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Joseph Heavner Honors Complex Analysis Assignment 3 April 6, 2015

2.5 Reciprocal Function 1.) Find the image of the circle |z| = 5 under the reciprocal function mapping. This is classical inversion in a unit circle, in which the argument is invariant but the modulus is transformed to be the reciprocal of that of the domain. In other words, here our modulus becomes 1/5, making our image 1 The Circle |w| = 5

3.) Find the image of the semicircle |z| = 3 , −π/4 ≤ arg(z) ≤ 3π/4 under the reciprocal function mapping. Again, our function inverts the moduli to be 1/3, but here we too have reflection about the x-axis plays a non-trivial role. In particular, this inverts (additive) the arguments to be −3π/4 ≤ arg(w) ≤ π/4. Thus, we have The Semicircle |w| =

1 3π π , − ≤ arg(w) ≤ 3 4 4

5.) Find the image of the annulus 1/3 ≤ |z| ≤ 2 under the reciprocal function mapping. As with (1) reflection about the x-axis has no bearing on the image (fully circular regions are invariant under rotation). However, it remains that the moduli are inverted (multiplicative), thus we have The Annulus

1 ≤ |w| ≤ 3 2

7.) Find the image of the ray arg(z) = π/4 under the reciprocal function mapping. Suppose we let z = reiθ , then our domain is given by r > 0 , θ = π/4. (Note that r cannot be negative in any case, and the case where r is zero includes all arguments and so is ignored here.) If we invert r then we have 1/r > 0, but because this includes the set of all possible r, we have an invariant (intuition of the fact will here suffice, though a proof is accessible). Now, we also know that our mapping inverts arguments, in particular θ = π/4 7−→ θ 0 = −π/4. In conclusion, we have The Ray arg(w) = −

π 4

9.) Find the image of the line y = 4 under the reciprocal function mapping. This problem fits a general form that states that a line y = k, under the reciprocal function, maps to the 1 circle |w + 21 k| = | 2k |. With that in mind, the image is clearly 1 1 |w + i| = 8 8 1

10.) Find the image of the line x =

1 6

under the reciprocal function mapping.

Similar to (9) we have a problem of the form: find the image of the line x = k under the reciprocal mapping, which has been shown to be answered by

|w −

1 1 |=| | 2k 2k

Thus, we arrive at our answer

|w − 3i | = 3

11.) Find the image of the circle |z + 1| = 1 under the reciprocal mapping. As mentioned in the Remarks on page 96 and can be seen from the results regarding have lines as 1 1 i | = | 2k | to the line y = k. So, we domains, this reciprocal mapping maps a circle of the form |z + 2k identify k = 1/2, implying that we map to the line y=

1 2

15.) Find the image of the set S under the mapping w = 1/z on C ∪ {∞}

. Figure 1: The Set S We already know how to map lines to circles via the reciprocal function. So, we simply identify these as the lines x = −2 and x = −1, which map to 1 1 1 1 |w + | = ; |w + | = 4 4 2 2 We now check a point within the domain such as, say, z = −3/2, which maps to w = −2/3. Thus, we have the set containing the point w = −2/3 and bounded by the circles 1 1 1 1 |w + | = ; |w + | = 4 4 2 2

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23.) Show that the image of the line x = k, x 6= 0, under the reciprocal map defined on the extended complex plane is the circle 1 1 |w − | = | | 2k 2k The vertical line x = k consists of all points z = k + iy such that x, y ∈ R where k 6= 0. So we may write w=

1 y k − 2 i = 2 k + iy k + y2 k + y2

Thus our real and imaginary parts of w = f (u, v) are u= But observe that v =

−yu k ,

k −y , v= 2 , y∈R k 2 + y2 k + y2

which implies

vk u And so, upon substituting into our initial expression for u, we have y=−

u=

(1)

k k2

+ ( −uvk )2

It is here advantageous to us to simplify and complete the square as follows u=

k

k2 + ( −uvk )2   −vk 2 2 k = u k +( ) u v2 k 2 u k 2 u2 + v2 k 2 u u2 + v2 − k   u 1 2 2 u − + − + v2 k 2k   1 2 u− + v2 2k

k = k2 u + uk =



0= 2 1 − = 2k 1 = 4k2

Finally, by recognition, we have 1 1 |w − k| = | k| 2 2 Note that we never made the restriction in (1) that u 6= 0. This is because here we are working in the extended complex plane.

