1.5 Sets of Points in the Complex Plane -...

Joseph HeavnerHonors Complex Analysis

Assignment 2January 25, 2015

1.5 Sets of Points in the Complex Plane

1.) Sketch the graph of|z− 4 + 3i| = 5

.

Figure 1: Circle with radius 5 centered at 4− 3i

3.) Sketch the graph of|z + 3i| = 2

.

Figure 2: Circle with radius 2 centered at −3i

5.) Sketch the graph ofRe(z) = 5

.

Figure 3: Vertical line at x = 5

1

7.) Sketch the graph ofIm(z + 3i) = 6

Unlike the previous problems, which were obvious enough as to call for no explanation, here we willwrite the equation in a more graph-able form. In particular, for z = a + bi

I(z + 3i) = 6I(a− bi + 3i) = 6

−b + 3 = 6b = −3

.

Figure 4: Horizontal line at y = −3

9.) Sketch the graph of|Re(1 + iz)| = 3

|R(1 + iz̄)| = 3|R(1 + i(x− iy))| = 3

|1 + y| = 3

(1 + y)2 = 9y = 2 , −4

.

Figure 5: Horizontal lines at y = 2 and y = −4

2

11.) Sketch the graph ofRe(z2) = 1

R((a + bi)2) = 1

R(a2 + 2bi− b2) = 1

a2 − b2 = 1

.

Figure 6: Hyperbola x2 − y2 = 1

13.) Sketch the set S of the points in the complex plane satisfying the given inequality. Determinewhether the set is (a) open, (b) closed, (c) a domain, (d) bounded, or (e) connected.

Re(z) < −1

a.) yes b.) no c.) yes d.) no e.) yes

Some explanation is that any neighborhood in the region R is entirely contained (i.e. any ε-ball isentirely contained), therefore it is open. Also, not all of the boundary points of R are contained in R,therefore it is closed. Because any two points z0 , z1 ∈ R can be connected by a collection of connectedlines, the set is connected; by definition, an open, connected set is a domain. The set is not bounded,however, as the modulus of a point z0 ∈ R can become arbitrarily large (i.e. the set extends indefinitelytowards x = −∞).

.

Figure 7: Half-plane x < −1

3


Im(z) > 3


The justification here is identical to that in (13).

.

Figure 8: Half-plane y > 3


2 < Re(z− 1) < 4

To make this set easier to plot, we simplify the expression describing it a bit.

2 <R(z− 1) < 42 < x− 1 < 43 < x < 5


The justification here is identical to that in (13) and (15).

.

Figure 9: 3 < x < 5

4


Re(z2) > 0

Again, we’ll simplify this a bit.

R(x2 + 2xy− y2) > 0

x2 − y2 > 0

a.) yes b.) no c.) no d.) no e.) no

The justifications for (a), (b), and (d) are the same as in previous questions. This set is not connectedbecause a line cannot possibly pass through the origin without leaving the set. Lack of connected-nessthen implies that the set is not a domain.

.

Figure 10: 3 < x < 5


|z− i| > 1

We know the general form of a circle from our text and the last homework assignment.


The justification here is analogous to (13).

.

Figure 11: The region outside the open disk x2 + (y− 1)2 < 1

5


1 ≤ |z− 1− i| < 2

This is an annulus like that described in our text.

a.) no b.) no c.) no d.) yes e.) yes

For (b) and (e) the explanation is analogous to (13). This set is not open because, for the inner boundary,we could find a neighborhood that contains some point outside the set (in fact, every neighborhoodon the boundary contains points not in the set). The set is not a domain, as it is not open. The set isbounded, because there exists a number r ∈ R : r > 0 such that |z| < r ∀z ∈ R. In particular, for thisregion, there exists no point in the region with modulus greater than 2 (with new coordinates such thatthe origin is the same as the center of the annulus).

.

Figure 12: Circular annulus centered at (1,1) with inner boundary |z| = 1 and outer boundary |z| < 2

25.) Give the boundary points of the sets in Problems 13-24.

