ECE 6382 y x −12 1 Fall 2020 David R. Jackson
Notes 8 Analytic Continuation
Notes are adapted from D. R. Wilton, Dept. of ECE
1 Analytic Continuation of Functions
We define analytic continuation as the process of continuing a function off of the real axis and into the complex plane such that the resulting function is analytic.
More generally, analytic continuation extends the representation of a function in one region of the complex plane into another region, where the original representation may not have been valid.
For example, consider the Bessel function Jn (x).
How do we define Jn (z) so that it is computable in some region and agrees with Jn (x) when z is real?
2 Analytic Continuation of Functions (cont.)
One approach to extend the domain of a function is to use Taylor series.
We start with a Taylor series that is valid in some region. We extend this to a Taylor series that is valid in another region.
Note: This may not be the easiest way in practice, but it always works in theory.
3 Analytic Continuation of Functions (cont.)
Original geometric series two alternative representations Example y 1 ∞ f ( z) = = ∑ z,n z <1 1− z n=0 1 3 • Expand about z = − . Since both series are Rc = 2 2 valid there, find coefficients of a new series : ∞ x m −12 = + 1 +<1 3 1 fz( ) ∑ bm ( z 2 ) , z 2 2 m=0 by using use the above series representation 1 dfm 11d m ∞∞ = = n = − − −+ nm− bm m m ∑∑z nn( 12)( n)( n m 1) z m! 1 mm!! dz z=− dz n=0 zz=−=11 nm= − 2 22 ∞ m 1!ndnm− = −1 n = >= − ∑ ( 2 ) ()Note m z0 , m n : n 01 , , ,m 1 m!!nm= ( nm− ) dz
The coefficients of the new series--- with extended region of convergence --- are determined from the coefficients of the original series, even though that series did not converge in the extended region. The information to extend the convergence region is contained in the coefficients of the original series --- even if it was divergent in the new region! 4 Analytic Continuation of Functions (cont.)
Example (cont.)
Another way to get the Taylor series expansion:
∞ m m 11d = +11 +<3 = fz( ) ∑ bmm( z2) , z 22 b m m=0 mz!1dz − z/=−12
so that m+1 1 12 = = bmm (123)( )( )( ) m+1 m!31− z ( ) z/=−12 Note: ∞ m+1 2 m This is sort of “cheating” in the ⇒ = +11 +<3 sense that we assume we fz( ) ∑ ( z2) , z 22 m=0 3 already know a closed form expression for the function.
5 Analytic Continuation of Functions (cont.) Example (cont.)
∞ fz( ) ≡ ∑ zn n=0
Here we show the continuation of fz( ) from its power series y representation in the region z < 1 into the entire complex plane using Taylor series. x −12 1
If the singularities are isolated, we can continue any function into the entire complex plane via a sequence of continuations using Taylor and / or Laurent series ! 6 The Zeros of an Analytic Function are Isolated
(The zeros cannot be arbitrarily close together.)
Proof of theorem:
Assume that fz() is analytic in a connected region A , and suppose fz(0 )= 0 . 2 Then fz() has a Taylor series fz()0= az 1 ( −+ z 02 ) a ( z − z 0 ) + with a0 = fz()0 0 =at z 0 . More generally, we may have a zero with multiplicity N such that fz() has a Taylor series as : NN+1 fz()()= azzNN −+0 a+ 10 () zz − + 1 =−()zzNN a + a () zz −+1 () a = f()()0 z ≠ 0NN+ 10 NN ! 0 N =(z−≠ z00 ) gz (), gz ( ) 0,and gz () is analytic inA (it is represented by a converging Taylor series).
Since gz() is analytic it is also continuous. Since gz(0 )≠ 0 , gz ()cannot vanish within a sufficiently small neighborhood of z0. That is, the zeros of fz() must be isolated.
The only exception is if the function f (z) is identically zero.
7 The Zeros of an Analytic Function are Isolated
◊ Example: 1 The function fz( )= sin( 1 z) has zeros at z=, n = 1, 2, nπ
This function cannot be analytic at z = 0 since the zeros accumulate there and hence are not isolated there.
y
x 1 / π
The origin is an “accumulation point” for the zeros.
