Analytic Functions II

Lecture 7

Analytic Functions II 43

Example 7.1.

Consider the exponential functionf(z) =e^z=e^x(cosy+

isiny).Then, u(x, y) =e^xcosy, v(x, y) =e^xsiny,and

∂u

∂x = ∂v

∂y = e^xcosy, ∂u

∂y = −∂v

∂x = −e^xsiny

everywhere, and these derivatives are everywhere continuous. Hence,f(z) exists and

f(z) = ∂u

∂x+i∂v

∂x = e^xcosy+ie^xsiny = e^z =f(z).

Example 7.2.

The functionf(z) =z³ =x³−3xy²+i(3x²y−y³) is an entire function andf(z) = 3z².

Example 7.3.

Consider the functionf(z) =x²+y+i(2y−x).We have u(x, y) =x²+y, v(x, y) = 2y−x,andux= 2x, uy= 1, vx=−1, vy = 2.

Thus, the Cauchy-Riemann equations are satisfied whenx = 1. Since all partial derivatives off are continuous, we conclude that f(z) exists only on the linex= 1 and

f(1 +iy) = ∂u

∂x(1, y) +i∂v

∂x(1, y) = 2−i.

Example 7.4.

Let f(z) = ze^−|z|2. Determine the points at which f(z) exists, and findf(z) at these points. Sincef(z) = (x−iy)e^−(x2+y2), u(x, y) =xe^−(x2+y2), v(x, y) =−ye^−(x2+y2),and

∂u

∂x = e^−(x2+y2)−2x²e^−(x2+y2), ∂u

∂y = −2xye^−(x2+y2),

∂v

∂x = 2xye^−(x2+y2), ∂v

∂y = −e^−(x2+y2)+ 2y²e^−(x2+y2). Thus, ∂u

∂y =−∂v

∂x is always satisfied, and ∂u

∂x =∂v

∂y holds, if and only if 2e^−(x2+y2)−2x²e^−(x2+y2)−2y²e^−(x2+y2) = 0,

or

2e^−(x2+y2)(1−x²−y²) = 0,

or x²+y² = 1. Since all the partial derivatives of f are continuous, we conclude thatf(z) exists on the unit circle|z|= 1.Furthermore, on|z|= 1,

f(x+iy) = ∂u

∂x(x, y) +i∂v

∂x(x, y)

= e^−(x2+y2)−2x²e^−(x2+y2)+ 2ixye^−(x2+y2)

= e⁻¹(1−2x²+ 2xyi).

Now recall the following result from calculus.

Theorem 7.1.

Supposeφ(x, y) is a real-valued function defined in a domainS. Ifφx=φy= 0 at all points inS,thenφis a constant inS.

Analogously, for an analytic function, we have the following theorem.

Theorem 7.2.

If f(z) is analytic in a domain S and if f(z) = 0 everywhere inS, thenf(z) is a constant inS.

Proof.

Sincef(z) = 0 in S, all first-order partial derivatives ofuand v vanish in S; i.e.,ux =uy =vx=vy = 0.Now, since S is connected, we haveu= a constant andv= a constant in S.Consequently,f =u+iv is also a constant inS.

Remark 7.1.

The connectedness property of S is essential. In fact, if f(z) is defined byf(z) =

1 if |z|<1

2 if |z|>2, thenf is analytic andf(z) = 0 on its domain of definition, butf is not a constant.

Theorem 7.3.

If f is analytic in a domain S and if |f| is constant there, then f is constant.

Proof.

As usual, let f(z) =u(x, y) +iv(x, y). If |f| = 0,then f = 0.

Otherwise, |f|² = u²+v² ≡ c = 0. Taking the partial derivatives with respect to xandy,we have

uux+vvx = 0 and uuy+vvy = 0.

Using the Cauchy-Riemann equations, we find

uu_x−vu_y = 0 and vu_x+uu_y = 0 so that

(u²+v²)ux = 0 and (u²+v²)uy = 0,

and hence ux = uy = 0. Similarly, we have vx = vy = 0. Thus, f is a constant.

