MAT389 Fall 2016, Problem Set 6

Transcription

1 MAT389 Fall 016, Problem Set 6 Trigonometric and hperbolic fnctions 6.1 Show that e iz = cos z + i sin z for eer comple nmber z. Hint: start from the right-hand side and work or wa towards the left-hand side. Simpl sing the definition of the sine and cosine fnctions in terms of the eponential ields cos z + i sin z = eiz + e iz + i eiz e iz i = e iz. 6. Show that sin z = sin cosh + i cos sinh. Dedce from it the formla sin z = sin + sinh Note: I otlined the proof of the first formla in class. Here I am asking o to finish the calclation. First of all, notice that mch as in the last problem. Then, cosh z + sinh z = ez + e z + ez e z = e z, sin z = eiz e iz = ei(+i) e i(+i) = ei e e i e i i i (cos + i sin )(cosh sinh ) (cos i sin )(cosh + sinh ) = i = sin cosh + i cos sinh Now that we hae separated sin z in its real and imaginar parts, its modls is eas to compte: sin z = sin cosh + cos sinh = sin (1 + sinh ) + (1 sin ) sinh = sin + sinh where we hae sed the fndamental trigonometric and hperbolic identities sin z + cos z = 1, cosh z sinh z = 1 I mentioned in class that there is an alternatie deriation of the formla sin z = sin cosh + i cos sinh sing the identit sin(z 1 ± z ) = sin z 1 cos z ± cos z 1 sin z

2 and the relationship between trigonometric and hperbolic fnctions, sinh z = i sin(iz), cosh z = cos(iz), Here it is: sin z = sin( + i) = sin z cos(i) + cos sin(i) = sin cosh + i cos sinh. 6.3 Dedce from the formla sin z = sin + sinh that sin z sin. Since sinh 0, sin z = sin + sinh sin. Taking the (positie!) sqare root, sin z sin sin. 6.4 Use Problem 5.11 to check that sin z = sin z and cos z = cos z. Note: as in Problem 5.11, o can interpret this geometricall. ( e sin z = iz e iz ) = eiz e iz i i = e i z e i z i = ei z e i z i = sin z ( e cos z = iz + e iz ) = eiz + e iz = e i z + e i z = ei z + e i z = cos z Here is an illstration of the geometric significance of the statement sin z = sin z: points in the red (resp., green) regions in the z-plane get mapped to the red (resp., green) regions in the w-plane. The pictre can also be interpreted as an illstration of the identit sin( z) = sin z. z sin z π/ π/ w = sin z 1 1 z z sin( z) sin z

3 6.5 Use the identities sin(z 1 ± z ) = sin z 1 cos z ± cos z 1 sin z, cos(z 1 ± z ) = cos z 1 cos z sin z 1 sin z, and the relationship between trigonometric and hperbolic fnctions, sinh z = i sin(iz), cosh z = cos(iz) to dedce epressions for sinh(z 1 ± z ) and cosh(z 1 ± z ). For the hperbolic sine, we hae sinh(z 1 ± z ) = i sin [ i(z 1 ± z ) ] = i [ sin(iz 1 ) cos(iz ) ± cos(iz 1 ) sin(iz ) ] = i [ i sinh z 1 cosh z ± i cosh z 1 sinh z ] For the hperbolic cosine, = sinh z 1 cosh z ± cosh z 1 sinh z. cosh(z 1 ± z ) = cos [ i(z 1 ± z ) ] = cos(iz 1 ) cos(iz ) sin(iz 1 ) sin(iz ) = cosh z 1 cosh z (i sinh z 1 )(i sinh z ) = cosh z 1 cosh z ± sinh z 1 sinh z. 6.6 The fnctions sin z and cos z are π-periodic: sin(z + π) = sin z, cos(z + π) = cos z Moreoer, shifting their argment b π changes their sign: sin(z + π) = sin z, cos(z + π) = cos z What do these identities impl for sinh z and cosh z? The fnctions sinh z and cosh z are πi-periodic: sinh(z + πi) = i sin[i(z + πi)] = i sin(iz π) = i sin(iz) = sinh z cosh(z + πi) = cos[i(z + πi)] = cos(iz π) = cos(iz) = cosh z Shifting the argments of these fnction b πi reslts in a sign change: sinh(z + πi) = i sin[i(z + πi)] = i sin(iz π) = i sin(iz) = sinh z cosh(z + πi) = cos[i(z + πi)] = cos(iz π) = cos(iz) = cosh z These periodicit (and anti-periodicit) conditions can be easil read off in the following pictres.