8 Analytic Continuation Principle
Theorem of analytic continuation:
Assume that f(z) and g(z) are analytic in a connected region A, and f (zn ) = g(zn) on a set of points zn in A that converge to a point z0 in A.
In other words, there can be only one function Then g(z) = f (z) in A. that is analytic in A and has a defined set of values at the converging points zn.
Proof: z . Construct the difference function g(z)-f(z), 0 A which in analytic in A. This function must have a Taylor series at z0. . This function has all zero coefficients and zn g(z) thus be zero; otherwise, the analytic difference function must have isolated zeros because it is analytic – which it does not, by assumption. . By continuing the Taylor series that has Note: Defining zn on a contour in A also suffices. zero coefficients (analytic continuation), the difference function must be zero throughout A. 9 Analytic Continuation Principle (cont.)
Corollary (extending a domain from A to A ∪ B) Assume that f (z) is analytic in A and g(z) is analytic in B, and the two domains overlap in a region A ∩ B, and f(z) = g(z) in A ∩ B.
f( z,z) ∈ A Define hz( ) = g( z,z) ∈ B
Then h(z) is the only analytic function in A ∪ B that equals f (z) on A.
AB∪ f( z) = gz( ) The function h(z) uniquely extends the domain of f (z) from A to A ∪ B. A AB∩ fz( ) B gz( )
10 Analytic Continuation Principle (cont.)
Then h(z) is the only analytic function in A ∪ B that equals f (z) on A.
Proof:
. The function h is analytic in the region A ∪ B and also equals f(zn) on any set of converging points in the intersection region. . The theorem of analytic continuation thus ensures that h is unique in A ∪ B.
AB∪ Converging points
A fz( ) AB∩ B gz( )
11 Analytic Continuation Principle (cont.)
Example (example of theorem) The function sin(x) is continued off the real axis.
eeiz− − iz y gz( ) =sin ( z) ≡ 2i
x Line segment
eeix− − ix fx( ) =sin ( x) = 2i
The function g(z) is the only one that is analytic in the blue region of the complex plane and agrees with sin(x) on any segment of the real axis. 12 Analytic Continuation Principle (cont.)
Example (example of theorem)
The Bessel function Jn(x) is continued off the real axis.
k nk+2 ∞ (−1) z = = y gz( ) Jn ( z) ∑ k=0 kn!( + k) !2
x Line segment
k nk+2 ∞ (−1) x fx( ) = Jn ( x) = ∑ k=0 kn!( + k) !2
The function g(z) is the only one that is analytic in the blue region of the
complex plane and agrees with Jn(x) on any segment of the real axis. 13 Analytic Continuation Principle (cont.)
Example (of corollary) The function Ln(z) (principal branch) is continued beyond a branch cut. f( z) ≡Ln( z) = ln ( r) + i,θ −<< πθπ
y Note: “Ln” denotes the principal branch.
Region of overlap (second quadrant)
Branch cut for f x
gz( ) ≡=+ln( z) ln ( r) iθ −π//2 << θπ 32 Branch cut for g In the blue region (second and third quadrants), g(z) is analytic. Also, g(z) agrees with the function f (z) in the second quadrant. The original function f (z) is not analytic in the entire blue region. 14 Analytic Continuation Principle (cont.)
Example (cont.)
The function g(z) is the only function that is analytic in the entire left-half plane and agrees with Ln(z) in the second quadrant. y
x
gz( ) ≡ln( z) = ln ( r) + i,θ − π /2 << θπ 3 / 2
15 Analytic Continuation Principle (cont.)
Example (of corollary)
The function Yn (z) (Bessel function of the second kind) is continued beyond a branch cut. y
Branch cut x
Ln( z) = ln ( ri) + θ −<<πθπ
2kn− 21zzn−1 (nk−−1!) = +−γ Yznn( ) J( z) Ln ∑ ππ2 k=0 k!2 ∞ 2kn+ γ 0.577216 11k z −∑( −1) Φ(k) +Φ( nk + ) (Euler’s constant) π k=0 knk!( + ) !2 11 1 Φ( pp) =+++1 +( > 0) 23 p 16 Analytic Continuation Principle (cont.) Example (cont.)
