$ of $\frac1{z-z_0}$ in its Laurent series about $z_0$.

Key points.

Simple pole: $\operatorname{Res}(f,a)=\lim_{z\to a}(z-a)f(z)$.
If $f=\frac{p(z)}{q(z)}$ with $q(a)=0,\ q'(a)\ne0$: $\operatorname{Res}=\frac{p(a)}{q'(a)}$.
Pole of order $m$: $$\operatorname{Res}(f,a)=\frac1{(m-1)!}\lim_{z\to a}\frac{d^{m-1}}{dz^{m-1}}\big[(z-a)^mf(z)\big].$$
Find the pole and its order from the denominator, and remember that a numerator zero can lower the order.
For an essential singularity expand in a Laurent series and read $a_{-1}$.

Example 1. $\frac{ze^z}{(z-1)^3}$: pole $z=1$, $m=3$. $\frac{d}{dz}(ze^z)=(1+z)e^z$, $\frac{d^2}{dz^2}(ze^z)=(2+z)e^z$. $\operatorname{Res}=\frac12(3)e=$ $\dfrac{3e}{2}$.

Example 2. $\frac{e^{iz}}{z^2+1}$: simple poles $z=\pm i$. At $i$: $\frac{e^{i\cdot i}}{2i}=\frac{e^{-1}}{2i}$; at $-i$: $\frac{e^{-i\cdot i}}{-2i}=-\frac{e}{2i}$.

Example 3. $\frac1{(z-1)(z-2)^4}$: at $z=1$ (simple): $\frac1{(1-2)^4}=1$. At $z=2$ ($m=4$): $\frac1{3!}\frac{d^3}{dz^3}(z-1)^{-1}=\frac16\cdot\frac{-6}{(z-1)^4}=-1$.

For a pole of order $m$ at $a$, the residue is $\frac1{(m-1)!}\lim_{z\to a}\frac{d^{m-1}}{dz^{m-1}}[(z-a)^mf(z)]$.

Answer frame. Open by defining residue; factorise the denominator to list poles and orders; state the matching formula; differentiate and substitute; close with the residue at each pole.

Asked: [7 marks] (Dec 2023, Dec 2024, Jun 2025) Find the residue of $\frac{ze^z}{(z-1)^3}$ at its pole; find the poles and residues of $\frac{e^{iz}}{z^2+1}$; if $f=\frac1{(z-1)(z-2)^4}$, find the residue at all poles.

Residue Theorem

Definition. If $f(z)$ is analytic inside and on a simple closed curve $C$ except at finitely many singular points $z_1,\dots,z_n$ inside $C$, then $$\oint_Cf(z)\,dz=2\pi i\sum_{k=1}^n\operatorname{Res}(f,z_k).$$

Key points.

First find all singular points and keep only those inside $C$; outside ones are ignored.
Compute the residue at each retained pole by the formulas of the previous topic.
Multiply the sum of residues by $2\pi i$.
It contains Cauchy's integral formula as the case of a simple pole, and reduces to Cauchy-Goursat when no singularity is inside.
A zero of the denominator that is cancelled is not a pole.

Example 1. $\oint_{|z|=2}\frac{dz}{(z+4)z^8}$: $z=-4$ is outside; $z=0$ is a pole of order 8. Since $\frac1{z+4}=\frac14\sum(-1)^k\left(\frac z4\right)^k$, the residue is the coefficient of $z^7$: $\frac{(-1)^7}{4^8}=-\frac1{4^8}$. Value $=$ $-\dfrac{2\pi i}{4^8}=-\dfrac{\pi i}{2^{15}}$.

Example 2. $\oint_{|z|=2}\frac{\tan z}{z^2-1}dz$: poles $\pm1$ and $\pm\frac\pi2\approx\pm1.571$ (zeros of $\cos z$), all inside. At $\pm1$: residue $\frac{\tan1}2$ each. At $\pm\frac\pi2$: $\tan z$ has residue $-1$, so $f$ has $\frac{-1}{\pi^2/4-1}=-\frac4{\pi^2-4}$ each. Sum $=\tan1-\frac8{\pi^2-4}$, so value $=$ $2\pi i\left[\tan1-\frac8{\pi^2-4}\right]$.

==$\oint_Cf(z)dz=2\pi i\sum\operatorname{Res}$ over the singular points lying inside $C$.==

Answer frame. Open by stating the theorem; list poles, tick those inside $|z|=r$; compute each residue; sum and multiply by $2\pi i$; close with the value.

Pitfall: For $\tan z$ forgetting the poles at $\pm\frac\pi2$, which lie inside $|z|=2$.

Asked: [7 marks] (Jun 2022) Evaluate $\int_C\frac1{(z+4)z^8}dz$, $C$: $|z|=2$. Asked: [7 marks] (Nov 2022) By Residue theorem, evaluate $\oint_C\frac{\tan z}{z^2-1}dz$, $C$: $|z|=2$.

Application of Residues theorem for Evaluation of Real Integral (Unit Circle)

Definition. An integral $\int_0^{2\pi}F(\cos\theta,\sin\theta)\,d\theta$ is converted to a contour integral over $|z|=1$ using $z=e^{i\theta}$, so $\cos\theta=\frac{z+z^{-1}}2$, $\sin\theta=\frac{z-z^{-1}}{2i}$, $d\theta=\frac{dz}{iz}$.