Next, letf(z) =u(x, y)+iv(x, y) be an analytic function in a domainS, so that the Cauchy-Riemann equationsux=vy anduy=−vxare satisfied.

Differentiating both sides of these equations with respect to x (assuming that the functions uand v are twice continuously differentiable, although we shall see in Lecture 18 that this assumption is superfluous), we get

∂²u

∂x² = ∂²v

∂x∂y and ∂²u

∂x∂y = −∂²v

∂x². Similarly, differentiation with respect toy yields

∂²u

∂y∂x = ∂²v

∂y² and ∂²u

∂y² = − ∂²v

∂y∂x.

Analytic Functions II 45 Hence, it follows that

∂²u

∂x²+∂²u

∂y² = 0 (7.1)

and

∂²v

∂x² +∂²v

∂y² = 0. (7.2)

The partial differential equation (7.1) ((7.2)) is called theLaplace equation.

It occurs in the study of problems dealing with electric and magnetic fields, stationary states, hydrodynamics, diffusion, and so on.

A real-valued functionφ(x, y) is said to be harmonic in a domain S if all its second-order partial derivatives are continuous in S and it satisfies φxx+φyy= 0 at each point ofS.

Theorem 7.4.

Iff(z) =u(x, y) +iv(x, y) is analytic in a domain S, then each of the functionsu(x, y) andv(x, y) is harmonic inS.

Example 7.5.

Does there exist an analytic function on the complex plane whose real part is given byu(x, y) = 3x²+xy+y²? Clearly, uxx= 6, uyy= 2,and henceuxx+uyy= 0; i.e.,uis not harmonic. Thus, no such analytic function exists.

Letu(x, y) andv(x, y) be two functions harmonic in a domain S that satisfy the Cauchy-Riemann equations at every point of S. Then, u(x, y) andv(x, y) are called harmonic conjugatesof each other. Knowing one of them, we can reconstruct the other to within an arbitrary constant.

Example 7.6.

Construct an analytic function whose real part is u(x, y) =x³−3xy²+ 7y.Sinceuxx+uyy= 6x−6x= 0, uis harmonic in the whole plane. We have to find a functionv(x, y) so thatuandvsatisfy the Cauchy-Riemann equations; i.e.,

vy = ux = 3x²−3y² (7.3)

and

v_x = −u_y = 6xy−7. (7.4)

Integrating (7.3) with respect toy,we get

v(x, y) = 3x²y−y³+ψ(x).

Substituting this expression into (7.4), we obtain v_x = 6xy+ψ(x) = 6xy−7,

and henceψ(x) =−7,which implies thatψ(x) =−7x+a,whereais some constant. It follows that v(x, y) = 3x²y−y³−7x+a.Thus, the required analytic function is

f(z) = x³−3xy²+ 7y+i(3x²y−y³−7x+a).

Example 7.7.

Find an analytic function f whose imaginary part is given by e^−ysinx. Let v(x, y) = e^−ysinx. Then it is easy to check that v_xx+v_yy= 0.We have to find a function u(x, y) such that

ux = vy = −e^−ysinx, (7.5) uy = −vx = −e^−ycosx. (7.6) From (7.5), we getu(x, y) =e^−ycosx+φ(y).Substituting this expression in (7.6), we obtain

−e^−ysinx+φ(y) = −e^−ysinx.

Hence, φ(y) = 0; i.e., φ(y) = c for some constant c. Thus, u(x, y) = e^−ycosx+c and

f(z) = e^−ycosx+c+ie^−ysinx = e^−y+ix+c.

Problems

7.1. Find f(z) when (a).f(z) = z−1

2z+ 1 (z=−1/2), (b).f(z) =e^z3, (c). f(z) =(1 +z²)⁴

z⁴ (z= 0), (d).f(z) =z³+z.

7.2. Use the definition to findf(z) when (a).f(z) = 1

z (z= 0), (b).f(z) =z²−z.