The “extended Bessel function” is analytic within the blue region. y Note: In the third quadrant, this “extended” function will not be the same as the usual Bessel function there.
x In the third quadrant : ext Ynn() z= Yz () 2 + Jz( )(2π i) ln(z) = ln ( ri) + θ π n −π//2 << θπ 32
n−1 2kn− ext 21zz(nk−−1!) Ynn( z )≡ Jz( ) ln +−γ ∑ ππ2 k =0 k!2 ∞ 2kn+ 11k z −∑( −1) Φ(k) +Φ( nk + ) π k =0 knk!( + ) !2 17 Analytic Continuation Principle (cont.) Example
How do we extend F(x) to arbitrary z?
∞ ≡≥−xt F( x) ∫ e J0 ( t) dt, x 0 0 y Note: The integral does not converge for x < 0.
Original domain: x ≥ 0 x
Identity : ∞ −xt 1 e J0 ( t) dt= , x ≥ 0 ∫ 2 0 x +1
18 Analytic Continuation Principle (cont.) Example (cont.) 1 ≡ Here we define: Fz( ) 12/ (z2 +1)
This is defined everywhere in the complex plane (except on the branch cuts). y
12/ ( xx22+=+11) i x −i Note: The shape of the branch cuts is arbitrary here.
19 Analytic Continuation Principle (cont.) Example (cont.) 1 Fz≡ Here we define: ( ) 2 +≥ z2 +1 Note: Rez 1 0
This corresponds to using vertical branch cuts.
With these branch cuts, y 12/ Rez2 +≥ 1 0 ( ) 12/ ( xx22+=+11) The derivation (omitted) is similar to that of the Sommerfeld branch cuts. i x 12/ ⇒+( zz2211) =+ −i
(with these branch cuts)
20 Analytic Continuation Principle (cont.) Example
How do we extend F(x) to arbitrary z?
x 1 F( x) ≡ dt Im(t) ∫ (real valued for real x > 0) 1 t
Original domain: x > 0 × Re(t) 1 x
Fx( ) =ln( x) −= ln( 1) Ln( x) − Ln( 1)
Recall: ln( z) = Ln( zi) + ( 2π n) Note : Ln( zz) means −<ππarg ( ) < Fx( ) = Ln ( x) 21 Analytic Continuation Principle (cont.)
Example (cont.)
Analytic continuation:
Fz( ) ≡ Ln ( z)
(This agrees with F(x) on the real axis.)
y
x
Branch cut
22 Analytic Continuation Principle (cont.) Example (cont.)
From another point of view: z 1 F( z) ≡ ∫ dt 1 t Im(t)
z z C z × Re(t) z 1
z
We analytically continue F(z) from the real axis.
We require that the path is varied continuously as z leaves the real axis.
23 Analytic Continuation Principle (cont.) Example (cont.)
z 1 Im(t) F( z) ≡ ∫ dt 1 t z C + z × Re(t) + Fz( ) z 1
z − − C Fz( ) z
As we encircle the pole at the origin, we get a different result for the function F(z).
111 dz= dz −=−= dz F+−( z) F( z) 2π i ∫∫∫zzz C CC+−
This is why we need a branch cut in the z plane, with a branch point at z = 0. 24 Analytic Continuation Principle (cont.) Example
How do we interpret cosh-1(z) for arbitrary z?
−12 − cosh( x) = Ln( x +− x 1) ,x > 1 Note : cosh1 ( 1) = 0
Note : Ln( xx+−22 1) is real (arg( xx+−= 1) 0)
y
Original domain: x > 1 x 1 x
Recall: Ln( zz) means −<ππarg ( ) <
25 Analytic Continuation Principle (cont.)
Example (cont.)
Analytic continuation:
cosh−12( z) ≡ Ln( zz +− 1)
y
z x 1 x
cosh−12( x) = Ln( xx +− 1)
26 Analytic Continuation Principle (cont.)
Example (cont.)
y −12 Where would branch cuts be? cosh( z) ≡ Ln( zz +− 1)
12/ ( zz22−=−11) Sommerfeld branch cuts 1 x −1
Note: The function cosh-1(z) is discontinuous as we cross the branch cuts. F(z) is analytic in any one quadrant.
fz( ) ≡ z2 −⇒1 Refz( ) > 0for all z ⇒ Sommerfeld branch cuts
(Recall: Rew > 0) 27 Analytic Continuation Principle (cont.) Example (cont.) y
−12 Do we need another branch cut cosh( z) ≡ Ln( zz +− 1) due to the Ln function? 1 x −1
Examine argument of Ln function:
2 11 wz=+ z −⇒=1 z w + (derivation omitted) 2 w
The branch cut for the Ln(w) function corresponds to w being a negative real number.
