ANALYTIC FUNCTIONS
23. POLAR COORDINATES
Assuming that z0=0, we shall in this section use the coordinate transformation x=rcosθ , y=rsinθ
(1)
to restate the theorem in Sec. 22 in polar coordinates.
Depending on whether we write
z=x+iy or z=reiθ (z=0)
sec. 23 Polar Coordinates 69
when w=f (z), the real and imaginary components ofw=u+iv are expressed in terms of either the variables x and y or r and θ. Suppose that the first-order partial derivatives of u and v with respect to x and y exist everywhere in some neighborhood of a given nonzero pointz0and are continuous at z0. The first-order partial derivatives of u and v with respect to r and θ also have those properties, and the chain rule for differentiating real-valued functions of two real variables can be used to write them in terms of the ones with respect toxandy. More precisely, since
∂u
∂r = ∂u
∂x
∂x
∂r +∂u
∂y
∂y
∂r, ∂u
∂θ = ∂u
∂x
∂x
∂θ +∂u
∂y
∂y
∂θ, one can write
ur =uxcosθ+uysinθ , uθ= −uxrsinθ+uyrcosθ . (2)
Likewise,
vr =vxcosθ+vysinθ , vθ= −vxrsinθ+vyrcosθ . (3)
If the partial derivatives ofu and v with respect to x and y also satisfy the Cauchy–Riemann equations
ux =vy, uy= −vx
(4)
atz0, equations (3) become
vr = −uycosθ+uxsinθ , vθ =uyrsinθ+uxrcosθ (5)
at that point. It is then clear from equations (2) and (5) that rur =vθ, uθ = −rvr
(6) atz0.
If, on the other hand, equations (6) are known to hold atz0, it is straightforward to show (Exercise 7) that equations (4) must hold there. Equations (6) are, therefore, an alternative form of the Cauchy–Riemann equations (4).
In view of equations (6) and the expression forf(z0)that is found in Exercise 8, we are now able to restate the theorem in Sec. 22 usingr andθ .
Theorem. Let the function
f (z)=u(r, θ )+iv(r, θ )
be defined throughout some ε neighborhood of a nonzero point z0=r0 exp(iθ0), and suppose that
(a) the first-order partial derivatives of the functions uand vwith respect to r and θ exist everywhere in the neighborhood;
(b) those partial derivatives are continuous at(r0, θ0)and satisfy the polar form rur =vθ, uθ = −rvr
of the Cauchy–Riemann equations at (r0, θ0).
Then f(z0)exists, its value being
f(z0)=e−iθ(ur +ivr), where the right-hand side is to be evaluated at (r0, θ0).
EXAMPLE 1. Consider the function f (z)= 1
z = 1 reiθ = 1
re−iθ = 1
r(cosθ−isinθ ) (z=0).
Since
u(r, θ )= cosθ
r and v(r, θ )= −sinθ r ,
the conditions in this theorem are satisfied at every nonzero point z=reiθ in the plane. In particular, the Cauchy–Riemann equations
rur = −cosθ
r =vθ and uθ = −sinθ
r = −rvr
are satisfied. Hence the derivative of f exists when z=0; and, according to the theorem,
f(z)=e−iθ
−cosθ
r2 +isinθ r2
= −e−iθ e−iθ
r2 = − 1
(reiθ)2 = −1 z2.
EXAMPLE 2. The theorem can be used to show that whenαis a fixed real number, the function
f (z)=√3
reiθ/3 (r >0, α < θ < α+2π ) has a derivative everywhere in its domain of definition. Here
u(r, θ )=√3 rcosθ
3 and v(r, θ )=√3 rsinθ
3. Inasmush as
rur = 3
√r 3 cosθ
3 =vθ and uθ = − 3
√r 3 sinθ
3 = −rvr
sec. 23 Exercises 71
and since the other conditions in the theorem are satisfied, the derivativef(z)exists at each point wheref (z) is defined. The theorem tells us, moreover, that
f(z)=e−iθ 1 3(√3
r)2cosθ
3+i 1 3(√3
r)2sinθ 3
, or
f(z)= e−iθ 3(√3
r)2eiθ/3= 1 3(√3
reiθ/3)2 = 1 3[f (z)]2.