I think this is straightforward enough that it requires no explanation.

13: x = −1 15: y = 3 17: x = 3, 5 19: y = ±x 21: |z− i| = 123: |z− 1− i| = 1 & |z− 1− i| = 2

27.) Sketch the set of points in the complex plane satisfying

0 ≤ arg(z) ≤ π/6

This question is most easily answered with a geometric explanation. Because the modulus of our com-plex number is not specified, we need only concern ourselves with the argument. We generally measureangles from the positive x-axis. Therefore, this region in polar coordinates would be (r, θ ≤ π/6). Tome, this is already intuitive, the figure will be all z0 ∈ C in the area sweeped out by a ray of indefinitelength with endpoint at the origin such that the angle between the ray and the positive x-axis is lessthan or equal to π/6 = 30◦.

.

Figure 13: A triangular region extending indefinitely

6

29.) Describe the shaded set in the given figure using arg(z) and an inequality.

As a reference, below is the figure in question.

.

Figure 14: Relevant image from the text

With similar reasoning to (27) we know this is the concatenation of the regions 0 ≤ arg(z) ≤ 2π/3 and−2π/3 ≤ arg(z) ≤ 0. By combining the inequalities we get −2π/3 ≤ arg(z) ≤ 2π/3 or

|arg(z)| ≤ 2π

3

31.) Solve the simultaneous equations

|z| = 2 , |z− 2| = 2

By the transitive property we have

|z| = |z− 2|√x2 + y2 =

√(x− 2)2 + y2

x2 + y2 = (x− 2)2 + y2

x2 + y2 = x2 − 4x + 4 + y2

−4x + 4 = 0x = 1

Now, we have to consider the second equality for a bit more information (i.e. a restriction).

(x− 2)2 + y2 = 4

x2 − 4x + 4 + y2 = 4

x2 − 4x + y2 = 0

(1)2 − 4(1) + y2 = 0

y = ±√

3

Now, we know that z = x + iy, therefore

1±√

3i

7

33.) On page 28 we stated that if ρ1 > 0, then the set of points satisfying ρ1 < |z− z0| is the exterior tothe circle of radius ρ1 centered at z0. In general, describe the set if ρ1 = 0. In particular, describe theset defined by

|z + 2− 5i| > 0√(x + 2)2 + (y− 5)2 > 0

(x + 2)2 + (y− 5)2 > 0

Now, let us consider some cases to figure out what this set is. Consider the case where x = −2, then wehave that y 6= 5; similarly, if we let x 6= 2 then all y are valid; finally, y 6= 5 =⇒ ∀x (note the loose useof notation), which is the same as the first case. Therefore, we have that our set describes the complexplane with the point (−2, 5) deleted, i.e. C− {−2− 5i}. This can be thought of as the exterior of a circleof radius 0. To sum that up we have

{z : z 6= −2− 5i}

37.) Suppose z0 and z1 are distinct points. Using only the concept of distance, describe in words theset of points z in the complex plane that satisfies

|z− z0| = |z− z1|

The word ”concept” to me implies that one should avoid algebra in solving this problem. With thatbeing said, this describes the set of points in the complex plane such that their distance to an arbitrarypoint (i.e. a variable) is equivalent. For instance, for the origin the points z0 = 1 and z1 = −i wouldsatisfy this, as their moduli are equivalent, i.e. they are the same distance from the origin. In otherwords, this describes the set of all equivalent circles (for either a value of z or the values of z0 and z1);in particular, the intersecting points on the circles are described, so the result is a line passing throughthe midpoint of z0 and z1 and orthogonal to the segment joining z0 and z1.

39.) Describe the shaded set in the given figure by the filling in the two blanks in set notation, usingcomplex notation for equations and inequalities and one of the words and or or.

.

Figure 15: Relevant image from the text

This is the combination of the exterior of a circle centered at 4i with radius 4 and the interior of anothercircle with center (1, 3i) and radius 3.