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2.6 Limits and Continuity Note: In order to save time, the extreme pedantism and wordiness that is seen in the text has here been avoided. For instance, substitution to evaluate a limit may take one line rather than half a page. 1.) Use Theorem 2.6.1 and the properties of real limits on page 104 to evaluate lim (z2 − z)

z→2i

Let us simply substitute in z = 2i as follows:

(2i )2 − 2i = 4i2 + 2i = −4 + 2i

3.) Use Theorem 2.6.1 and the properties of real limits on page 104 to evaluate lim (|z|2 − i z¯ )

z →1− i

Again, let us try substitution

|1 − i |2 − i (1 − i ) = 2 − i + i 2 = 2 + 1 − i = 3 − i

5.) Use Theorem 2.6.1 and the properties of real limits on page 104 to evaluate lim (ez )

z→πi

Again, we need only substitute. eiπ = −1

9.) Use Theorem 2.6.2 and the basic limits (15) and (16) to compute lim (z2 − z)

z →2− i

We try substitution as follows

(2 − i )2 − 2 + i = 4 − 4i + i2 + i − 2 = 4 − 2 − 1 − 3i = 1 − 3i

11.) Use Theorem 2.6.2 and the basic limits (15) and (16) to compute 1 lim (z + ) z

z→eiπ/4

Yet again, we need only substitution for the following limit. In particular,

√ 1 lim (z + ) = eiπ/4 + e−iπ/4 = cos(π/4) + i sin(π/4) + cos(−π/4) + i sin(−π/4) = 2 iπ/4 z z→e

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13.) Use Theorem 2.6.2 and the basic limits (15) and (16) to compute lim

z→−i

z4 − 1 z+i

For the first time thus far, substitution will not suffice, for, obviously, it yields an indeterminate form (note that L’Hopital’s rule applies only to the reals). However, if we simply recognize that z4 − 1 = (z2 − 1)(z2 + 1) = (z − 1)(z + 1)(z + i )(z − i ), then we have that lim

z→−i

z4 − 1 = lim (z + 1)(z − 1)(z − i ) = (−i + 1)(−i − 1)(−2i ) = 4i z+i z→−i

19.) Consider the limit lim

 z 2 z

z →0

a.) What value foes the limit approach as z approaches 0 along the real axis? b.) What value does the limit approach as z approaches along the imaginary axis? c.) Do the answers from (a) and (b) imply that the limit exists? Explain. d.) What value does the limit approach as z approaches along the line y = x? e.) What can you say about the limit in general? a.) Along the x axis we have z = x + 0y = x, thus the limit is lim

x →0

 x 2

= lim 1 = 1

x

x →0

b.) Along the imaginary axis we have x = 0, implying that z = iy, thus our limit becomes  lim

y →0

yi −yi

2

= lim (−1)2 = 1 y →0

c.) No. The limit must be the same along any of the infinitely (uncountably) many paths in the complex plane for it to exist in general. In other words, two, three, or even a trillion paths, while perhaps suggesting that the limit may exist and equal some value c, do not actually demonstrate that the limit is c. This must be shown using more general methods. However, if the limit along one path does not equal the limit along another, then we may say that the limit does not exist. d.) If we approach along y = x then z = x + ix and  lim

x →0

x + xi x − xi

2



= lim

x →0

1+i 1−i

2

=

2i = −1 −2i

e.) As discussed in part (c), the limit does not exist because the limit is dependent on path.

