## 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)
 ⎛⎝ ℜ 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.
(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.
(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}.
(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.
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 = e―z.
More generally, is
 - w(z) = w ⎛⎝ - z ⎞⎠ ?
(Solution)
Yes as
 ez = ex (cos y + i sin y) = ex (cos y − i sin y),
and
 e―z = ex − i y = ex (cos y − i 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) + e− i (i y)) = 1 2 (e−y + ey) = cosh y,
and

 sin (i y) = 1 2 i ( ei (i y) − e− i (i y)) = i 2 (ey − e−y) = i sinh y,
so
 cos (x + i y) = cos x cosh y − i 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 + e−z 2 ⎞⎠ 2 − ⎛⎝ ez − e−z 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+e−z1 2 ez2+e−z2 2 + ez1−e−z1 2 ez2−e−z2 2
separately and verify that the both are the same.
(g)
 ⎛⎝ ez + e−z 2 ⎞⎠ 2 − ⎛⎝ ez − e−z 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)−e− i(3 + πi) 2 i
 =
 e3 i e−π − e−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 = ( eiw − e−iw)/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 5π 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 e−iθ ⎛⎝ u(r+∆r, θ)−u(r) ∆r +i v(r+∆r, θ)−v(r) ∆r ⎞⎠
(9)
 =
 e−i 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 e−i θ  vθ − i r e−iθ 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 2y − i cos 2y)
 =
 e2x (−i2 sin 2y − i 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.

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On 13 Jan 2019, 22:24.