{z : |z− 4i| ≥ 4∪ |z− 1− 3i| ≤ 3}

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48.) Find a point (x0, y0, z0) on the unit sphere that corresponds to the complex number 2 + 5i.

Using parametric equations of lines as vectors, we can solve this quite easily. The line is as followed:x = 2t , y = 5t , z = 1− t or

(2t, 5t, 1− t)

Now, we know that (x, y, z) must be one unit from the origin (by virtue of it lying on the unit sphere),so we use the distance formula as follows

(2t)2 + (5t)2 + (1− t)2 = 1

30t2 − 2t = 0

t =115

Therefore, our point is(2/15 , 1/3 , 14/15)

50.) Express the coordinates of the point (x0, y0, z0) on the unit sphere in Figure 1.5.10(b) in terms ofthe coordinates of the point (a, b, 0) in the complex plane. Use these formulas to verify your answerto Problem 48.[ Hint: First show that all points on the line containing (0, 0, 1) and (a, b, 0) are of the form(ta, tb, 1− t).]

I have changed my slightly more abstract geometric proof to the suggested one using parametrizations.

We have that d = e = c = 0 and f = 1, so our parametrization is (at, bt, 1− t). now, similar to before,we will find a point, given our parametrization (ta, tb, 1− t), that lies on the sphere and line. We do thisas follows

(at)2 + (bt)2 + (1− t)2 = 1

a2t2 + b2t2 + t2 − 2t = 0

t(a2t + b2t + t− 2) = 0

t =2

a2 + b2 + 1By that fact we can now explicitly write our general mapping.

a = 2Re(z)|z|2+1

b = 2Im(z)|z|2+1

c = |z|2−1|z|2+1

or(2a

a2 + b2 + 1,

2ba2 + b2 + 1

,a2 + b2 − 1a2 + b2 + 1

)

For z = 2 + 5i we have(2 · 2

22 + 52 + 1,

2 · 522 + 52 + 1

,22 + 52 − 122 + 52 + 1

)⇐⇒ (2/15 , 1/3 , 14/15)

9

2.1 Complex Functions

3.) Evaluatef (z) = loge |z|+ iArg(z)

at (a) 1 (b) 4i (c) 1+i

a.)f (1) = ln(|1|) + iArg(1) = ln(1) + i(0) = 0 + 0 = 0

b.)

f (4i) = ln(|4i|) + iArg(4i) = ln(4) +π

2i

c.)

f (1 + i) = ln(|1 + i|) + iArg(i + 1) = ln(√

2) +π

4i =

12

ln(2) +π

4i

6.) Evaluatef (z) = ez

at (a) 2− πi (b) π3 i (c) loge 2− 5π

6 i

a.)

f (2− πi) = e2−πi =e2

eiπ = −e2

b.)

f (π

3i) = e

π3 i = cos(π/3) + i sin(π/3) =

12+

√2

2i

c.)

f (loge 2− 5π

6i) = eloge 2− 5π

6 i =2

e5πi

6= 2e−

5πi6 = 2cis(−5π/6) = −

√3− i

9.) Find the real and imaginary parts u and v of

f (z) = 6z− 5 + 9i

Let z = x + iy, then

f (z) = 6(x + iy)− 5 + 9i = 6x + 6iy− 5 + 9i = 6x− 5 + i(6y + 9)

Thus,u = 6x− 5 v = 6y + 9

15.) Find the real and imaginary parts u and v of

f (z) = e2z+1

Similar to in (9), we let z = x + iy, then

e2x+2iy+1 = e2x · (cos(2iy + 1) + i sin(2iy + 1))

Now, we already have real and imaginary parts split, thus

u = e2x cos(2y + 1) v = e2x sin(2y + 1)

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17.) Find the real and imaginary parts u and v as a function of r and θ of

f (z) = z

Let z = x + iy, thenz = x− iy = r (cos(θ) + i sin(θ))

This then implies thatu = r cos θ v = r sin θ


f (z) = z4

Let z = x + iy, thenz4 = r4 (cis(4θ))