4 3π/ π/ π/ 3π/ w = sin z 1 1 π π 0 π π w = cos z 1 1 3πi/ πi/ w = sinh z i πi/ i 3πi/

5 πi πi 0 w = cosh z 1 1 πi πi 6.7 Let f(z) = tanh z = sinh z/ cosh z. Find the domain of holomorphicit of f(z), as well as all of its zeroes. Since f(z) is a qotient of entire fnctions, it is holomorphic awa from the zeroes of the denominator, which occr at z = (k + 1)πi/ for k Z. The zeroes of f(z) are those of the nmerator. We saw in class that the soltions to the eqation sinh z = 0 are those comple nmbers of the form z = πki for k Z. πi 3πi/ πi/ 0 πi/ πi 3πi/

6 6.8 Find all roots of the eqations (i) cosh z = 1/, (ii) sinh z = i, (iii) cosh z =, (i) cosh z = i. Hint: notice that sinh z and cosh z are linear combinations of e z and e z. Letting w = e z (so e z = w 1 ) trns the eqations aboe into qadratic eqations for w. (i) After the change of ariables w = e z, the eqation cosh z = ez + e z = 1 e z + e z = 1 becomes w w + 1 = 0, whose soltions are w = (1 ± i 3)/ = e ±iπ/3. For the pls sign, we obtain ( π ) z = i 3 + πk, k Z. For the mins sign, it is z = i ( π ) 3 + πk, k Z. 7πi/3 5πi πi/3 πi 5πi/3 7πi (ii) Performing the sbstittion w = e z as in (i), we obtain the eqation w iw 1 = 0, with has a niqe soltion: w = i. Ths, ( π ) z = i + kπ.

7 9πi/ 5πi/ πi/ 3πi/ 7πi/ (iii) We take w = e z et again to obtain w + 4w + 1 = 0. The soltions to the latter are w = ± 3 = ( 3)e iπ (since > 3) and hence z = Log( 3) + (k + 1)πi, k Z. log( 3) + 3πi log( + 3) + 3πi log( 3) + πi log( + 3) + πi log( 3) πi log( + 3) πi log( 3) 3πi log( + 3) 3πi (i) With w = e z, we get w iw + 1 = 0, with soltions w = i(1 ± { (1 + )e iπ/ ) = ( 1)e iπ/ (carefl: > 1). Ths, Log(1 + ( π ) ) + i + kπ, k Z z = Log( 1) + i ( π ) + kπ, k Z

8 log( 1) + πi log(1 + ) + 3πi/ log(1 + ) + πi/ log( 1) log( 1) πi log(1 + ) πi/ log(1 + ) 3πi/ The transformation w = sin z The net two problems work ot the images of ertical and horizontal lines in the z-plane nder the transformation w = sin z which I otlined in class. Once o know how those work, o can find the image of an (possibl infinite) rectangle in the z-plane with sides parallel to the real and imaginar aes for eample, that in Problem 6.10 below not onl nder w = sin z, bt also nder other transformations like w = cos z, w = sinh z and w = cosh z (after all, these three are related to sin z b translations and rotations). Recall that, nder the transformation w = sin z, the inerse image of the real ais = 0 consists of the real ais = 0 and the lines = π/+kπ, where k Z; and the inerse image of the imaginar ais = 0 consists of the lines = kπ, where k Z. 6.9 Show that the image of the line gien b = c 1 for some fied 0 < c 1 < π/ nder the transformation w = sin z is a branch of the hperbola gien b the eqation ( ) ( ) = 1. sin c 1 cos c 1 Pa close attention to the orientation: if o moe along the line = c 1 from to + in, how do o traerse that hperbola branch? Show also that, if π/ < c 1 < 0, the image of = c 1 is the other branch of the same hperbola. Hint: remember that sin z = sin cosh + i cos sinh. See Figre 1. Let z(t) = c 1 + it, t R, be a parametrization of the gien line. Then, w(t) = sin z(t) = sin(c 1 + it) = sin c 1 cosh t + i cos c 1 sinh t