Note that when a specific pointzis taken in the domain of definition off,the valuef (z)is one value ofz1/3(see Sec. 9). Hence this last expression forf(z) can be put in the form
d
dzz1/3= 1 3(z1/3)2
when that value is taken. Derivatives of such power functions will be elaborated on in Chap. 3 (Sec. 33).
EXERCISES
1. Use the theorem in Sec. 21 to show thatf(z)does not exist at any point if (a)f (z)=z; (b)f (z)=z−z;
(c)f (z)=2x+ixy2; (d)f (z)=exe−iy.
2. Use the theorem in Sec. 22 to show thatf(z) and its derivativef(z)exist every- where, and findf(z)when
(a)f (z)=iz+2; (b)f (z)=e−xe−iy;
(c)f (z)=z3; (d)f (z)=cosxcoshy−isinxsinhy.
Ans.(b)f(z)=f (z); (d)f(z)= −f (z).
3. From results obtained in Secs. 21 and 22, determine wheref(z)exists and find its value when
(a)f (z)=1/z; (b)f (z)=x2+iy2; (c)f (z)=zImz.
Ans.(a)f(z)= −1/z2(z=0); (b)f(x+ix)=2x; (c)f(0)=0.
4. Use the theorem in Sec. 23 to show that each of these functions is differentiable in the indicated domain of definition, and also to findf(z):
(a) f (z)=1/z4 (z=0);
(b) f (z)=√
reiθ/2 (r >0, α < θ < α+2π );
(c) f (z)=e−θcos(lnr)+ie−θsin(lnr) (r >0,0< θ <2π ).
Ans.(b)f(z)= 1
2f (z); (c)f(z)=if (z) z .
5. Show that whenf (z)=x3+i(1−y)3, it is legitimate to write f(z)=ux+ivx=3x2
only whenz=i.
6. Letuandvdenote the real and imaginary components of the functionf defined by means of the equations
f (z)=
z2/z when z=0, 0 when z=0.
Verify that the Cauchy–Riemann equationsux=vy anduy = −vx are satisfied at the originz=(0,0). [Compare with Exercise 9, Sec. 20, where it is shown thatf(0) nevertheless fails to exist.]
7. Solve equations (2), Sec. 23 forux anduy to show that ux =urcosθ−uθ
sinθ
r , uy=ursinθ+uθ
cosθ r .
Then use these equations and similar ones for vx and vy to show that in Sec. 23 equations (4) are satisfied at a pointz0 if equations (6) are satisfied there. Thus com- plete the verification that equations (6), Sec. 23, are the Cauchy–Riemann equations in polar form.
8. Let a functionf (z)=u+iv be differentiable at a nonzero pointz0=r0exp(iθ0).
Use the expressions forux andvx found in Exercise 7, together with the polar form (6), Sec. 23, of the Cauchy–Riemann equations, to rewrite the expression
f(z0)=ux+ivx
in Sec. 22 as
f(z0)=e−iθ(ur+ivr), whereur andvr are to be evaluated at(r0, θ0).
9. (a) With the aid of the polar form (6), Sec. 23, of the Cauchy–Riemann equations, derive the alternative form
f(z0)= −i z0
(uθ+ivθ) of the expression forf(z0)found in Exercise 8.
(b) Use the expression forf(z0)in part(a)to show that the derivative of the function f (z)=1/z (z=0)in Example 1, Sec. 23, isf(z)= −1/z2.
10. (a) Recall (Sec. 5) that ifz=x+iy, then x= z+z
2 and y= z−z 2i .
sec. 24 Analytic Functions 73
Byformally applying the chain rule in calculus to a functionF (x, y)of two real variables, derive the expression
∂F
∂z = ∂F
∂x
∂x
∂z +∂F
∂y
∂y
∂z =1 2
∂F
∂x +i∂F
∂y
. (b) Define the operator
∂
∂z = 1 2
∂
∂x+i ∂
∂y
,
suggested by part (a), to show that if the first-order partial derivatives of the real and imaginary components of a function f (z)=u(x, y)+iv(x, y) satisfy the Cauchy–Riemann equations, then
∂f
∂z = 1
2[(ux−vy)+i(vx+uy)]=0.
Thus derive thecomplex form ∂f/∂z=0of the Cauchy–Riemann equations.