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21.) Use (24) or (25), Theorem 2.2, and the basic limits (15) and (16) to compute lim

z→∞

z2 + iz − 2 (1 + 2i )z2

Let us attempt to simplify our expression, as its current form yields an indeterminate form and there is no natural idea of the top ”approaching infinity faster” than the bottom as there is in R (this is more formally shown by dividing by the largest power of the variable in question), at least not when we have an imaginary number in question. So, upon simplification we have

(1 − 2i )z2 + (2 + i )z − (2 − 4i ) 5z2 Now that all factors are properly separated with the denominator a real expression of a complex variable, we may use our familiar laws of rational functions as apply in R. In particular, it is found that lim f (z) =

z→∞

1 2 − i 5 5

23.) Use (24) or (25), Theorem 2.2, and the basic limits (15) and (16) to compute lim z →i

z2 − 1 z2 + 1

By substitution we have that, noting the nature of complex infinity (e.g. its lack of sign) z2 − 1 −2 → → ∞ 2 0 z +1

27.) Show that f is continuous at z0 f (z) = z2 − iz + 3 − 2i ; z0 = 2 − i We have that f (2 − i ) = 5 − 8i and similarly lim (z2 − iz + 3 − 2i ) = (2 − i )2 − i (2 − i ) + 3 − 2i = 4 − 4i − 1 − 2i − 1 + 3 − 2i = 5 − 8i

z →2− i

Therefore, by definition f is continuous at z0 . 28.) Show that f is continuous at z0 f ( z ) = z3 −

1 ; z0 = 3i z

Similar to (27) we know that f (z0 ) = − 80i 3 and the limit at that point is the same by virtue of substitution, i.e. 1 1 1 −80 lim f (z) = (3i )3 − = 27i3 − = −27i − = z → z0 3i 3i 3i 3i Thus f is continuous at z0 .

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29.) Show that f is continuous at z0 f (z) =

z3 ; z0 = i + 3z2 + z

z3

Substitution here works for computing the limit, and substitution is equivalent in computation to evaluating f (z0 ), thus the two are equal. Here is the explicit calculation: lim f (z) =

z → z0

i3 −i −i 1 = = = i 3 2 −i − 3 + i −3 3 i + 3( i ) + i

And so f is continuous at z0 .

31.) Show that f is continuous at z0 ( f (z) =

z3 −1 z −1

3

: |z| 6= 1 : |z| = 1

where z0 = 1 Clearly, f (1) = 3 and so we simply try the limit as follows (note the expression we chose to evaluate – this was done because it is equivalent to the value obtained along the other possible path) z3 − 1 (z − 1)(z2 + z + 1) = lim = lim (z2 + z + 1) = 3 ( z − 1) z →1 z − 1 z →1 z →1 lim

Therefore, it has been demonstrated that f is continuous at z0 = 1.

35.) Show that f is discontinuous at z0 f (z) =

z2 + 1 ; z0 = − i z+i

Evaluating f (−i ) yields

(−i )2 + 1 =∅ −i + 1 Thus, the criteria for continuity are not met and f is discontinuous at z0 .

37.) Show that f is discontinuous at z0 f (z) = Arg(z) ; z0 = −1 The problem here is not that f does not exist but that in fact the limit does not exist, thus not meeting the criteria for continuity, and so making f discontinuous at z0 . Let z be a point on the negative real axis, then Arg(z) = π but we have that there are points {zn } such that they are arbitrarily close to z but have their image under f has some nonzero imaginary part and so the argument becomes arbitrarily close to −π. Thus, we have that the limit does not exist. (An e − δ argument would make this more precise — Also, note that we must be careful with substitution in the case of poorly behaved functions like this)

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39.) Show that f is discontinuous at z0 ( f (z) =

z3 −1 z −1

3

: |z| 6= 1 : |z| = 1

where z0 = i If we consider the approach along the imaginary axis then we have lim z → i f (z) =

i3 − i =i i−1

However, trivially we know that along the unit circle we have that the limit evaluates to 3. But, 3 6= i, thus f is discontinuous at the point in question.