9 π/ π/ w = sin z 1 1 Figre 1: Problems 6.9 and 6.10 Notice that (t) Re w(t) = = cosh t, sin c 1 sin c 1 (t) Im z(t) = = sinh t. cos c 1 cos c 1 The fndamental hperbolic identit cosh t sinh t = 1 then implies the eqation ( (t) sin c 1 ) ( ) (t) = 1. cos c 1 If 0 < c 1 < π/, then sin c 1 > 0. Since cosh t is alwas positie too, we hae (t) = sin c 1 cosh t > 0. That is, the image of the line = c 1 is contained in the branch sitting in the half-plane > 0. On the other hand, π/ < c 1 < 0 implies sin c 1 < 0 and (t) = sin c 1 cosh t < 0 landing then in the other branch of the same hperbola. That the images of these lines are the entiret of the branches aboe follows from the fact that, as t moes from to +, so does sinh t; hence (t) = cos c 1 sinh t sweeps the whole real line, from to + (since cos c 1 > 0 for π/ < c 1 < π/) Show that the image of the line segment gien b π π, and = c for some fied c > 0 nder the transformation w = sin z is the top half of the ellipse with eqation ( ) ( ) + = 1. cosh c sinh c What happens if c < 0? And if is otside the interal ( π/, π/)?

10 See Figre 1. Let z(t) = t + ic, t [ π, π], be a parametrization of the line segment gien. Then w(t) = sin z(t) = sin(t + ic ) = sin t cosh c + i cos t sinh c is a parametrization of its image nder the transformation w = sin z. Notice that (t) Re w(t) = = sin t, cosh c cosh c (t) Im z(t) = = cos t. sinh c sinh c The fndamental trigonometric identit sin t + cos t = 1 then implies the eqation ( (t) cosh c ) ( ) (t) + = 1. sinh c Ths the image of the segment is contained in the ellipse described b the eqation aboe. To see that the image is the top half of the ellipse, notice that c > 0 implies sinh c > 0 and (t) = cos t sinh c > 0 for t in the interal ( π/, π/). On the other hand, if c < 0 then sinh c < 0 and (t) = cos t sinh c < 0. As t moes otside of the interal [ π/, π/], the π-periodicit (and the π-antiperiodicit) of w = sin z implies that we keep circling arond the ellipse Find a conformal transformation w = f(z) that takes the semi-infinite strip 0 < < π/, > 0 onto the pper half-plane H. Hint: start b considering the image of the domain gien nder Z = sin z. Do o know of a conformal transformation w = g(z) that takes the reslting domain to the entire pper half-plane? Note: b a conformal transformation I simpl mean a trasformation that is conformal at all points of the set of interest. From the last three problems we dedce that the image of the gien semi-infinite strip in the z-plane nder Z = sin z is the (open) first qadrant in the Z-plane, and that the transformation is conformal at eer point in it since it is holomorphic and its deriatie anishes nowhere on it. Now compose with the transformation w = Z, which takes the first qadrant to the whole pper half-plane. The latter is also conformal eerwhere on the first qadrant, and hence the composed fnction w = sin z is conformal on the strip.

