\(\blacksquare \) Complex identities
\(\blacksquare \) A complex function \(f\left ( z\right ) \) is analytic in a region \(D\) if it is defined and differentiable at all points in \(D\). One way to check for analyticity is to use the Cauchy Riemann (CR) equations (this is a necessary condition but not sufficient). If \(f\left ( z\right ) \) satisfies CR everywhere in that region then it is analytic. Let \(f\left ( z\right ) =u\left ( x,y\right ) +iv\left ( x,y\right ) \), then these two equations in Cartesian coordinates are
Sometimes it is easier to use the polar form of these. Let \(f\left ( z\right ) =r\cos \theta +i\sin \theta \), then the equations become
To remember them, think of the \(r\) as the \(x\) and \(\theta \) as the \(y\).
Let us apply these on \(\sqrt {z}\) to see how it works. Since \(z=re^{i\theta +2n\pi }\) then \(f\left ( z\right ) =\) \(\sqrt {r}e^{i\frac {\theta }{2}+n\pi }\).This is multi-valued function. One value for \(n=0\) and another for \(n=1\). The first step is to make it single valued. Choosing \(n=0\) gives the principal value. Then \(f\left ( z\right ) =\sqrt {r}e^{i\frac {\theta }{2}}\). Now we find the branch points. \(z=0\) is a branch point. We can pick \(-\pi <\theta <\pi \) and pick the negative real axis as the branch cut (the other branch point being \(-\infty \)). This is one choice.
We could have picked \(0<\theta <2\pi \) and had the positive \(x\) axis as the branch cut, where now the second branch point is \(+\infty \) but in both cases, origin is still part of the branch cut. Let us stick with \(-\pi <\theta <\pi \).
Given all of this, now\(\sqrt {z}=\sqrt {r}e^{i\frac {\theta }{2}}=\sqrt {r}\left ( \cos \left ( \frac {\theta }{2}\right ) +i\sin \left ( \frac {\theta }{2}\right ) \right ) \), hence \(u=\sqrt {r}\cos \left ( \frac {\theta }{2}\right ) \) and \(v=\sqrt {r}\sin \left ( \frac {\theta }{2}\right ) \). Therefore \(\frac {\partial u}{\partial r}=\frac {1}{2}\frac {1}{\sqrt {r}}\cos \left ( \frac {\theta }{2}\right ) ,\) and \(\frac {\partial v}{\partial \theta }=\frac {1}{2}\sqrt {r}\cos \left ( \frac {\theta }{2}\right ) \) and \(\frac {\partial u}{\partial \theta }=-\frac {1}{2}\sqrt {r}\sin \left ( \frac {\theta }{2}\right ) \) and \(\frac {\partial v}{\partial r}=\frac {1}{2}\frac {1}{\sqrt {r}}\sin \left ( \frac {\theta }{2}\right ) \). Applying Cauchy-Riemann above gives
Satisfied. and for the second equation
so \(\sqrt {z}\) is analytic in the region \(-\pi <\theta <\pi \), and not including branch points and branch cut.
\(\blacksquare \) We can’t just say \(f\left ( z\right ) \) is Analytic and stop. Have to say \(f\left ( z\right ) \) is analytic in a region or at a point. When we say \(f\left ( z\right ) \) analytic at a point, we mean analytic in small region around the point.
If \(f\left ( z\right ) \) is defined only at an isolated point \(z_{0}\) and not defined anywhere around it, then the function can not be analytic at \(z_{0}\) since it is not differentiable at \(z_{0}\). Also \(f\left ( z\right ) \) is analytic at a point \(z_{0}\) if the power series for \(f\left ( z\right ) \) expanded around \(z_{0}\) converges to \(f\left ( z\right ) \) evaluated at \(z_{0}\). An analytic complex function mean it is infinitely many times differentiable in the region, which means the limit exist \(\lim _{\Delta z\rightarrow 0}\frac {f\left ( z+\Delta z\right ) -f\left ( z\right ) }{\Delta z}\) and does not depend on direction.
\(\blacksquare \) Before applying the Cauchy Riemann equations, make sure the complex function is first made to be single valued.
\(\blacksquare \) Remember that Cauchy Riemann equations as necessary but not sufficient condition for function to be analytic. The extra condition needed is that all the partial derivatives are continuous. Need to find example where CR is satisfied but not the continuity on the partial derivatives. Most of the HW problems just needs the CR but good to keep an eye on this other condition.