41.) Use Theorem 2.3 to determine the largest region in the complex plane on which the function f is continuous. z0 f (z) = Re(z) Im(z) Let z = x + iy, then f (z) = Re( x + iy) Im( x + iy) = xy But, we know that the function f (z) = xy is continuous for arbitrary z, thus f is continuous on all of C. 43.) Use Theorem 2.3 to determine the largest region in the complex plane on which the function f is continuous. z−1 f (z) = zz¯ − 4 Let z = x + iy, then f (z) =

x + iy − 1 x−1 y +i 2 = 2 ( x + iy)( x − iy) − 4 x + y2 − 4 x + y2 − 4

Therefore, u( x, y) =

x2

x−1 y ; v( x, y) = 2 2 +y −4 x + y2 − 4

These expressions are continuous on their domains. In particular, they are continuous for all ( x, y) : x2 + y2 6= 4, thus f is continuous for all z : |z| 6= 2.

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3.1 Differentiability and Analyticity 3.) Use definition 3.1 to find f 0 (z) where f (z) = iz3 − 7z2 The definition f 0 (z) = lim

∆z→0

f (z + ∆z) − f (z) ∆z

becomes i (z + ∆z)3 − 7(z + ∆z)2 − iz3 + 7z2 ∆z ∆z→0 i (z3 + 3z2 ∆z + 3z + ∆z3 ) − 7(z2 + 2z∆z + ∆z2 ) − iz3 + 7z2 = lim ∆z ∆z→0 ∆z(3iz2 + 3iz∆z + i (∆z)2 − 14z − 7∆z) = lim ∆z ∆z→0 = lim 3iz2 + 3iz∆z + i (∆z)2 − 14z − 7∆z

f 0 (z) = lim

∆z→0 2

= 3iz − 14z Thus, f 0 (z) = 3iz2 − 14z

5.) Use definition 3.1 to find f 0 (z) where f (z) = z − f 0 (z) = lim

1 z

z + ∆z −

1 z+∆z

∆z z∆z + z2 + 1 = lim ∆z→0 z ( ∆z + z ) ∆z→0

z2 + 1 z2 1 = 1+ 2 z

=

Thus, f 0 (z) = 1 +

9

1 z2

−z+

1 z

9.) Use alternate definition (12) to find f 0 (z) where f (z) = z4 − z2 The definition mentioned is restated below for completeness: f 0 (z0 ) = lim

z → z0

f ( z ) − f ( z0 ) z − z0

With that, we identify the pieces and substitute, then solve as follows on the next page: z4 − z2 − z40 + z20 z → z0 z − z0 z4 − z40 z2 − z2 = lim + 0 z → z0 z − z 0 z − z0 2 2 (z + z0 )(z − z0 )(z + z0 ) (z + z0 )(z0 − z) − = lim z → z0 z − z0 z0 − z

f 0 (z0 ) = lim

= lim z30 + z20 z + z0 z2 + z3 − z − z0 z → z0

= z30 + z20 (z0 ) + z0 (z0 )2 + (z0 )3 − (z0 ) − z0 = 4z30 − 2z0 Note, however, that this is a general point, so we may replace z0 , and so we arrive at our solution f 0 (z) = 4z3 − 2z

19.) The function f (z) = |z|2 is continuous at the origin. (a) Show that f is differentiable at the origin. (b) Show that f is not differentiable at any point z 6= 0. (a) If we take |z| to be zz then we can easily demonstrate this

(z + ∆z)(z + ∆z) − zz ∆z ∆z→0 ∆z = lim z + ∆z + z ∆z→0 ∆z

f 0 (z) = lim

At this point we may stop, noting that if z = 0, then all terms vanish and so the function is differentiable with derivative equal to zero. (b) Given the final limit of (a) we see that the term ∆z ∆z is problematic. Noting that the expression can be rewritten as an exponential. If this is done we note that the angle φ could be anything, regardless of how arbitrarily close ∆z is to zero. Thus, this portion of the limit justifies the lack of differentiability of f (z) at any point z0 ∈ C : z0 6= 0.

