21.2

5(b). (Problem) Show that
| zn | = | z |n     and    
1

zn

= 1

| z |n
   for     n = 1, 2, …
(Solution) (b) Use polar form, i.e. let z = r ei θ, then |zn| = rn = |z|n. Note |eiθ|=|cos θ+ i sin θ| = 1.
9.(Problem) Evaluate each of the following, that is, express each in standard Cartesian form x + i y.

    (a)
    (2 − i)3

    (e)

    1 + i

    2 − i

    3

     

    (g)
    ℑ(1 + i)3

    (h)

    1

    1 + i

    3

     
(Solution)
(a) (2 − i)3 = 2 − 11 i.
(e)

1 + i

2 − i

3
 
= − 26

125
18

125
i.
(g) Note 1 + i = √2 eπi/4 and (1 + i)3 = 23/2 ei/4 so ℑ(1 + i)3 = 23/2 sin (3 π/4) = 2.
(h) 1/8.

21.3

1.(Problem) With a labeled sketch, show the point sets defined by the following.

    (a) | z − 1 | ≤ 4
    (c) | z + 2 − i | < 2
    (i) | z + 1 | = | z | + 1
(Solution)
(a) |z − 1| ≤ 4 can be read as a set of points inside a circle with a radius of 4 centered at z = 1.
(c) Read it as | z − (−2 + i)| < 2, i.e. a set of points inside the circle with a radius 2 centered at −2 + i.
(i) Square the both sides to get |z + 1|2 = |z|2 +2 |z| + 1 which is expanded to (x + 1)2 + y2 = x2 + y2 + 2 √{x2 + y2} + 1 from which one gets x = √{x2 + y2}, i.e. y = 0 and x > 0 (don't forget this condition !).
2.(Problem) Determine the range R for the given function. Include sketches of both the domain D and the range R, and give the equations of any curved parts of the boundary of R.

    (a) w(z) = z + 2 + i on 0 < x < 1, 0 < y < 1
    (d) w(z) = z2 on −∞ < x < 0, 0 < y < ∞
    (f) w(z) = i z2 on 0 < x < 1, 0 < y < 1
    (g) w(z) = z3 on 0 < x < ∞, 0 < y < ∞
(Solution)
(a) Shift the rectangle (0 < x < 1, 0 < y < 1) to the right by 2 and up by 1.
hw_text_21_sol_21_3_1.gif
(d) Note that with w = z2, the argument on z is doubled. So the positive y axis (θ = π/2) is mapped to the negative v axis (ϕ = π) and the negative x axis (θ = π) is mapped to the positive u axis (ϕ = 2 π), i.e. the area expressed by v ≤ 0.
hw_text_21_sol_21_3_2.gif
(f) w = i z2 is equivalent to u + i v = i (x + i y)2. Comparing both the real part and the imaginary part gives u = −2 x y and v = x2y2. Using this relationship, one can transform the boundary of the rectangle (0 < x < 1, 0 < y < 1). i.e. { x = 0, 0 < y < 1} → { u = 0, −1 < v < 0}, { y = 0, 0 < x < 1} → { u = 0, 0 < v < 1}, { x = 1, 0 < y < 1 } → { u = − 2y, v = 1 − y2 } → { v = 1 − u2/4}, { y = 1, 0 < x < 1 }→ {u = − 2 y, v = y2 − 1 }→ {v = u2/4 − 1}.
hw_text_21_sol_21_3_3.gif
(g) Note that with w = z3, the argument is tripled. So the positive x axis (θ = 0) remains the positive u axis while the positive y axis (θ = π/2) is mapped to the negative v axis (θ = 3 π/2). So the first quadrant is tripled on the w plane.
hw_text_21_sol_21_3_4.gif
3.(Problem) Show whether or not | ez | = e| z | . More generally, is | w ( z ) | = w ( | z | )?
(Solution)

ez = e(x + i y) = ex ei y = ex (cos y + i sin y).
Thus,
|ez| = ex |cos y + i sin y| = ex.
On the other hand,
e|z| = e√{x2 + y2},
so

|ez| ≠ e|z|.
Counterexample: z = i y. |ez| = |cos y + i sin y| = 1 while e|z| = ey ≠ 1. In general, |w(z)| ≠ w(|z|).
4.(Problem) Show whether or not
-
ez
 