7.3. Show that

(a).f(z) =x−iy² is differentiable only aty=−1/2 and f(z) = 1, (b).f(z) =x²+iy² is differentiable only whenx=yand f(z) = 2x, (c). f(z) =yx+iy²is differentiable only atx=y= 0 and f(z) = 0, (d).f(z) =x³+i(1−y)³is differentiable only atx= 0, y= 1 andf(z) = 0.

7.4. For each of the following functions, determine the set of points at which it is (i) differentiable and (ii) analytic. Find the derivative where it exists.

(a).f(z) = (x³+ 3xy²−3x) +i(y³+ 3x²y−3y), (b).f(z) = 6z²−2z−4i|z|²,

(c). f(z) = (3x²+ 2x−3y²−1) +i(6xy+ 2y),

Analytic Functions II 47

(d).f(z) = 2z²+ 6 z(z²+ 4),

(e).f(z) =e^y2−x2(cos(2xy)−isin(2xy)).

7.5. Finda, b, cso that the functionw= (ay³+ix³) +xy(bx+icy) is analytic. Ifz=x+iy,expressdw/dz in the form φ(z).

7.6. Show that there are no analytic functions of the formf =u+iv withu=x²+y².

7.7. Letf(z) be an analytic function in a domain S.Show that ∂²

∂x² + ∂²

∂y²

|f(z)|² = 4|f(z)|².

7.8. Letf :C→Cbe defined by f(z) =

⎧⎨

⎩ z²

z if z= 0 0 if z= 0.

(a). Verify that the Cauchy-Riemann equations forf are satisfied atz= 0.

(b). Show thatf(0) does not exist.

This problem shows that satisfaction of the Cauchy-Riemann equations at a point alone is not enough to ensure that the function is differentiable there.

7.9. Let D andS be domains, and letf : D→C and g: S →C be analytic functions such thatf(D)⊆S.Show thatg◦f :D→Cis analytic.

7.10.In polar coordinatesx=rcosθ, y=rsinθ,the function f(z) = u(r, θ) +iv(r, θ).Show that the Cauchy-Riemann conditions can be written

as ∂u

∂r = 1 r

∂v

∂θ, 1 r

∂u

∂θ = −∂v

∂r, (7.7)

and

f(z) = e^−iθ(ur+ivr). (7.8) In particular, show that f(z) =√

re^iθ/2is differentiable at allz except z= 0 andf(z) = (1/2√

r)e^−iθ/2.

7.11.Show that the Cauchy-Riemann conditions are equivalent tow_z= 0.Hence, deduce that the functionw=f(z) =ze^−|z|2 is not analytic.

7.12. Let w = f(z) be analytic in a neighborhood of z₀, and w₀ = f(z₀), f(z₀) = 0. Show that f defines a one-to-one mapping of a neigh- borhood ofz₀ onto a neighborhood ofw₀.

7.13.Use L’Hˆopital’s Rule to find the following limits:

(a). lim

z→i

z⁷+i

z¹⁴+ 1, (b). lim

z→3i

z⁴−81

z²+ 9 , (c). lim

1+i√3

z⁶−64 z³+ 8 .

7.14.Show that iff =u+ivis analytic in a regionSanduis a constant function (i.e., independent ofxandy), thenf is a constant.

7.15.Show that ifh: IR²→IR andf = 2h³+ihis an entire function, thenhis a constant.

7.16.Suppose f is a real-valued function defined in a domain S ⊆C.

Iff is complex differentiable atz0∈S, show thatf(z0) = 0.

7.17. Suppose that f = u+iv is analytic in a rectangle with sides parallel to the coordinate axes and satisfies the relationu_x+v_y = 0 for all xandy.Show that there exist a real constantcand a complex constantd such thatf(z) =−icz+d.

7.18.Show that the functions

(a).u(x, y) =e^−xsiny, (b).v(x, y) = cosxcoshy,

are harmonic, and find the corresponding analytic function u+iv in each case.

7.19.Suppose thatuis harmonic in a domainS.Show that:

(a). Ifv is a harmonic conjugate ofu,then−uis a harmonic conjugate of v.

