Analytic Functions
We will cover following topics
Analytic Functions
A function f(z) is said to be analysic at a point z0 if there exists a ndghbothood |z−z0|<δ at all points of which f′(z) exists.
Cauchy-Riemann Equations
A necessary condition that w=f(z)=u(x,y)+ic(x,y) be analytic in a region R is that, in R, u and v satisfy the Cauchy-Riemann equations
a∂x=∂v∂y,∂u∂y=−∂v∂xIf 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 hxx+hyy=0.
If f(z)=u+iv 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)+iv(x,y) be the required function.
Case 1: u(x,y) is given. Then, f(z) is obtained by integrating f′(z)=ux(z,0)−iuy(z,0).
Case 2: v(x,y) is given. Then, f(z) is obtained by integrating f′(z)=vy(z,0)+ivx(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 ∫Cf(z)dz=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)=12πi∫f(z)z−adzLiouville’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.
PYQs
Analytic Functions
1) Use Cauchy integral formula to evaluate ∫ce3z(z+1)4dz, where c is the circle |z|=2.
[2012, 15M]
From Cauchy Integral Formula, we have
fn(a)=n!2πi∫Cf(z)dz(z−a)n+1Put a=−1, n=3
f3(−1)=3!2πi∫Cf(z)dz(z+1)4⋯(i)Take f(z)=e3z, then fn(z)=3ne3z
∴(3)(−1)=33e3=27e3∴ from (i)
27e3=3!2πi∫e3zdz(z+1)4 ⇒∫ce3zdz(z+1)4=9πie32) Prove that the function f defined by f(z)={z5|z|4,z≠00,z=0 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)=∏nj=1(z−αj) where αj=xj+iyj,xj<0
If Rez≥0, then P(z)≠0 and
P′(z)P(z)=n∑j=11z−αj=n∑j=11x−xj+i(y−yj)=n∑j=1x−xj−i(y−yj)(x−xj)2+(y−yj)2Since xj<0,1≤j≤n, it follows that
Re(P′(z)P(z))=n∑j=1x−xj(x−xj)2+(y−yj)2>0whenever Re z=x≥0. Thus x′γP(z) and therefore P′(z) has no zeros in the right half plane Re z≥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(1n)=g(1n) for n=1, 2, 3, …. Then show that f(z)=g(z) for each z in ϕ.
[2002, 12M]
Let G(z)=f(z)−g(z), then G(1n)=0 for n=1,2,…. We shall show that G(z)≡0 for z∈C which would prove the result.
Let G(z)=∑∞n=0anzn be the power series of G(z) with center 0 and radius of convergence R, clearly R>0. We shall now prove that an=0 for every n If an≠0 for some n, let ak be the first non-zero coefficient. Then G(z)=zk(ak+ak+1z+…)=zkH(z) Clearly H(z) is analytic in |z|<R, and H(0)≠0. We now claim that H(z)≠0 in a neighborhood |z|<δ of 0. Let ϵ=|H(0)|2, then continuity of H(z) at z=0 implies that there exists a δ>0 such that |z|<δ⇒|H(z)−H(0)|<ϵ or |H(0)|−ϵ<|H(z)|<|H(0)|+ϵ for |z|<δ.
Thus |H(z)|>|H(0)|2>0 for |z|<δ. Consequently, G(z)≠0 for any z in 0<|z|<δ. But this is not possible, as |z|<δ contains all but finitely many 1n, at which G(z) vanishes. Thus our assumption that an≠0 for some n is false, thus G(z)≡0 in |z|<R
Let z′ be any point in C, and let r(t),a≤t≤b be a continuous curve joining 0 and z′. Using uniform continuity of r(t), we get a partition a=t0<t1<…<tn=b of [a,b] such that r(t0)=0,r(t1)=z1,…,r(tn)=r(b)=z′, and |zj−zj−1|<R.
Now the disc K0=|z−0|<R contains z1, the center of disc K1=|z−z1|<R. since G(z1)=0 as z1∈K0∩K1, and K0∩K1 contains a sequence of points yn such that yn→z1 and G(yn)=0, we can prove as before that G(z)≡0 in K1. Proceeding in this way, in n steps we get G(z)≡0 in Kn, or G(z′)=0. since z′ is an arbitrary point of C, we get G(z)≡0 in C.
5) Prove that the Riemann zeta function ζ defined by ζ(z)=∑∞n=1n−z converges for Rez>1 and converges uniformly for Rez≥1+ε where ε>0 is arbitrary small.
[2001, 12M]
|1nz|=|1nx⋅niy|=|1nx|∵|1niy|=|1eiylogn|=1 since ∑∞n=11nx converges for x>1, it follows that ∑∞n=1n−z converges absolutely for Rez>1. If Re z≥1+ϵ, then 1nx≤1n1+ϵ and ∑∞n=1|n−z|≤∑∞n=11n1+ϵ for Re z≥1+ϵ. Weierstrass’ M-test gives that the given series converges uniformly and absolutely for Rez≥1+ϵ
Cauchy-Riemann Equations
1) Suppose f(z) is analytic function on a domain D∈C and satisfies the equation Imf(z)=(Ref(z))2, Z∈D. Show that f(z) is constant in D.
[2019, 10M]
Hint: Take f(z)=u+iu2 and apply CR equations.
In the final step, we get,
ux=0, uy=0, vx=0, vy=0
Hence, u=c1 and v=c2
⟹ f(z)=c1+ic2is constant.
2) Prove that the function: u(z,y)=(x−1)3−3xy2+3y2 is harmonic and find its harmonic conjugate and the corresponding analytic function f(z) in terms of z.