= ez.
More generally, is
-
w(z)
 
= w
-
z
 

?
(Solution)
Yes as

ez
 
=

ex (cos y + i sin y)
 
= ex (cos yi sin y),
and
ez = exi y = ex (cos yi sin y).
This is not true in general. Counterexample: w(z) = i.
7.(Problem) Show that

    (a) sin (−z ) = −sin z and cos (−z ) = cos z
    (b) cos (z1 + z2 ) = cos z1 cos z2 − sin z1 sin z2
    (e) cos (x + i y ) = cos x cosh yi sin x sinh y
    (f) sin (x + i y ) = sin x cosh y + i cos x sinh y
    (g) cosh2 z − sinh2 z = 1
(Solution)
(a)(b) Both are proven by analytical continuation. If you are not satisfied by this, use the definition of equations (14) and (15).
(e)
cos (x + i y) = cos x cos i y − sin x sin i y,
Note that

cos (i y) = 1

2
( ei (i y) + ei (i y)) = 1

2
(ey + ey) = cosh y,
and

sin (i y) = 1

2 i
( ei (i y)ei (i y)) = i

2
(eyey) = i sinh y,
so
cos (x + i y) = cos x cosh yi sin x sinh y.
(f) sin (x + i y) = sin x cos i y + cos x sin i y = sin x cosh y + i cos x sinh y.
(g)
cosh 2 z − sinh 2 z
=

ez + ez

2

2

 

ezez

2

2

 
(1)
=
e2 z + 2 + e−2 z

4
e2 z − 2 + e−2 z

4
(2)
=
1.
(3)
8.(Problem)
Show that
(b)
cosh (z1 + z2) = cosh z1 cosh z2 + sinh z1 sinh z2.
(g)
cosh2  z − sinh2  z = 1.
Solution
(b) Expand
e(z1 + z2) + e−(z1 + z2)

2
and
ez1+ez1

2
ez2+ez2

2
+ ez1ez1

2
ez2ez2

2
separately and verify that the both are the same.
(g)

ez + ez

2

2
 

ezez

2

2
 
= e2z+2+e−2z − ( e2z −2 + e2z)

4
= 1.
9. (Problem) Evaluate each of the following in standard Cartesian form.

    (a) e2 + πi
    (d) sin ( 3 + πi )
(Solution)
(a) e2 eπi = e2 (cos π+ i sin π) = − e2.
(d)
sin (3 + πi)
=
ei (3 + πi)ei(3 + πi)

2 i
=
e3 i e−πe−3 i eπ

2 i
=
=
cosh π sin 3 + i cos 3  sinh π.
11.Prove that

    (a) ez = 1 if and only if z = 2 n πi, where n is any integer.
    (b) ez1 = ez2 if and only if z1 = z2 + 2 n πi, where n is any integer.
(Solution)
(a) This comes from 1 = e2 n πi (polar form).
(b) ez1 = ez2 is equivalent to ez1z2 = 1. From the result of (a), z1z2 must be equal to 2 n π.
13.(Problem) Show that the range R of the function sin z on the semi-infinite strip −π/2 < x < π/2, 0 < y < ∞ is shown in the accompanying figure.
(Solution)
sin z = sin (x + i y) = sin x cosh y + i cos x sin h y, i.e. u = sin x cosh y and v = cos x sin h y. The bottom (y = 0, −π/2 < x < π/2) is mapped to u = sin x, v = 0 or −1 < u < 1, v = 0. The right edge (x = π/2, y > 0) is mapped to u = cosh y, v = 0 or 1 < u < ∞, v = 0. The left edge (x = −π/2, y > 0) is mapped to u = −cosh y, v = 0 or −∞ < u < 1, v = 0. Combining them, it is found that the rectangle is mapped to the upper half plane.

