Complex analysis formulas

Conjugate $\bar{z}=x-jy=r{e}^{-j\theta }$, with $z\bar{z}=|z{|}^2$, $\mathrm{Re}z=\frac{z+\bar{z}}{2}$ and $\mathrm{Im}z=\frac{z-\bar{z}}{2j}$. The argument is fixed only up to adding $2k\pi$.

Here ${\tan }^{-1}$ returns a value in $(-\pi /2,\pi /2)$, and $\mathrm{Arg}0$ is undefined. Some texts take the principal range as $[0,2\pi )$; these notes use $(-\pi ,\pi ]$ throughout, so $\mathrm{Arg}(-1)=\pi$.

The argument rules hold only up to multiples of $2\pi$: $\mathrm{Arg}(z_1{z}_2)$ can differ from $\mathrm{Arg}{z}_1+\mathrm{Arg}{z}_2$ by $\pm 2\pi$. Triangle inequality: $||z_1|-|z_2||\le |z_1+z_2|\le |z_1|+|z_2|$.

For every integer $n$. For a fractional power it gives only one of the values; use the roots formula below.

Every $z\ne 0$ has exactly $n$ distinct $n$-th roots. They lie on the circle of radius $r^{1/n}$ (the positive real root), at angles spaced $2\pi /n$ apart; for $n\ge 3$ they are the vertices of a regular $n$-gon.

With $\omega =e^{j2\pi /n}$ the roots are $1,\omega ,{\omega }^2,\dots ,{\omega }^{n-1}$ and ${\omega }^n=1$. Cube roots: $1+\omega +{\omega }^2=0$ with $\omega =-\frac{1}{2}+j\frac{\sqrt{3}}{2}$.

$|e^z|=e^x$, and $y$ is one value of $\arg e^z$; $e^z\ne 0$ for every $z$, and $e^z$ has period $2\pi j$. $\sin z$ and $\cos z$ are entire but unbounded, so $|\sin z|$ and $|\cos z|$ can exceed 1, e.g. $\cos (j)=\cosh 1\approx 1.543$. The zeros of $\sin z$ are $z=n\pi$ and those of $\cos z$ are $z=(n+\frac{1}{2})\pi$, with $n$ an integer; all of them are real.

Defined for $z\ne 0$ and multi-valued; $\mathrm{Log}$ is the principal value ($k=0$). The principal branch is analytic everywhere except on its branch cut, the non-positive real axis (where $\mathrm{Arg}$ jumps by $2\pi$), and off the cut $(\mathrm{Log}z{)}^{\prime }=1/z$. Example: $\mathrm{Log}(-1)=j\pi$. The rule $\mathrm{Log}(z_1{z}_2)=\mathrm{Log}{z}_1+\mathrm{Log}{z}_2$ can fail by $\pm 2\pi j$.

Defined for $z\ne 0$. Multi-valued when $c$ is not an integer: for rational $c=p/q$ in lowest terms with $q\ge 2$ there are $q$ values; for irrational or non-real $c$ there are infinitely many. All values of $j^j$ are $e^{-\pi /2-2k\pi }$ with $k\in \mathbb{Z}$, every one of them real.

Necessary at every point where $f$ is differentiable. Sufficient for differentiability at $z_0$ when, in addition, $u_x,u_y,v_x,v_y$ exist in a neighbourhood of $z_0$ and are continuous at $z_0$. $f$ is analytic in a domain where these conditions hold at every point. Subscripts denote partial derivatives.

For $z=r{e}^{j\theta }$ with $r\ne 0$ and $f=u(r,\theta )+jv(r,\theta )$. Use it when $f$ is naturally written in $r$ and $\theta$: for $\mathrm{Log}z=\ln r+j\theta$ with $-\pi <\theta <\pi$ (off the branch cut), $u_r=1/r=v_{\theta }/r$ and $v_r=0=-u_{\theta }/r$, and $f^{\prime }=e^{-j\theta }/r=1/z$.