(b). If v₁ andv₂ are harmonic conjugates ofu,thenv₁and v₂ differ by a real constant.

(c). Ifv is a harmonic conjugate ofu,thenv is also a harmonic conjugate ofu+c,wherecis any real constant.

7.20. Show that a necessary and sufficient condition for a function f(z) =u(x, y) +iv(x, y) to be analytic in a domain S is that its real part u(x, y) and imaginary partv(x, y) be conjugate harmonic functions inS.

7.21.Letf(z) =u(r, θ) +iv(r, θ) be analytic in a domain S that does not include the point z = 0. Use (7.7) to show that both uand v satisfy the Laplace equation in polar coordinates

r²∂²φ

∂r² +r∂φ

∂r +∂²φ

∂θ² = 0.

Analytic Functions II 49 7.22.Usef(t) = t

t²+ 1 = Re 1

t−i

to show that

f⁽ⁿ⁾(t) = (−1)ⁿn!(n+ 1)!

(t²+ 1)ⁿ⁺¹

[(n+1)/2]

k=0

(−1)^kt^n+1−2k (2k)!(n+ 1−2k)!.

7.23(a). Iff(x+iy) =Re^iφ is analytic, show that ∂R

∂x =R∂φ

∂y. (b). Consider the two annulus domains Da = {z : a < |z| < 1} and D_b ={z:b <|z|<1}.Definef :D_a →D_b by

f re^iθ

=

1−b 1−a

r+ b−a 1−a

e^iθ.

Show that (i)f is bijective and (ii)f is analytic if and only ifa=b.

7.24.Letf(z) =u(x, y) +iv(x, y) be an analytic function. Show that level curves of the family u(x, y) =c are orthogonal to the level curves of the familyv(x, y) =d; i.e., the intersection of a member of one family with that of another takes place at a 90^oangle, except possibly at a point where f(z) = 0.Verify this result for the functionf(z) =z².

Answers or Hints

7.1. (a). 3/(2z+ 1)²,(b). 3z²e^z3,(c). 4(1 +z²)³(z²−1)/z⁵,(d). 3z²+ 1.

7.2. (a). f(z) = limΔz→0

z+Δz1 −¹_z

Δz = limΔz→0 z−(z+Δz)

(Δz)(z+Δz)z = −_z¹2, (b). f(z) = limΔz→0(z+Δz)²−(z+Δz)−(z²−z)

Δz = limΔz→0 2zΔz+(Δz)²−Δz

Δz =

2z−1.

7.3. (a). Since u = x, v = −y², ux = 1, uy = 0, vx = 0, vy = −2y, the function is differentiable only when 1 = −2y or y =−1/2,and f = ux+ivx= 1.(b). Sinceu=x², v=y², ux= 2x, uy= 0, vx= 0, vy = 2y, the function is differentiable only when 2x= 2y andf(z) = 2x.(c). Since u = yx, v = y², ux = y, uy = x, vx = 0, vy = 2y, the function is differentiable only wheny = 2y, x = 0 or x= 0, y = 0, and f(z) = 0.

(d). Sinceu=x³, v= (1−y)³, ux= 3x², uy= 0, vx= 0, vy =−3(1−y)² the function is differentiable only when 3x²=−3(1−y)² orx= 0, y= 1, andf(z) = 0.

7.4. (a). f is differentiable at every point on the x and y axes, f(x+ i0) = 3x²−3, f(0 +iy) = 3y²−3, not analytic anywhere. (b). f is differentiable only at the pointz = ₁₆³ −₁₆¹i, f(₁₆³ − ₁₆¹i) = ¹₄ −³₄i, not analytic anywhere. (c). f is differentiable everywhere,f(z) = (6x+ 2) + i(6y),analytic everywhere (entire). (d). fis differentiable for allz= 0,±2i,

f(z) =−2(z⁴+ 5z²+ 12)/[z²(z²+ 4)²],analytic inC− {0,±2i}.(e). f is differentiable everywhere,f(z) =−2ze^−z2,analytic everywhere (entire).