21.4

1.(Problem) Determine r and the principal argument θ0 (in radians and in degrees) for each of the following values of z.

    (a) −3 i
    (f) 2 − 12 i
(Solution)
(a)
−3 i = 3 e3/2 πi.

(f)
2 − 12 i = 2

 

37
 
e−1.405 i.

4.(Problem) Obtain z10 and z20, in both polar and Cartesian form, for each given z.

    (a) −1 + i
(Solution)
(a)
z = √2 e3 π/4 i.

z10 = 25 e7.5 πi = −32 i.


z20 = 210 e15 πi = −1024.

5.(Problem) Find all values of z1/2 and z1/5 for each given z. Express those values in polar form, and show their location in the z plane, as we have done in Fig. 5.

    (a) i
(Solution)
(a) z=eπ/2 i so z1/2 = eπ/4 i and z1/5 = eπ/10 i.
6.(Problem) Obtain, in Cartesian form, all values of log z for each given z.

    (a) −2
(Solution)
(a) −2 = 2 eπi + 2 n πi so ln (−2) = ln 2 + ( 2 n + 1)πi.
8.(Problem) Obtain, in polar form, all values of z2/3, z3/2, and zπ for each given z.

    (a) 2 i
(Solution)
(a) z = 2 e (π/2 + 2 n π) i so z2/3 = 22/3 e2/3 (π/2 + 2 n π) i = 22/3 eπ/3 i e4 n/3 πi.
11.(Problem) Obtain, in Cartesian form, the principal values of log z and √z for each given z.

    (a) −3 i
(Solution)
(a) ln (−3 i) = ln (3 e(3 π/2 + 2 n π)i) = ln 3 + i (3 π/2 + 2 n π). By setting n = 0, the principal value of ln (−3) is ln 3 + 3 πi/2.
√{−3 i} = √( 3 e (3 π/2 + 2 n π)i) = √3 e (3 π/2 + 2 n π) i /2. By setting n = 0, the principal value of √{− 3 i} is
√3 e3 πi/4 = √3
−1

√2
+ i

√2

.
13.(Problem) (Inverse of sine function) We define the inverse of the sine function

w(z) = arcsin z
(4)
such that z = sin w

    (a) Writing the latter as
    z = ( eiweiw)/2i,
    (5)
    show that eiw = iz + ( 1 − z2 )1/2, and hence that

    arcsin z = −ilog
    iz +

     

    1−z2
     

    .
    (6)

    (b) Observe that arcsin z is multi-valued because of the ( 1−z2 )1/2 and also because of the log[   ]. Specifically, for each value of z ( ≠ ±1 ), the ( 1− z2 )1/2 gives two values. Then, for each of these values the log gives an infinite set of values. To illustrate this point, show that