If $f=u+jv$ is analytic in a domain, both $u$ and $v$ are harmonic there (the derivative of an analytic function is again analytic, so the second partials exist and are continuous). $v$ is called a harmonic conjugate of $u$. Where $f^{\prime }(z)\ne 0$, the level curves $u=c_1$ and $v=c_2$ cut at right angles. In polar form Laplace's equation is $u_{rr}+\frac{1}{r}{u}_r+\frac{1}{r^2}{u}_{\theta \theta }=0$.

Builds $f$ directly from $u$ without finding $v$: compute $u_x$ and $u_y$, put $x=z$ and $y=0$, and integrate in $z$. If $v$ is given instead, $f^{\prime }(z)=v_y(z,0)+j{v}_x(z,0)$. $C$ is a complex constant. When $u$ is given, $\mathrm{Re}C$ is fixed by $u$ itself and $\mathrm{Im}C$ stays arbitrary unless a value such as $f(0)$ is also given (and the other way round when $v$ is given).

Conformal wherever $cz+d\ne 0$, since $w^{\prime }(z)=\frac{ad-bc}{(cz+d{)}^2}$. For $c\ne 0$, $z=-d/c$ maps to $w=\infty$ and $z=\infty$ maps to $w=a/c$. It maps circles and straight lines (a line counts as a circle through $\infty$) onto circles and straight lines, and it is fixed completely by the images of three distinct points. Its inverse $z=\frac{dw-b}{-cw+a}$ is also bilinear.

Fixed points solve $z=\frac{az+b}{cz+d}$; there are at most two unless the map is the identity (for $c=0$ and $a\ne d$ there is one finite fixed point, and $\infty$ is also fixed). The cross-ratio equation gives the unique map sending $z_1,z_2,z_3$ to $w_1,w_2,w_3$. If one of the given points is $\infty$, delete the two factors that contain it.

$z=z(t)$, $a\le t\le b$, traces $C$ in the given direction. Circle $|z-z_0|=\rho$ counter-clockwise: $z=z_0+\rho e^{j\theta }$, $dz=j\rho e^{j\theta }d\theta$, $0\le \theta \le 2\pi$. Segment from $z_1$ to $z_2$: $z=z_1+t(z_2-z_1)$, $0\le t\le 1$.

Counter-clockwise, any radius $\rho >0$. Every term of a Laurent series integrates to zero except the $1/(z-z_0)$ term, which is why residues work.

Used to bound an integral, and to show that the integral over a large semicircular arc tends to zero when real integrals are evaluated by residues.

Conditions: $C$ is a simple closed contour and $f$ is analytic at every point on $C$ and inside it (more generally, in a simply connected domain containing $C$). Analyticity on $C$ alone is not enough: $1/z$ is analytic on $|z|=1$, yet ${\oint }_{|z|=1}dz/z=2\pi j$.

$f$ analytic on and inside the simple closed contour $C$ (counter-clockwise), and $z_0$ strictly inside $C$. If $z_0$ is outside $C$ the integral is $0$ by Cauchy's theorem; if $z_0$ lies on $C$ the formula does not apply (in general the integral then does not exist as an ordinary integral).

Same conditions, $n=0,1,2,\dots$ It shows that an analytic function has derivatives of every order, all analytic. Mind the power $n+1$: for $(z-z_0{)}^3$ in the denominator use $f^{\prime \prime }(z_0)$ and divide by $2!$.

Only a test for divergence: if $\lim a_n\ne 0$, or the limit does not exist, the series diverges. The converse is false: $\sum 1/n$ diverges although $1/n\to 0$.

The geometric series (real or complex $r$) diverges for $|r|\ge 1$. $\sum 1/n$ ($p=1$) diverges, while $\sum 1/n^2={\pi }^2/6$. These two families are the usual benchmarks for the comparison tests.