7.5. a= 1, b=−3, c=−3, y= 0 givesf(x) =ix³,and hencef(z) =iz³ anddw/dz= 3iz².

7.6. ux = 2x, uy = 2y ⇒vy = 2x andvx =−2y⇒v = 2xy+f(y) and v=−2xy+g(x), which is impossible.

7.7. LHS = 2(u²_x+u²_y) + 2(v_x²+v_y²).

7.8. (a). u(x, y) =

(x³−3xy²)/(x²+y²), (x, y)= (0,0)

0, (x, y) = (0,0) andv(x, y) =

(y³−3x²y)/(x²+y²), (x, y)= (0,0)

0, (x, y) = (0,0). u_x(0,0) = 1, u_y(0,0) = 0, v_x(0,0) = 0, v_y(0,0) = 1.(b). For z= 0, (f(z)−f(0))/(z−0) = (z/z)². Now see Problem 5.6 (a).

7.9. Use rules for differentiation.

7.10.Sinceur=uxxr+uyyr, uθ=uxxθ+uyyθ,we have

ur=uxcosθ+uysinθ, uθ=−uxrsinθ+uyrcosθ (7.9) and

v_r=v_xcosθ+v_ysinθ, v_θ=−v_xrsinθ+v_yrcosθ, which in view of (6.5) is the same as

vr=−uycosθ+uxsinθ, vθ=uyrsinθ+uxrcosθ. (7.10) From (7.9) and (7.10), the Cauchy-Riemann conditions (7.7) are immediate.

Now, sincef(z) =ux+ivxandux=urcosθ−uθsinθ

r =urcosθ+vrsinθ, vx=vrcosθ−vθsinθ

r =vrcosθ−ursinθ,it follows thatf(z) =ur(cosθ− isinθ) +ivr(cosθ−isinθ) =e^−iθ(ur+ivr).

Sinceu=√

rcos^θ₂, v=√

rsin^θ₂, u_r=₂√¹rcos^θ₂, u_θ=−₂¹√

rsin^θ₂, v_r

= ₂√¹rsin^θ₂, vθ=¹₂√

rcos^θ₂,and hence conditions (7.7) are satisfied. Thus, f is differentiable at all z exceptz= 0.Furthermore, from (7.8) it follows that f(z) =e^−iθ₂√¹r(cos^θ₂ +isin^θ₂) =₂√¹z.

7.11.Sincex= (z+z)/2 andy = (z−z)/2i, uandvmay be regarded as functions ofzandz.Thus, conditionw_z= 0 is the same as ^∂u_∂x^∂x_∂z+^∂u_∂y^∂y_∂z+ i

∂v

∂x

∂x

∂z +^∂v_∂y^∂y_∂z

= 0,which is the same as ¹₂^∂u_∂x−_2i^1∂u_∂y+₂ⁱ_∂x^∂v−¹₂^∂v_∂y = 0.

Now compare the real and imaginary parts.

7.12. Let f = u+iv. From calculus, it suffices to show that Jacobian J(x0, y0) =

ux vx

uy vy

(x0, y0) = 0. Now use (6.5) and (6.3) to obtain J(x0, y0) =u²_x(x0, y0) +v²_x(x0, y0) =|f(z0)|²= 0.

7.13.(a). i/2,(b). −18,(c). −16.

7.14. f = u+iv, u =c ⇒ux =uy = 0 ⇒ vx =vy = 0 ⇒ v is also a constant, and hencef is a constant.

7.15.Iff =u+iv,thenu= 2h³andv=h.Now, by the Cauchy-Riemann conditions, we have 6h²h_x =h_y and −6h²h_y =h_x.Thus, −12h⁴h_x=h_x, or h_x(12h⁴+ 1) = 0,which implies thath_x= 0,and from this h_y = 0.

Analytic Functions II 51 7.16. In (6.1), if we allow Δz → 0 along the x-axis, then f(z₀) is real.

However, if we allow Δz→0 along they-axis, thenf(z₀) is purely imagi- nary.