Where the result is by by de Moivre’s theorem. It then follows that

u = r4 cos 4θ r4 sin 4θ


f (z) = ez

Let z = x + iy, then

ez = ex+iy = ex · eiy = er cos θ · eri sin θ = er cos θ [cos(r sin θ) + i sin(r sin θ)]

Therefore,

u = er cos θ cos(r sin θ) v = er cos θ sin(r sin θ)

24.) Find the natural domain of

f (z) =3z + 2i

z3 + 4z2 + z

We factor the denominator as follows

3z + 2iz(z2 + 4z + 1)

=3z + 2i

z(z(z + 4) + 1)

It is now clear that the domain is all z such that z 6= 0,√

3− 2,−√

3− 2, where the last two factors areobtained by solving z2 + 4z = −1.

25.) Find the natural domain of

f (z) =iz|z− 1|

In order for this to be undefined we must have that |z− 1| = 0, therefore, for z = x + iy, we solve

|x + iy− 1| = 0 =⇒√(x− 1)2 + y2 = 0 =⇒ x2 − 2x + 1 + y2 = 0 =⇒ x = 1 , y = 0

Therefore, the natural domain is all z such that z 6= 1.

11

2.2 Complex Functions As Mappings

3.) Find the image S′ of the set S under

f (z) = 3z S : Im(z) > 2

If the points in the domain must have imaginary part greater than two, then the image under multipli-cation by 3 means that the image’s imaginary part must be greater than 6. Thus, the image is the halfplane with imaginary part greater than 6. In other words,

S′ : Im(w) > 6


f (z) = (1 + i)z S : x = 2

We know that z must be of the form 2 + iy, so we find that

f (2 + iy) = (1 + i)(2 + iy) = 2 + iy + 2i− y = (2− y) + i(y + 2)

Now, we have a system of equations to solve. Solving for y in terms of u we find that y = 2 − u,therefore the image is

the line v = 4− u


f (z) = iz + 4 S : Im(z) ≤ 1

Break the domain into −∞ < x < ∞ , y ≤ 1. Our mapping is w = i(x + iy) + 4 = (4− y) + ix,therefore its parts are u = 4− y , v = x. Now, we combine this to 4− u ≤ 1 =⇒ u ≥ 3 and, of course,v is still a variable. Therefore, the image is the half plane

S′ : Re(w) ≥ 3

9.) Find the image of the line y = 1 under w = z2.

For z = x + iy we have w = (x + iy)2 = x2− y2 + 2ixy, therefore u = x2− y2 and v = 2xy, but for y = 1we have u = x2 − 1 and v = 2x. So, we have that x = v/2, implying that u = v2

4 − 1. In other words,the image is the parabola

S′ : u =14

v2 − 1

13.) Find the image of the line y = x under w = z2.

z = x + iy we have w = x2 − y2 + 2ixy, but x = y, therefore we have w = 2ix2. Now, we know that x2

has domain of all real numbers, thus 2ix2 has domain C+ −R, in other words we have that the imageis set of all z such that the y component varies indefinitely but the real part is zero, in other words thisis the ray u = 0, with the restriction, due to the square, that 0 ≤ v < ∞. In short,

the ray u = 0 , 0 ≤ v < ∞

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15.) (a) Plot the parametric curve C given by z(t) and describe the curve in words, (b) find aparametrization of the image, C′, of C under the given complex mapping w = f (z), and (c) plotC′ and describe this curve in words.

z(t) = 2(1− t) + it , 0 ≤ t ≤ 1; f (z) = 3z

a.) Because z(0) = 2 and z(1) = i, we have that this curve C is the line segment from 2 to i.

b.)w(t) = f (z(t)) = 3(2(1− t) + it) = 6(1− t) + 3it

Thus, the parametrization isw(t) = 6(1− t) + 3it , 0 ≤ t ≤ 1

c.) Similar to (a) but under the mapping in (b) we have that z0 = 6 and z1 = 3i, thus the image is theline segment from 6 to 3i.

.