Use whichever limit exists (the ratio form needs $a_n\ne 0$ for all large $n$); $R=\infty$ if the limit is $0$ and $R=0$ if it is $\infty$; in general $1/R=\mathrm{limsup}|a_n{|}^{1/n}$. The series converges absolutely for $|z-z_0|<R$ and diverges for $|z-z_0|>R$; points on the circle $|z-z_0|=R$ must be tested separately. Inside the disc it may be differentiated or integrated term by term without changing $R$.

The series equals $f$ throughout the largest open disc about $z_0$ in which $f$ is analytic. When the singularity on the edge of that disc is a pole or an essential singularity, that disc is exactly the disc of convergence, so $R$ is the distance from $z_0$ to that singularity (a removable singularity does not limit $R$). $z_0=0$ gives the Maclaurin series. The expansion is unique, so any method that produces a power series (substitution, multiplication, term-by-term differentiation or integration) gives the Taylor series.

$\sinh z$ and $\cosh z$ have the same series as $\sin z$ and $\cos z$ without the alternating signs. Binomial series: $(1+z{)}^{\alpha }=1+\alpha z+\frac{\alpha (\alpha -1)}{2!}{z}^2+\cdots$ for $|z|<1$ (principal value).

Valid in an annulus $r_1<|z-z_0|<r_2$ in which $f$ is analytic ($r_1$ may be $0$ and $r_2$ may be $\infty$); $C$ is any closed counter-clockwise contour around $z_0$ inside the annulus. The negative-power part is the principal part. In a given annulus the expansion is unique, but each annulus bounded by the singularities has its own series. In practice the coefficients come from known series, not from the integral.

Here $p$ and $q$ are analytic at $z_0$; take $k=0$ when $p(z_0)\ne 0$. Find orders from series: $1-\cos z=\frac{z^2}{2}-\cdots$ has a zero of order 2 at $0$, so $\frac{1-\cos z}{z^5}$ has a pole of order 3 there, not 5.

$a_{-1}$ is the coefficient of $(z-z_0{)}^{-1}$ in the Laurent series valid in $0<|z-z_0|<r$; $C$ is a small counter-clockwise circle around $z_0$ enclosing no other singularity. At a removable singularity the residue is $0$. At an essential singularity it must be read from the Laurent series: ${\mathrm{Res}}_{z=0}{e}^{1/z}=1$ and ${\mathrm{Res}}_{z=0}{z}^2{e}^{1/z}=\frac{1}{3!}=\frac{1}{6}$.

Here $p$ and $q$ are analytic at $z_0$. The $p/q^{\prime }$ form avoids factorising: ${\mathrm{Res}}_{z=n\pi }\frac{1}{\sin z}=\frac{1}{\cos n\pi }=(-1{)}^n$, ${\mathrm{Res}}_{z=0}\cot z=\frac{\cos 0}{\cos 0}=1$, and for $\frac{1}{z^4+1}$ at a root $z_k$ the residue is $\frac{1}{4z_k^3}=-\frac{z_k}{4}$, since $z_k^4=-1$.

Using a larger $m$ than the true order still gives the right answer (with more work); using a smaller one does not. For high orders, or when $f$ contains $\sin$, $\cos$ or $e^z$, reading $a_{-1}$ from a series is usually faster: ${\mathrm{Res}}_{z=0}\frac{\sin z}{z^4}$ is the coefficient of $z^3$ in $\sin z$, which is $-\frac{1}{6}$.

Equivalently ${\mathrm{Res}}_{z=\infty }f=-b_1$, where $b_1$ is the coefficient of $1/z$ in the expansion of $f$ valid for large $|z|$. The sum rule holds when $f$ has finitely many singularities. So if a contour $C$ encloses every finite singularity, ${\oint }_{C}fdz=-2\pi j{\mathrm{Res}}_{z=\infty }f$. For a rational $f$ whose denominator degree is at least the numerator degree plus 2, ${\mathrm{Res}}_{z=\infty }f=0$ and the finite residues add to zero.