\(\blacksquare \) Cauchy-Goursat: If \(f\left ( z\right ) \) is analytic on and inside closed contour \(C\) then \({\displaystyle \oint \limits _{C}} f\left ( z\right ) dz=0\). But remember that if \({\displaystyle \oint \limits _{C}} f\left ( z\right ) dz=0\) then this does not necessarily imply \(f\left ( z\right ) \) is analytic on and inside \(C\). So this is an IF and not an IFF relation. For example \({\displaystyle \oint \limits _{C}} \frac {1}{z^{2}}dz=0\) around unit circle centered at origin, but clearly \(\frac {1}{z^{2}}\) is not analytic everywhere inside \(C\), since it has a singularity at \(z=0\).
proof of Cauchy-Goursat: The proof uses two main ideas. It uses the Cauchy-Riemann equations and also uses Green theorem. Green’s Theorem says
So Green’s Theorem transforms integration on the boundary \(C\) of region \(D\) by integration over the area inside the boundary \(C\). Let \(f\left ( z\right ) =u+iv\). And since \(z=x+iy\) then \(dz=dx+idy\). Therefore
We now apply (1) to each of the two integrals in (3). Hence the first integral in (2) becomes
But from CR, we know that \(-\frac {\partial u}{\partial y}=\frac {\partial v}{\partial x}\), hence the above is zero. And the second integral in (2) becomes
But from CR, we know that \(\frac {\partial u}{\partial x}=\frac {\partial v}{\partial y}\), hence the above is zero. Therefore the whole integral in (2) is zero. Therefore \({\displaystyle \oint \limits _{C}} f\left ( z\right ) dz=0\). QED.
\(\blacksquare \) Cauchy residue: If \(f\left ( z\right ) \) is analytic on and inside closed contour \(C\) except at some isolated points \(z_{1},z_{2},\cdots ,z_{N}\) then \({\displaystyle \oint \limits _{C}} f\left ( z\right ) dz=2\pi i\sum _{j=1}^{N}\operatorname {Res}\left ( f\left ( z\right ) \right ) _{z=z_{j}}\). The term \(\operatorname {Res}\left ( f\left ( z\right ) \right ) _{z=z_{j}}\) is the residue of \(f\left ( z\right ) \) at point \(z_{j}\). Use Laurent expansion of \(f\left ( z\right ) \) to find residues. See above on methods how to find Laurent series.
\(\blacksquare \) Maximum modulus principle: If \(f\left ( z\right ) \) is analytic in some region \(D\) and is not constant inside \(D\), then its maximum value must be on the boundary. Also its minimum on the boundary, as long as \(f\left ( z\right ) \neq 0\) anywhere inside \(D\). In the other hand, if \(f\left ( z\right ) \) happened to have a maximum at some point \(z_{0}\) somewhere inside \(D\), then this implies that \(f\left ( z\right ) \) is constant everywhere and will have the value \(f\left ( z_{0}\right ) \) everywhere. What all this really mean, is that if \(f\left ( z\right ) \) is analytic and not constant in \(D\), then its maximum is on the boundary and not inside.
There is a complicated proof of this. See my notes for Physics 501. Hopefully this will not come up in the exam since I did not study the proof.
\(\blacksquare \) These definitions from book of Joseph Bak
\(\blacksquare \) Some important formulas.
If \(f\left ( z\right ) \) is analytic on and inside \(C\) then
If \(f\left ( z\right ) \) is analytic on and inside \(C\) then and \(z_{0}\) is a point in \(C\) then
From the above, we find, where here \(f\left ( z\right ) =1\)
On Finding coefficient of the principle part of the Laurent series expansion around \(z_{0}\). Let
The goal is to determine all the coefficients \(b_{1},b_{2},\cdots ,b_{N}\) in Laurent series expansion. This assumes the largest order of the pole is finite. To find \(b_{1}\), we multiply both side of the above by \(\left ( z-z_{0}\right ) ^{N}\) which gives
Differentiating both sides \(N-1\) times w.r.t. \(z\) gives
Evaluating at \(x=x_{0}\) the above gives
To find \(b_{2}\) we differentiate both sides of (2) \(N-2\) times which gives
Hence
We keep doing the above to find \(b_{3},b_{4},\cdots ,b_{N}\). Therefore the general formula is
And for the special case of the last term \(b_{N}\) the above simplifies to
Where in (3) \(n\) is the coefficient \(b_{n}\) needed to be evaluated and \(N\) is the pole order and \(z_{0}\) is the expansion point. The special value \(b_{1}\) is called the residue of \(f\left ( z\right ) \) at \(z_{0}\).