Figure 16: The mapping from the line segment from 2 to i to the line segment from 6 to 3i

17.) (a) Plot the parametric curve C given by z(t) and describe the curve in words, (b) find aparametrization of the image, C′, of C under the given complex mapping w = f (z), and (c) plotC′ and describe this curve in words.

z(t) = 1 + 2eit , 0 ≤ t ≤ 2π; f (z) = z + 1− i

a.) From our list of common parametrizations on page 57, we see this to be a circle, for it is of the formz(t) = z0 + reit, 0 ≤ t ≤ 2π. In particular, this is a circle of radius 2 centered at (1, 0).

.

Figure 17: C, i.e. a circle of radius 2 centered at (1, 0)

b.) This mapping is fairly straightforward; upon substitution of z into f (z) we have that

w(t) = 2− i + 2eit

c.) This is then just another circle, but this time the set is a circle with center 2− i and radius 2.

.

Figure 18: C′, i.e. a circle of radius 2 centered at (2,−i)

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21.) Use parametrizations to find the image, C′, of the curve C under

f (z) = z3 ; C is the positive imaginary axis

We have the parametrization z = it , 0 ≤ t ≤ ∞ for the curve, thus z = (it)3 = −it3. From this, wehave that, because t is a positive real number and only scales the imaginary unit, this forms a line on thenegative imaginary axis.

23.) Use parametrizations to find the image, C′, of the curve C under

f (z) = 1/z ; C : |z| = 2

If the modulus of an element of the domain of the mapping must be equal to 2, then the multiplicativeinverse of that element must have modulus equal to the multiplicative inverse of the element’s modulus(i.e. 1/2). In other words, because the multiplicative identity z = 1 has modulus 1, for some z0 : |z0| =2, we must have that |z−1

0 | = 1/2, as to ensure we reach the identity, which has modulus 1, undermultiplication. In terms of parametrizations we have the parametrization z(t) = 2eit , 0 ≤ t < 2π.Furthermore,

f (z) =1

2eit =12

e−it , 0 ≤ t < 2π

This is the circle |w| = 1/2.

28.) Consider the parametrization z(t) = i(1− t) + 3t, 0 ≤ t ≤ 1.(a) Describe in words this parametric curve.(b) What is the difference between the curve in part (a) and the curve defined by z(t) = 3(1− t) +it , 0 ≤ t ≤ 1(c) What is the difference between the curve in (a) and the curve define by the parametrization z(t) =32 t + i(1− 1

2 t) , 0 ≤ t ≤ 2(d) Find the parametrization of the line segment from 1 + 2i to 2 + i where the parameter satisfies0 ≤ t ≤ 3.

a.) By page 57, this is the line segment from i to 3.

b.) Both are line segments between i and 3, but this one is from 3 to i rather than the reverse. However,line segments typically have no sense of directionality between points, so the two are practically thesame.

c.) Again, this is practically the same, but here the parameter takes on different values to produce thesame geometry.

d.)

(1 + 2i)(

1− 13

t)+

(2 + i

3

)t

29.) Use the parametrizations to find the image of the circle |z− z0| = R under the mapping f (z) =iz− 2.

This mapping rotates by arg(i) = π/2 and translates by −2. For a parametrization we have

z(t) = z0 + Reit , 0 ≤ t ≤ 2π

This is then mapped by iz− 2, meaning that

w(t) = f (z(t)) = i(z0 + Reit)− 2 , 0 ≤ t ≤ 2π

Again, this describes the circle C shifted 2 units to the left (note that circles are geometrically invariantunder rotation).

14

2.3 Linear Mappings

1) (a) Find the closed disk |z| ≤ 1 under the linear mapping f (z) and (b) represent the linear mappingwith a sequence of plots.

f (z) = z + 3i

This is equivalent to shifting a disk of radius 1 by 3 units vertically, so the image is

|w− 3| ≤ 1

3.) (a) Find the closed disk |z| ≤ 1 under the linear mapping f (z) and (b) represent the linear mappingwith a sequence of plots.

f (z) = 3iz

By equation (6) we have that f (z) = |3i|(

3i|3i z)= 3(iz) + 0, thus the mapping rotates by arg(i) = π/2

and magnifies by 3. Disks are invariant under rotation, but the magnifying factor still affects it. Thus,the function maps to a similar closed disk and is completely described as

f : |z| ≤ 1 7→ |w| ≤ 3

.