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21.) Show that f (z) = z is nowhere differentiable. We consider the definition of the derivative, here letting ∆z = w. lim

w →0

x − yi z+w−z w = lim = lim w w →0 w x,y→0 x + yi

This is clearly non-existent, for along the line x = 0 it evaluates to −1, whereas it is 1 along y = 0.

23.) Use L’Hopital’s rule to compute lim z →i

z7 + i z14 + 1

If we rewrite the numerator as P(z) and the denominator as Q(z) then we see that f (z) must be analytic at z = i, because it is the quotient of analytic functions (polynomials). Thus, we differentiate both P(z) and Q(z). lim z →i

i 7z6 7( i )6 −7 = = = 13 13 14i 2 14z 14(i )

27.) Determine the points at which f (z) =

iz2 − 2z is not analytic. 3z + 1 − i

We can take the derivative of the function to begin. To save paper, the work here will be somewhat ignored. Regardless, the conclusion is that f 0 (z) =

(−2 + 2i )(2 + 2i )z + 3iz2 ((1 − i ) + 3z)2

Now, we see if this is undefined anywhere, and indeed ((1 − i ) + 3z)2 = 0 at z = − 31 + 13 i, thus the function cannot be analytic there. But, because the function’s derivative exists at all other points z ∈ C we have the function is analytic everywhere except at z = − 13 + 13 i.

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3.2 Cauchy–Riemann Equations 1.) Given that f is analytic, show that the Cauchy-Riemann equations are satisfied for f (z) = z3 . By expanding and grouping we rewrite f (z) = z3 as f (z) = ( x3 − 3xy2 ) + i (3x2 y − y3 ) we take the partial derivatives: ∂u ∂v ∂v ∂u = 3x2 − 3y2 ; = −6xy ; = 6xy ; = 3x2 − 3y2 ∂x ∂y ∂x ∂y Thus the equations

∂u ∂v ∂u ∂v = and =− are satisfied. ∂x ∂y ∂y ∂x

3.) Show that f (z) = Re(z) is nowhere analytic. If we let z = x + iy then f (z) = x, thus u( x, y) = x and v( x, y) = 0. Therefore, ∂u ∂u ∂v ∂v =1; =0; =0; =0 ∂x ∂y ∂x ∂y Because the Cauchy-Riemann equations are not satisfied (in particular ∂u/∂x 6= ∂v/∂y) for no points in the complex plane. So, we may conclude that f (z) is nowhere analytic. 7.) Show that f (z) = x2 + y2 is nowhere analytic. u( x, y) = x2 + y2 and v( x, y) = 0. Computing the partials of f : ∂u ∂u ∂v ∂v = 2x ; = 2y ; =0; =0 ∂x ∂y ∂x ∂y Therefore the Cauchy-Riemann equations are only satisfied at the point z = 0 in the complex plane. However, there does not exist a domain R with z = 0 in R in which there all points in R satisfy the Cauchy-Riemann equations. Therefore, f is nowhere analytic.

8.) Show that f (z) = u( x, y) =

x x 2 + y2

x y +i 2 is nowhere analytic. x 2 + y2 x + y2

and v( x, y) =

y , x 2 + y2

which means that

∂u y2 − x 2 ∂u −2xy ∂v −2yx ∂v x 2 − y2 = 2 ; = ; = ; = ∂x ( x + y2 )2 ∂y ( x2 + y2 )2 ∂x ( x2 + y2 )2 ∂y ( x 2 + y2 )2 We observe that ∂u/∂x = ∂v/∂y if and only if y = ± x where x 6= 0. However, ∂u/∂y = −∂v/∂x if and only if x = 0 where y 6= 0 or y = 0 where x 6= 0. Clearly, these two conditions are mutually exclusive, ensuring that the Cauchy-Riemann equations are not satisfied, implying that f is nowhere analytic.