Note : w∈( −∞ ,01) ⇒ z, ∈( − −∞)
28 Analytic Continuation Principle (cont.)
Example (cont.) y
−12 Final Picture cosh( z) ≡ Ln( zz +− 1)
1 x −1
F(z) is an analytic continuation of cosh-1(x) off of the real axis.
F(z) is analytic in any one quadrant.
29 EM Example
Assume a radiating phased line source on the z axis.
z Note: j is used here instead of i.
− jkz z Iz()= Ie0
kz is real y
x 221/2 kρ =( kk0 − z )
22 kk00−
30 EM Example (cont.)
z ( xyz,,) We can also write
− ∞ jkz z − jk0 R R Iz()= Ie − ′ e 0 = jkz z µ ′ Az ∫ I00 e dz −∞ 4π R (0,0, z′) y Note: The integral converges for real kz.
x Hence
∞ − jk0 R −−′ e µ I = jkzz z µρ′ = 00 (2) jk z Az ∫ I00 e dz H0() kρ e −∞ 44π Rj
Exists only for real kz Exists for complex kz
The second form is the analytic continuation of the first form off of the real axis. 31 EM Example (cont.)
µ z 00I (2) − jkz z Az = Hke0 ()ρ ρ 4 j − jkz z Iz()= Ie0
221/2 kz is complex kρ =( kk0 − z ) y 221/2 =−−jk( z k0 )
x
In order for this to be the analytic continuation off of the real axis of the integral form, we must chose the branch of the square root function correctly so that it changes smoothly and it is correct when kz = real > k0.
221/2 kρ =−−= jk( zz k00) negative imaginary number for k=> real k
32 EM Example (cont.)
221/2 kρ =−− jk( z k0 )
Im(kz ) 22 kρ =−− jkz k0
Imkρ < 0 Imkρ < 0
k0 Re(kz )
−k0
Imkρ < 0 Imkρ < 0
The Sommerfeld branch cuts are a convenient choice. (but not necessary) 33 EM Example (cont.)
221/2 kρ =−− jk( z k0 )
22 kρ =−− jkz k0
Im(kz ) (Top sheet)
Riemann surface
k0 Re(kz )
−k0 Note: The function is continuous on the Riemann surface.
We can now let kz wander anywhere we wish on the Riemann surface, and we know how to calculate the square root. (We analytically continue to the entire Riemann surface.) 34 EM Example (cont.)
Example 221/2 kρ =−− jk( z k0 )
What is kρ at the final 22 indicated point? kρ =−− jkz k0
Im(kz ) (Top sheet)
Riemann surface
Imkρ < 0
k0
Re(kz )
−k0 (Bottom sheet)
Final point Imkρ > 0
At the indicated final point, the imaginary part of kρ is chosen to be positive.
35 EM Example (cont.)
Example 221/2 kρ =−− jk( z k0 ) What type of wave is at the indicated points? "Surface wave" : kj=−+( ) Im(kz ) ρ
"Phased array" : kρ =( +)
(θ12= πθ,0 = ) Riemann surface
k0
Re(kz )
−k0
Leaky wave : kjρ =++( ) ( +) 1/2 1/2 Note : kρ =− jk( − k) k −−( k) zz00( ) (θπ1= + ∆ 12,; θ = −∆ 2) ∆ 1 > ∆ 2
Note: The leaky wave field is the analytic continuation of the phased array field when kz becomes complex. 36 Schwarz Reflection Principle
Assume that f (z) is the analytic continuation of a real function f (x) off the real axis (or a segment of the real axis).
* * Then within the analytic region, we have fz( ) = fz( ) y (proof omitted)
fz( ) x Line segment: fx( )
Examples: f (z) is assumed analytic in this region. z sin( zeJz) , , n ( )
ln( zz) , Re( ) > 0 (assuming branch cut on negative real axis)
37