Figure 19: f : |z| ≤ 1 7→ |w| ≤ 3

5.) (a) Find the closed disk |z| ≤ 1 under the linear mapping f (z) and (b) represent the linear mappingwith a sequence of plots.

f (z) = 2z− i

This is the mapping which magnifies the disk by a factor of 2 and shifts it vertically down i unit. Inother words, the radius of our disk doubles and it’s center shifts by negative i.

|w + i| ≤ 2

15

7.) (a) Find the image of the triangle with vertices 0, 1, and i under the linear mapping f (z) and (b)represent the linear mapping with a sequence of plots.

f (z) = z + 2i

The image of the triangle is not magnified or rotated, simply translated by 2i. Thus, the image is

the triangle with vertices 2i, 1 + 2i, and 3i

.

Figure 20: Triangle with vertices 2i, 1 + 2i, 3i

9.) (a) Find the image of the triangle with vertices 0, 1, and i under the linear mapping f (z) and (b)represent the linear mapping with a sequence of plots.

f (z) = eiπ/4z

This mapping quite clearly rotates the triangle by π/4, so we just see what happens when we rotatethe vertices, written in polar form, by this. In particular, 0 maps to itself, 1 = e0 maps to eiπ/4 =√

2/2 (1 + i), and similarly i = eiπ 7−→ e5π/4 =√

22 (1− i). So, we have the following result

triangle with vertices 0 ,

√2

2+

√2

2i , −√

22

+

√2

2i

13.) Express the given linear mapping f (z) as a composition, rotation, magnification, and translationas in (6). Then describe the action of the linear mapping in words.

f (z) = 3iz + 4

f (z) = |3i|(

3i|3i| z

)+ 4 = 3(iz) + 4

So, the linear mapping rotates by arg i = π/2, magnifies by 3, and translates by 4. In other words,R(z) = eiπ/2z , T(z) = z + 4 , M(z) = 3z.

15.) Express the given linear mapping f (z) as a composition, rotation, magnification, and translationas in (6). Then describe the action of the linear mapping in words.

f (z) = −12

z + 1−√

3i

f (z) = | − 12|(−1/2| − 1/2| z

)+ 1−

√3i =

12(−1z) + 1−

√3i

So, this mapping rotates by arg(−1) = π, magnifies by 12 , and translates by 1−

√3i.

16

17.) Find a linear mapping that maps S onto S′ when S is the triangle with vertices 0, 1, and 1 + i. S′

is the triangle with vertices 2i, 3i, and −1 + 3i.

We know that our mapping must include T(z) = z + 2i, because 0 is invariant under rotation andmagnification. With that in mind, we need a mapping with R(z) = π/2z, as that is how we can send1 + 2i (which is 1 under translation) to the final 3i. Finally, we need no magnification as our last vertexmaps correctly as is. So, we have

f (z) = iz + 2i

19.) Find a linear mapping that maps S onto S′ when S is the imaginary axis and S′ is the line throughthe points i and 1 + 2i.

I think of this as two parts. One is that we need to rotate our vertical line by −π/4. Second, we need toshift our line upward to have it pass through 1 + 2i rather than 1 + i, so we do so by 1. Thus,

f (z) = e−iπ/4 z + 1 =

(√2

2−√

22

i

)z + i

21.) Find the two different linear mappings that map the square with vertices 0, 1, 1 + i, and i, ontothe square with vertices -1, 0, i, -1+i.