Key points.

Then $\int=\oint_{|z|=1}\dots=2\pi i\sum\operatorname{Res}$ over poles with $|z|<1$.
For $\cos n\theta$ take the real part of $z^n$ and finally the real part of the answer.

Example. $I=\operatorname{Re}\oint\frac{z^4}{5+2(z+1/z)}\frac{dz}{iz}=\operatorname{Re}\oint\frac{z^4\,dz}{i(2z+1)(z+2)}$. Only $z=-\frac12$ is inside; residue $\frac{1/16}{i\cdot2\cdot\frac32}=\frac1{48i}$. $I=2\pi i\cdot\frac1{48i}=\frac{\pi}{24}$.

Asked: [7 marks] (Jun 2023) Using complex integration, solve $\int_0^{2\pi}\frac{\cos4\theta}{5+4\cos\theta}d\theta$.

Last-minute revision

Analytic means differentiable in a neighbourhood; Cauchy-Riemann: $u_x=v_y,\ u_y=-v_x$ (necessary condition).
$f'(z)=u_x+iv_x$; polar form $u_r=\frac1rv_\theta,\ v_r=-\frac1ru_\theta$.
Harmonic: $u_{xx}+u_{yy}=0$; Milne-Thomson: $f'(z)=u_x-iu_y$ at $x=z,y=0$.
$e^x\cos y\Rightarrow f=e^z$; $e^{-2x}\sin2y\Rightarrow f=ie^{-2z}$.
For $f=\frac12\log(x^2+y^2)+i\tan^{-1}(px/y)$, $p=-1$; for $e^x(\cos ky+i\sin ky)$, $k=1$.
$\nabla^2|f|^2=4|f'|^2$; analytic with constant modulus is constant.
Line integral: $dz=dx+i\,dy$; horizontal $dz=dx$, vertical $dz=i\,dy$.
Cauchy-Goursat: $\oint_Cf\,dz=0$ if $f$ is analytic in and on $C$.
Cauchy: $\oint\frac{f}{(z-a)^{n+1}}dz=\frac{2\pi i}{n!}f^{(n)}(a)$; $\oint\frac{e^{2z}}{(z+1)^3}=4\pi ie^{-2}$.
Residue at pole of order $m$: $\frac1{(m-1)!}\frac{d^{m-1}}{dz^{m-1}}[(z-a)^mf]$; $\frac{ze^z}{(z-1)^3}$ has residue $\frac{3e}2$.
Residue theorem: $2\pi i\sum\operatorname{Res}$ inside $C$; unit circle: $z=e^{i\theta}$, $d\theta=\frac{dz}{iz}$.

Memory hooks

CR: "$u_x$ equals $v_y$, and $u_y$ is minus $v_x$": diagonals same, off-diagonals opposite.
Milne-Thomson: replace $x$ by $z$, $y$ by $0$, then integrate.
Cauchy derivative: the power in the denominator is $n+1$, so subtract one for $n$, divide by $n!$.
Residue of order $m$: differentiate $m-1$ times, divide by $(m-1)!$.
Inside gets counted, outside gets ignored; then multiply by $2\pi i$.

Coverage checklist

Functions of Complex Variables: Analytic Functions: $z\bar z$ (Nov 2022), find $p$ (Dec 2023), constant modulus (Jun 2023).
Harmonic Conjugate: $e^x\cos y$ and $e^{-2x}\sin2y$ (Jun 2022, Nov 2022, Dec 2023), $e^{-x}(x\sin y-y\cos y)$ (Dec 2024).
Cauchy-Riemann Equations (without proof): find $P$ and $K$ (Dec 2024, Jun 2025), $\nabla^2|f|^2$ (Jun 2022), short note with Stokes (Dec 2023).
Line Integral: both line integrals (Jun 2023, Jun 2025).
Cauchy-Goursat theorem (without proof): statement and consequences, not asked recently.
Cauchy Integral formula (without proof): $\frac{e^{2z}}{(z+1)^3}$, $\frac{e^{2z}}{(z-1)^2(z-3)}$, $\frac{z^3e^{-z}}{(z-1)^3}$, $\sin\pi z^2+\cos\pi z^2$ integral.
Singular Points: types of singularity, not asked recently.
Poles & Residues: $\frac{ze^z}{(z-1)^3}$, $\frac{e^{iz}}{z^2+1}$, $\frac1{(z-1)(z-2)^4}$.
Residue Theorem: $\frac1{(z+4)z^8}$ (Jun 2022), $\frac{\tan z}{z^2-1}$ (Nov 2022).
Application of Residues theorem for Evaluation of Real Integral (Unit Circle): $\int_0^{2\pi}\frac{\cos4\theta}{5+4\cos\theta}d\theta$ (Jun 2023).

Account

Mathematics II (BT-202) - Unit 4 Short Notes

How unit 4 is examined

Functions of Complex Variables: Analytic Functions

Harmonic Conjugate

Cauchy-Riemann Equations (without proof)

Line Integral

Cauchy-Goursat theorem (without proof)

Cauchy Integral formula (without proof)

Singular Points

Poles & Residues

Residue Theorem

Application of Residues theorem for Evaluation of Real Integral (Unit Circle)

Last-minute revision

Memory hooks

Coverage checklist