$C$ is a simple closed contour traversed counter-clockwise; $f$ is analytic on $C$ and inside it except at finitely many isolated singularities $z_k$ inside $C$. Only singularities strictly inside count. Cauchy's theorem and both integral formulas are special cases.

$C$ is a simple closed contour traversed counter-clockwise; $f$ is analytic on and inside $C$ except for poles inside it, and has no zeros or poles on $C$. $N$ and $P$ are the numbers of zeros and poles of $f$ inside $C$, each counted as many times as its order. Reason: at a zero of order $n$, $f^{\prime }/f$ has a simple pole with residue $n$, and at a pole of order $p$ one with residue $-p$. The left side also equals the number of times the image curve $f(C)$ winds counter-clockwise around $w=0$. Related (Rouche's theorem): if $f$ and $g$ are analytic on and inside $C$ and $|g(z)|<|f(z)|$ on $C$, then $f$ and $f+g$ have the same number of zeros inside $C$; e.g. on $|z|=2$, $|3z^2+1|\le 13<32=|z^5|$, so all five zeros of $z^5+3z^2+1$ lie in $|z|<2$.

Substitute $z=e^{j\theta }$, so $d\theta =\frac{dz}{jz}$ and $\theta :0\to 2\pi$ traces the unit circle once counter-clockwise. $F$ must be a rational function of $\cos \theta$ and $\sin \theta$, finite for all $\theta$. Keep only the poles with $|z|<1$. Also useful: $\cos n\theta =\frac{z^n+z^{-n}}{2}$ and $\sin n\theta =\frac{z^n-z^{-n}}{2j}$.

$a$ and $b$ real. If $a<-|b|$ the value is $-\frac{2\pi }{\sqrt{a^2-b^2}}$. If $|a|\le |b|$ the integrand is infinite somewhere on the interval and the integral does not exist as a proper integral. Over $0$ to $\pi$ the cosine version gives half, $\frac{\pi }{\sqrt{a^2-b^2}}$.

Conditions: $P$ and $Q$ polynomials, $\mathrm{deg}Q\ge \mathrm{deg}P+2$, and $Q$ has no real zeros. Close the real axis with a large semicircle in the upper half-plane; its contribution vanishes by the ML-inequality. Count only the poles with $\mathrm{Im}z>0$. For an even integrand, ${\int }_0^{\infty }$ is half of ${\int }_{-\infty }^{\infty }$.

Conditions: $\mathrm{deg}Q\ge \mathrm{deg}P+1$ and $Q$ has no real zeros. With $P$, $Q$ real, take real and imaginary parts to get $\int \frac{P}{Q}\cos mxdx$ and $\int \frac{P}{Q}\sin mxdx$. Use $e^{jmz}$, never $\cos mz$, on the contour: $|e^{jmz}|=e^{-my}\le 1$ in the upper half-plane, whereas $\cos mz$ grows there. Standard result: ${\int }_{-\infty }^{\infty }\frac{\cos mx}{x^2+a^2}dx=\frac{\pi }{a}{e}^{-ma}$ for $a>0$, $m\ge 0$.

Use this when $Q$ has real zeros $x_k$ and every one of them is simple. Degree conditions as above: $\mathrm{deg}Q\ge \mathrm{deg}P+1$ when $m>0$, and $\mathrm{deg}Q\ge \mathrm{deg}P+2$ when $m=0$. Each real pole is bypassed by a small semicircle above it, which contributes $-\pi j$ times its residue; that is where the half-weight $\pi j$ comes from. Example: $g=e^{jmz}/z$ with $m>0$ has residue $1$ at $z=0$ and no other pole, so $\mathrm{PV}{\int }_{-\infty }^{\infty }\frac{e^{jmx}}{x}dx=\pi j$; its imaginary part gives ${\int }_{-\infty }^{\infty }\frac{\sin mx}{x}dx=\pi$, and hence ${\int }_0^{\infty }\frac{\sin x}{x}dx=\frac{\pi }{2}$.