The obvious mapping is translation by one. The less obvious one is to simply rotate our points by π/4,which actually has the same effect in this scenerio. So, our mappings are

f (z) = z− 1 and g(z) = iz

23.) Consider the line segment parameterized by z(t) = z0(1− t) + z1t , 0 ≤ t ≤ 1(a) Find the parametrization of the image of the line segment under the translation T(z) = z + b , b 6=0. Describe the image in words.(b) Find the parametrization of the image of the line segment under the rotation R(z) = az , |a| = 1.Describe the image in words.(c) Find the parametrization of the line segment under the magnification M(z) = az , a > 0. Describethe image in words.

a.)

w(t) = T(z(t)) = z0(1− t) + z1t + b = z0(1− t) + z1t + b(1− t) + bt = (z0 + b)(1− t) + (z1 + b)t

This is the line segment between z0 + b and z1 + b (an analogous problem appears in our text).

b.)w = R(z(t)) = a(z0(1− t) + z1t) = (az0)(1− t) + (az1)t

is the line segment between az0 and az1 (again by (7) of 2.2 in our text).

c.)w = M(z(t)) = a(z0(1− t) + z1t) = (az0)(1− t) + (az1)t

This, virtually identical to the rotation previously described, is the line segment between az0 and az1.

17

2.4 Special Power Functions

1.) Find the image of the ray arg(z) = π/3 under the mapping w = z2. Represent the mapping bydrawing the set and its image.

Our set C can be rewritten as z = reπ/3i. From here, finding the image of the set under the providedmapping is simple.

w = (reπ/3i)2 = Re2π/3i =⇒ arg(w) =2π

3

where R = r2.

3.) Find the image of the line x = 3 under the mapping w = z2. Represent the mapping by drawingthe set and its image.

Using equation (3) from page 75 we let k = 3, thus our solution is the parabola

u = 9− v2

36−∞ < v < ∞

5.) Find the image of the line y = −1/4 under the mapping w = z2. Represent the mapping bydrawing the set and its image.

From the general formula found on page 76 of our text we know the mapping to be of the form u =v2

4k2 − k2. Here we have k = −1/4, therefore our image is

u =v2

4(−1/4)2 − (−1/4)2 =⇒ u = 4v2 − 116

, v ∈ R

9.) Find the image of the circular arc |z| = 1/2 , 0 ≤ arg(z) ≤ π under the mapping w = z2. Representthe mapping by drawing the set and its image.

Under the squaring function an arc has its modulus squared and both limits of its argument doubled,so we have that

|w| = 14

, 0 ≤ arg(w) ≤ 2π

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11.) Find the image of the triangle with vertices 0, 1, 1+i under the mapping w = z2. Represent themapping by drawing the set and its image.

If we consider the three line segments composing the triangle to be separately mapped by w, then wecan use know methods to find the image. In particular, the segment between 0 and 1 is mapped to itself(squaring the modulus and doubling the argument). The segment between 0 and 1 + i has argumentπ/4, therefore its image has argument π/2; similarly, the non-zero endpoint is squared, meaning thatthe segment maps to the segment from 0 to 2i. Finally, the segment from 1 to 1 + i lies on a vertical line,so, with k = 1, we have the segment mapping to u = 1− v2/4; however, for the bounds we look at ourtwo points, which shows us that 0 ≤ v ≤ 2.

Therefore, the image is the combination of the line segment from 0 to 1, the line segment from 0 to 2i, and theparabolic arc u = 1− v2

4 , 0 ≤ v ≤ 2.

.

Figure 21: The mapping f : C→ C described in (11)

15.) Find the image of the ray arg(z) = π/3 under f (z) = 2z2 + 1− i.

Let us consider this as a composition. In particular, f (z) = (h ◦ g)(z) where h(z) = 2z + 1 − i andg(z) = z2. Under squaring arg(z) = π/3 maps to arg(w) = 2π/3. The linear transformation thensimply translates and magnifies. The ray is invariant under magnification, but translation by 1 − imakes the image a ray from 1− i making angle 2π/3 with the line y = 1. Because zero cannot possiblyhave argument π/3 it is not in the original set, thus our image excludes 1− i. Finally, if we considerthe point 1

2 +√

32 i in our domain, then we see that it maps to

√3i− i. Therefore, our mapping is the ray

emanating from 1i, containing (√

3− 1)i, and excluding the point 1− i.