[2018, 10M]
Let, f(z)=u+iv u=(x−1)3−3xy2+3y2 ux=3(x−1)2−3y2 uxx=6(x−1)⋯(1) uy=−6xy+6y uyy=−6x+6=−6(x−1)⋯(2)
∴uxx+uyy=0 Hence, function u is harmonic.
We can use Milne’s method to find harmonic conjugate of u.
ϕ1(z,0)=(∂u∂x )x=z,y=0=3(z−1)2 ϕ2(z,0)=(∂u∂y )x=z,y=0=0 f(z)=∫[ϕ1(z,0)−iϕ2(z,0)]dz+c =∫3(z−1)2dz+c =(z−1)3+c
∴f(z)=(z−1)3+c f(z)=[(x−1)+iy]3+c=(x−1)3−iy3+3(x−1)iy[(x−1)+iy]=(x−1)3−3(x−1)y2−iy3+3i(x−1)2y=(x−1)3−3(x−1)y2+i[3(x−1)2y−y3]∴v(x,y)=3(x−1)2y−y33) Let f=u+iv be analytic function on the unit disc D={z∈C:|z|<1}. Show that ∂2u∂x2+∂2u∂y2=0=∂2v∂x2+∂2v∂y2 at all points of D.
[2017, 15M]
If f(z) is analytic in R, then the Cauchy-Riemann equations
∂u∂x=∂v∂y⋯(i)and
∂v∂x=−∂u∂y⋯(ii)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 (ii) with respect to y to obtain
∂2u∂x2=∂2v∂x∂yand
∂2v∂y∂x=−∂2u∂y2from which
∂2u∂x2=−∂2u∂y2 or ∂2u∂x2+∂2u∂y2=0Similarly, by differentiating both sides of (i) with respect to y and (ii) with respect to x, we get ∂2v∂x2+∂2v∂y2=0
4) Is v(x,y)=x3−3xy2+2y 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+iv Given: v=x3−3xy2+2y Therefore, vx=3x2−3y2,vxx=6x⋯(i)vy=−6xy+2,vyy=−6x⋯(ii) Now, vxx+vyy=+6x−6x=0 ⇒v is a harmonic function. Let u be harmonic conjugate of v. By Cauchy-Riemann equations, ux=vy⇒ux=−6xy+2∂u∂x=−6xy+2 ⇒ ∫du=∫(−6xy+2)dx ⇒u=−3x2y+2x+g(y), where g is an arbitrary function of y ⋯(iii) Also, uy=−vx⇒uy=−(3x2−3y2) ⇒ ∂u∂y=−3x2+3y2 ⇒ ∫du=∫(−3x2+3y2)dy ⇒u=−3x2y+y3+n(x), where n(x) is an arbitrary fucntion of x… (iv) From (iii) and (iv) u=−3x2y+2x+y3+c, where c is a constant So, f=(−3x2y+2x+y3+c)+i(x3−3xy2+2y)
5) Show that the function v(x,y)=ln(x2+y2)+x+y is harmonic. Find its conjugate harmonic function u(x,y). Also, find the corresponding analytic function f(z)=u+iv in terms of z.
[2015, 10M]
6) Prove that the function f(z)=u+iv, where f(z)=x3(1+i)−y3(1−i)x2+y2, z≠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 bea+1<1, where a and b are positive and real, then the function zne−a−bez has n zeros in the unit circle.
[2013, 10M]
8) Show that the function defined by f(z)={x3y5(x+iy)x6+y10,z≠00,z=0 is not analytic at the origin though it satisfies Cauchy Riemann-Equations at the origin.
[2012, 12M]
9) If f(z)=u+iv is an analytic function of z=x+iy and u−v=ey−cosx+sinxcoshy−cosx, find f(z) subject to the condition, f(π2)=(3−i)/2.
[2011, 12M]
10) Show that u(x,y)=2x−x3+3xy2 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 z7−5z3+12=0 lie between the circles |z|=1 and |z|=2.
[2006, 15M]
Let g(z)=z7−5z3,f(z)=12, then i) On |z|=1,|g(z)|≤|z7|+5|z3|=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)=z7−5z3+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 z7−5z3+12 has no zeros inside the unit circle. Now we take g(z)=12−5z3,f(z)=z7
i) On |z|=2,|g(z)|≤12+5|z3|=52<27=|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)=z7−5z3+12 and f(z) have the same number of zeros inside |z|=2. since f(z)=z7 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 z7−5z3+12 has no zeros inside and on |z|=1, therefore all zeros lie in the ring 1<|z|<2.
12) If f(z)=u+iv is an analytic function of the complex variable z and u−v=ex(cosy−siny), determine f(z) in terms of z.
[2005, 12M]
Let F(z)=(1+i)f(z)=(1+i)(u+iv)=(u−v)+i(u+v). Now ∂2(u−v)∂x2+∂2(u−v)∂y2=ex(cosy−siny)+ex(−cosy+siny)=0 Let F(z)=U+iV, where U=u−v is harmonic. If f(z) is analytic, then so is F(z) and dFdz=∂U∂x+i∂V∂x=∂U∂x−i∂U∂y as ∂V∂x=−∂U∂y by the Cauchy-Riemann equations. Thus F′(z)=ex(cosy−siny)−iex(−siny−cosy)=ex(cosy+isiny)+iex(cosy+isiny)=(1+i)ex⋅eiy=(1+i)ez Thus F(z)=(1+i)ez and f(z)=ez, which is the required function.