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9.) (a) Use Theorem 3.2.2 to show that f (z) = e− x cos y − ie− x sin y is analytic in an appropriate domain, and (b) find the derivative of f in said domain using (9) or (11). (a) u( x, y) = e− x cos y and v( x, y) = −e− x sin y, therefore ∂u ∂u ∂v ∂v = −e− x cos y ; = −e− x sin y ; = e− x sin y ; = −e− x cos y ∂x ∂y ∂x ∂y Clearly the Cauchy-Riemann equations are satisfied for all z ∈ C, making f analytic in all of the complex plane. (b) It is here advantageous to use ”Cartesian” (perhaps the term ”Argand” or rectangular is better here) coordinates, so we compute f 0 (z) by using (9), which states that f 0 (z) =

∂v ∂v ∂u ∂u +i = −i ∂x ∂x ∂y ∂y

Via substitution (note that here the book appears to have the incorrect answer) f 0 (z) = −e− x cos y + ie− x sin y

2

2

2

2

11.) (a) Use Theorem 3.2.2 to show that f (z) = e x −y cos(2xy) + ie x −y sin(2xy) is analytic in an appropriate domain, and (b) find the derivative of f in said domain using (9) or (11). (a) u( x, y) = e x

2 − y2

cos(2xy) and v( x, y) = 2x

2 − y2

sin(2xy), so

2 2 2 2 ∂u ∂u = 2e x −y ( x cos(2xy) − y sin(2xy)) ; = −2e x −y (y cos(2xy) + x sin(2xy)) ∂x ∂y 2 2 2 2 ∂v ∂v = 2e x −y (y cos(2xy) + x sin(2xy)) ; = 2e x −y ( x cos(2xy) − y sin(2xy)) ∂x ∂y

It is now clear that f is an entire function, i.e. it is analytic throughout all of the complex plane. (b) By formula (9) f 0 (z) = f (z) and in particular f 0 (z) = e x cos(y) + ie x sin(y)

17.) Find real constants a, b such that f (z) = 3x − y + 5 + i ( ax + by − 3) is analytic. u( x, y) = 3x − y + 5 and v( x, y) = ax + by − 3, so ∂u ∂u ∂v ∂v =3; = −1 ; =a; =b ∂x ∂y ∂x ∂y It is therefore clear that a = 1, b = 3

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19.) Show that f (z) = x2 + y2 + 2ixy is not analytic at any point but is differentiable along the x-axis, and (b) use (9) or (11) to compute the derivative along the axis. (a) u( x, y) = x2 + y2 and v( x, y) = 2xy, thus ∂u ∂u ∂v ∂v = 2x ; = 2y ; = 2y ; = 2x ∂x ∂y ∂x ∂y The first Cauchy-Riemann equation is universally satisfied, whereas the second is satisfied if and only if y = 0. Thus, while f is differentiable on the x-axis, it is not analytic, for there exists no neighborhood in which the Cauchy-Riemann equations are satisfied. (b) Via (9), f 0 (z) = 2x + i2y, which, along the x-axis where y = 0 means that f 0 (z) = 2x

21.) (a) Show that f (z) = x3 + 3xy2 − x + i (y3 + 3x2 y − y) is not analytic at any point but is differentiable along the coordinate axes, and (b) use (9) or (11) to compute the derivative along the axes. (a) u( x, y) = x3 + 3xy2 − x and v( x, y) = y3 + 3x2 y − y, meaning that ∂u ∂u ∂v ∂v = 3x2 + 3y2 − 1 ; = 6xy ; = 6xy ; = 3y2 + 3x2 − 1 ∂x ∂y ∂x ∂y The first Cauchy-Riemann equation is satisfied for all z ∈ C, whereas the second is satisfied only if x = 0, y = 0, or x = y = 0. Similar to (19), there exists no neighborhood in which the Cauchy-Riemann equations are satisfied, therefore f is not analytic, but the Cauchy-Riemann equations are satisfied on the coordinate axes, making them differentiable there. (b) In particular, f 0 (z) = 3x2 − 1 along the x −axis f 0 (z) = 3y2 − 1 along the y−axis

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