7.17. u_x+v_y = 0 and u_x = v_y imply that u_x = v_y = 0. Thus, u = φ(y), v =ψ(x).Now, u_y =−v_x implies that φ(y) =−ψ(x) =c.Hence, u = φ(y) = cy+d₁, v = ψ(x) = −cx+d₂, where d₁ and d₂ are real constants. Thus,f =u+iv=−icz+d,whered=d₁+id₂.

7.18.(a). ie^−z+ia,(b). sinxsinhy+icosxcoshy+a.

7.19.(a). Sincef =u+iv is analytic, (−i)f =v−iuis analytic. Thus,

−uis a harmonic conjugate ofv.

7.20.The necessary part is Theorem 7.4. To show the sufficiency part, we note that ifu(x, y) and v(x, y) are conjugate harmonic functions, then, in particular, they have continuous first derivatives in S, and hence are dif- ferentiable inS. Sinceu(x, y) andv(x, y) also satisfy the Cauchy-Riemann equations in S,it follows thatf(z) is analytic inS.

7.21.Verify directly.

7.22.Usef⁽ⁿ⁾(t) = Re_dt^dn_n 1

t−i

and the binomial theorem.

7.23.(a). f =Re^iφ=Rcosφ+iRsinφ.By the Cauchy-Riemann equations

∂R

∂xcosφ−Rsinφ^∂φ_∂x = ^∂R_∂y sinφ+Rcosφ^∂φ_∂y,

∂R

∂y cosφ−Rsinφ^∂φ_∂y =−^∂R_∂xsinφ−Rcosφ^∂φ_∂x; i.e.,

∂R

∂x cosφ−^∂R_∂y sinφ=Rsinφ^∂φ_∂x+Rcosφ^∂φ_∂y, (7.11)

∂R

∂y cosφ+^∂R_∂x sinφ=Rsinφ^∂φ_∂y −Rcosφ^∂φ_∂x. (7.12) Now (7.11) cosφ+ (7.12) sinφgives the result. (b). Let z1 =r1e^iθ1, z2 = r2e^iθ2.Thenf(z1) =f(z2) implies that

1−b 1−a

r1+₁^b−₋^a_a e^iθ1 = 1−b

1−a

r2+₁^b−₋^a_a e^iθ2;

i.e., ₁¹₋⁻_a^br1+₁^b−₋^a_a = ₁¹₋^−b_ar2+ ^b₁⁻₋^a_a and e^iθ1 =e^iθ2, and hence r1 = r2 and θ₁ = θ₂. Therefore, f is one-to-one. Now, for any z ∈ Db, letz = ρe^iθ. Thenb < ρ <1. Considerρ= ₁¹₋⁻_a^br+₁^b−₋^a_a so that r= ¹₁⁻₋^a_bρ+^a₁⁻₋^b_b. Since

dr

dρ = ¹₁⁻₋^a_b > 0, r is an increasing function of ρ. Whenρ = b, r = a and whenρ= 1, r= 1,so a < r <1; hencere^iθ ∈Da andf

re^iθ

=ρe^iθ=z;

i.e.,f is onto. Now supposef is analytic then, from part (a), we have

∂

∂x

1−b

1−ar+^b₁⁻₋^a_a

=₁¹₋^−b_a^∂r_∂x= ^∂θ_∂y 1−b

1−ar+^b₁⁻₋^a_a

.

From x=rcosθ, y =rsinθ, we have ^∂r_∂x = cosθ, ^∂θ_∂y = ^cos_r^θ, and hence

1−b

1−acosθ = ^cos_r^θ 1−b

1−ar+₁^b−₋^a_a

, which implies that ₁^b−₋^a_a = 0, and hence b=a.Conversely, if b=a,thenf(re^iθ) =re^iθ and sof is analytic.

7.24.Since ux+uydy/dx = 0 and vx+vydy/dx = 0, we have dy/dx =

−ux/uy and dy/dx = −vx/vy. Thus, at a point of intersection (x₀, y₀), from the Cauchy-Riemann conditions and the fact thatf(x₀+iy₀)= 0 it follows that (−ux/uy)(−vx/vy) =−1.

Lecture 8