17.) Find the image of the line x = 2 under f (z) = iz2 − 3.

We use equation (3) from page 75, with k = 2. This shows that the squaring function maps to u = 4− v2

16 ,but this is then rotated by arg(i) = π/2 and translated by 3. Translation affects our imaginary part v,but rotation by that angle switches real and imaginary parts, making our image the parabola

v = 4− 116

(u + 3)2

19.) Find the image of the circular arc |z| = 2 , 0 ≤ arg(z) ≤ π/2 under f (z) = 1/4eiπ/4z2.

As with (15) we have a composition f (z) = (h ◦ g)(z), this time with h(z) = 14 eiπ/4z and g(z) = z2.

The squaring function maps our circular arc to another such arch with |w| = 4 , 0 ≤ arg(w) ≤ π. Thelinear function rotates by π/4 and translates by 0; therefore, the circular arc maps again onto the arc|w| = 4 , π/4 ≤ arg(w) ≤ 5π/4. Finally, magnification brings about the final result. The image is

the circular arc |w| = 1 , π/4 ≤ arg(w) ≤ 5π/4

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25.) Use (14) to find the value of z1/2 at z = −i.

By (14) we have

(−i)1/2 =√| − i|eiArg(−i)/2 = e(−π/4) i = cis(−π/4) =

√2/2−

√2/2i

27.) Use (14) to find the value of z1/3 at z = −1.

Similar to (25) we have

(−1)1/3 =√| − 1|eiArg(−1)/3 = eπi/3 =

12+

√3

2i

29.) Use (14) to find the value of z1/4 at z = −1 +√

3i.

Again,

(−1 +√

3i)1/4 =4√| − 1 +

√3i| eiArg(−1+

√3i)/4 =

4√

2eiπ/6 =4√

2

(√3

2+

12

i

)=

12

4√18 +12

4√

2i

31.) Find the image of the ray arg(z) = π/4 under the mapping w = z1/2. Represent the mapping bydrawing the set and its image.

If the argument of z is π/4, then z must be of the form reiπ/4, thus we have the following from ourformula

z1/2 = reiπ/8 = r(.925 + .383i)

Now, we see that this is equivalent to a linear mapping in which the complex number .925 + .383i ismagnified by r for varying r. The image is thus the ray r(.925 + .383i), or, in more simple terms, the rayarg(w) = π/8.

33.) Find the image of the positive imaginary axis under the mapping w = z1/2. Represent themapping by drawing the set and its image.

This set is equivalent to arg(z) = π/2. Therefore, the image is the ray

arg(z) =π

4

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35.) Find the image of the arc |z| = 9 , −π/2 ≤ arg(z) ≤ π under the mapping w = z1/2. Representthe mapping by drawing the set and its image.

Because the principal square root function roots the modulus and halves the argument of any number,we have that the image is simply the circular arc |w| = 3 , −π/4 ≤ arg(w) ≤ π/2.

39.) Find the image of the region shown in Figure 2.4.22 under the function w = z1/2. (Be careful nearthe negative real axis!)

This figure is boundary formed by the union of the ray arg(z) = π/2 and parabola x = 4− 1/16y2. Withthat being said, the principal square root function halves all arguments, so the ray maps to arg(z) = π/4(note that this is equivalent to u = v). By the inverse relationship of z2 and z1/2, as well as the formulafor the mapping of a line onto a parabola under the squaring function (in this case k = 2), we know theparabola to map to u = 2. Now, let us consider a point within our domain. For instance, 3 + 4i. This,upon squaring, yields −7 + 24i. Now, we may completely specify our region as the set bounded by thelines u = v and u = 2 containing the point 3 + 4i.

.

Figure 22: The union of the ray arg(z) = π/2 and parabola x = 4− 1/16y2.

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