    arcsin  1

    2
    = π

    6
    + 2kπ  or  

    6
    + 2kπ
    (7)
    for k=0, ±1, ±2, …
    (c) Determine all possible values of arcsin 2.
    (d) Determine all possible values of arcsin ( 2i ).
(Solution)
(a) z=(eiweiw)/(2i) is equivalent to 2iz = eiweiw or (eiw)2 −2 i z eiw −1 = 0. Solving this quadratic equation, one gets eiw=iz + (1−z2)1/2, i.e. w=arcsin z = −i ln(i z + (1−z2)1/2). Note that (1−z2)1/2 is multi-valued (±) without an appropriate branch cut which should take care of the ± sign when solving the quadratic equation.
(b) Note that z1/2 = (r eθi + 2 n πi1/2) = √r e[(θ)/2]i + n πi where the values of en πi is either 1 or -1 depending on whether n is even or odd. Therefore, (1−(1/2)2)1/2 = ±[( √3)/2]. arcsin  (1/2) = −i ln(i/2 ±√3/2) = −i ln(eπ/6 i + 2k πi) or−i ln(e5π/6 i + 2k πi) = [(π)/6] + 2k πor [(6π)/6] + 2k π.
(c) arcsin 2 = −i ln(2i + √{−3}) = −i ln((2±√3)i) = −iln( (2±√3)e(π/2 + 2kπ)i) = −i (ln(2±√3) + i (π/2 + 2kπ)) = (π/2 + 2kπ) − i ln (2±√3).
(d) arcsin (2 i) = −i ln( i (2i) + (1−(2i)2)1/2) = −i ln(−2 ±√5) = −2n π−i ln (√5−2) or−(2n+1)π− i ln(√5+2).
14. (Problem) (Inverse of other trigonometric functions) Proceeding as in Exercise 13, derive these formulas.

    (a) arccos z = −ilog [ z + √{z2 −1} ]
    (b) arctan z = −[(i)/2]log [(iz)/(i+z)]
    (c) cot −1 z = −[(i)/2]log[(z+1)/(z−1)]
(Solution)
(a) Solve z = [(eiw+eiw)/2] for eiw to get arccos  z = −i ln(z+√{z2−1}1/2).
(b) Solve z=[(eiweiw)/(i(eiw+eiw))] to get arctan z = −[(i)/2] ln[(iz)/(i+z)].
(c) Similar to (b).

21.5

10.(Problem) Given f ( z ), use (19) to obtain f′( z ). Express your answer in terms of z.

    (a) cos z
    (e) [1/(z)] ( z ≠ 1 )
(Solution)
(a) f′(z) = −sin z.
(e) f′(z) = −1/z2.
If f(z) is a regular function, you can differentiate it as if it were a real valued function.
11.(Problem) Given f ( z ), determine f′( z ), where it exists, and state where f is analytic and where it is not.

    (a)
    (1 − 2 z3 )5

    (b)
    x + i y

    x2 + y2

    (c)
    | z | sin z

    (d)
    1

    z2 + 3 i z − 2

    (e)
    1

    z3 + 1

    (f)
    x + i sin y
(Solution)
(a) f(z) is regular so f′(z) = 5(1−2z3)4 ×(−6z2).
(b) Note that
x + i y

x2 + y2
= z

z
-
z
 
= 1

-
z
which is NOT analytic (contains z).
(c) This is not analytic as it contains
|z| =   ⎛


z
-
z
 
 
.

(d)
f′(z) = −(2 z + 3 i)

(−2 + 3 i z + z2)2
.

(e)
f′(z) = −3 z2

(1 + z3)2
.

(f) This is not an analytic function (use C-R to verify).
12.(Problem) (Cauchy-Riemann equations in polar coordinates) Derive the Cauchy-Riemann equations (30) in the manner indicated.

    (a) By carrying out the limit in (8). HINT: First let ∆z → 0 along the constant-θ line through z0, and then let ∆z → 0 along the constant-r line through z0. Pay careful attention to your expression for ∆z in each of these cases because these cases are trickier than the cases of a horizontal approach ( ∆z = ∆x ) and a vertical approach ( ∆z = iy ), used in (16) and (17).
    (b) By making the change of variables x = r cos θ, y = rsin θ in (18).
(Solution)
(a) On the constant θ line, ∆z = ∆r eiθ (r varies but θ remains constant) so

f′(z)
=

lim
r→ 0 
u(r+∆r, θ) + i v(r+∆r,θ)−u(r,θ) − i v(r, θ)

r eiθ
(8)
=

lim
r→ 0 
eiθ
u(r+∆r, θ)−u(r)

r
+i v(r+∆r, θ)−v(r)

r

(9)
=
ei θ (ur+i vr).
(10)
On the constant-r line, ∆z = i  r  ∆ θ eiθ (r is constant while θ varies) so

f′(z)
=

lim
∆θ→ 0 
u(r,θ+∆θ) + i v(r, θ+∆θ)−u(r,θ) − i v(r,θ)

i  r ∆θ eiθ
(11)
=
1

r
 ei θ  vθ i

r
 eiθ uθ.
(12)
By comparing the real and imaginary parts, one obtains

ur = 1

r
vθ
vr = − 1

r
uθ.
(13)
(b) Recall the chain differentiation rule, i.e.
ux
=
ur rx + uθ θx
(14)
vy
=
vr ry + vθ θy,
(15)
and

rx
=
r

x
=

x


 

x2 + y2
 
=
x




x2 + y2
=
rcos θ

r
=
cos θ
etc...
So ux = vyur rx + uθ θx = vr ry + vθ θx or

ur cos θ− uθ sin θ

r
= vr sin θ+ vθ cos θ

r
.
and
uy = −vxur ry + uθ θy = −vr rxvθ θy
or

ur sin θ+ uθ cos θ

r
= −vr cos θ+ vθ sin θ

r
.
Solving the above simultaneous equations for ur and vr yields ur = [1/(r)] vθ and vr = −[1/(r)] uθ.
13.(Problem) Determine where these functions are differentiable and where they are analytic, by checking for satisfaction of the Cauchy-Riemann equations and for continuity of u, v and their first-order partial derivatives.

    (a) f ( z ) = z100
    (b) f ( z ) = √z, defined by the branch cut shown in Fig. 6
    (c) f ( z ) = [1/(√z)], where √z is defined by the branch cut shown in Fig.6
(Solution)
(a) f(z) = z100 is analytic as it is a function of z alone. f′(z) = 100z99.
(b) f(z) = √z is analytic except for z=0 at which point f′(z) fails to exist.
f′(z) = 1

2√z
.

(c) f(z) = [1/(√z)] is analytic except for z=0 at which point f′(z) fails to exist.
f′(z) = − 1

2
z−3/2.

14.(Problem) (a) f=u+iv and f=uiv. The Cauchy-Riemann relation requires that ux = vy, uy = −vx, ux = −vy and uy = vx from which ux=uy=0 so u=const then, vx = vy = 0 so v=const too.
(b) By the fundamental theorem of calculus.
15. Determine whether or not the given function u is harmonic and, if so, in what region. If it is, find the most general conjugate function v and corresponding analytic function f (z). Express f in terms of z.

    (a) ex cos y
    (b) e2 x sin 2 y
    (c) x3 − 3 x y2
    (d) r3 sin 3 θ
    (e) r2 cos 2 θ +4
    (f) r
    (g) xcos 2x cosh 2y + y sin 2xsinh 2y
(Solution)
(In the following solution, add a constant C to f(z).)
(a) Yes. v = ex sin y so that f(z) = ez.
(b) Yes. v = −e2x cos 2y. Note that
u + i v
=
e2x (sin 2yi cos 2y)
=
e2x (−i2 sin 2yi cos 2y)
=
i e2x (cos 2y + i sin 2y)
=
i e2xe2yi
=
i e2 (x + i y)
=
i e2z.
(c) Yes. v = 3x2 yy3 so f(z) = z3.
(d) Yes. Take v=−r3 cos 3θ so that u+iv = r3 (sin 3θ− i cos 3θ) = r3 (−i2 sin 3θ−i cos 3θ) = r3 (−i) (cos 3θ+i sin 3θ) = −i (r3 ei) = −i z3.
(e) Yes. v = r2 sin θ so that f(z) = z2 + 4.
(f) No.
(g) Yes. v = y cos 2x cosh2yx sin 2x sin h 2y.



File translated from TEX by TTH, version 4.03.
On 13 Jan 2019, 22:24.