11 Y Z = sin z X w = sin z w = Z The logarithm 6.1 For each of the following comple nmbers z find the following: (1) all logarithms of z (i.e., log z as a mltialed fnction); () the principal logarithm Log z of z. (i) z = ei, (ii) z = 1 i, (iii) z = 1 + i 3. (i) For the mltialed logarithm, we hae ( π ) log( ei) = Log e i + πk The principal logarithm is the aboe ale for k = 0: (ii) Writing 1 i = e iπ/4, we hae log(1 i) = Log + i Log( ei) = 1 πi ( π ) = 1 i + πk. ( π 4 + πk ) = 1 Log + i ( π 4 + πk ). Once again, the principal ale of the logarithm is the one for k = 0: Log(1 i) = 1 Log πi 4

12 (iii) We hae 1 + i 3 = e πi/3. Hence, ( log 1 + i ) ( ) π 3 = Log + i 3 + πk and ( Log 1 + i ) 3 = Log + πi Find the image nder w = log (π/) z of the wedge {z C 0 < Arg z < π/4}. For z C, log (π/) z = Log z + i arg (π/) z, where π/ arg (π/) z < π/ + π = 5π/. The comple nmbers with principal argment between 0 and π/4 are those whose arg (π/) lie between π (= 0 + π) and 9π/4 (= π/4 + π). It follows that the image of the wedge nder log (π/) is the horizontal strip {w C π < Im w < 9π/4}. w = log (π/) z 9πi/4 πi 6.14 Determine the location of the branch ct of the fnction as well as the discontinit of f(z) across it. w = f(z) = Log(1 iz), Hint: break p the transformation aboe as Z = 1 iz followed b w = Log Z, as we did in class. Since Z = 1 iz is an affine linear transformation, we can inert it to get z = iz i. The branch point of w = Log Z is located at Z = 0, which gies z = i. The branch ct of w = Log Z lies along the negatie real ais in the Z-plane, and that corresponds to those points in the imaginar ais of the z-plane that are to the soth of i.

13 We can also write some formlas to see the location of the branch ct. In the Z-plane, it is located on the points Z = re iπ = r with r R >0 (remember that Log = log ( π) ). Using the formlas for the affine linear transformations aboe, we find that the branch bt in the z-plane lies on the points z = i ir with r R >0. Y i Z = 1 iz z = iz i πi 0 X πi 6.15 Consider the transformation w = f(z) = Log 1 + z 1 z (i) Determine the inerse image nder the transformation Z = 1 + z 1 z of the branch ct of w = Log Z. Use this information to determine where f(z) is holomorphic. (ii) Find a pair (α 1, α ) for which the following two conditions hold: the fnction g (α1,α )(z) = log (α1 )(1 + z) log (α )(1 z); is holomorphic eactl where f(z) is. f and g (α1,α ) coincide along the line segment joining the two branch points. These two conditions impl, in fact, that f and g coincide eerwhere. This follows from the Principle of Analtic Contination, which we will talk abot hopefll in Chapter 5. Hint: once o hae identified the line segment joining the two branch points, o can eplicitl calclate the ales of f and g (α1,α ) in terms of some real logarithms. Note: the fnctions f and g (α1,α ) are branches of the mltialed fnction argtanh z. (i) The inerse of the gien transformation is z = Z 1 Z + 1

14 It is of the form Z = az + b, a, b, c, d R, ad bc > 0 cz + d Remember that these presere the pper half-plane and hence the real line too. Hence the inerse image b Z = (1 + z)/(1 z) of the branch ct of w = Log Z mst be contained in the real line of the z-plane. It is clear that the Z = 0 corresponds to z = 1. On the other hand, the negatie real ais in the Z-plane can be thoght as the part of the real ais in the Z-plane between Z = 0 and Z =. Bt Z = corresponds to z = 1. Ths the branch ct in the z-plane can be one of the two following things: the line segment between z = 1 and z = 1 along the real ais, or the line segment the same two points going the other wa arond throgh z =. We see that it is, in fact, the second of these possibilities b looking at another point: Z = 1 comes from z =. So it looks like w = Log 1 + z 1 z has two branch points and two branch cts! Or mabe we shold jst iew it as one branch ct connecting two branch points. In an case, that second branch point comes from Z = a fact that we cold interpret as saing that w = Log Z has a branch point at (it s jst that we sall onl draw the finite comple plane). Y πi 1 1 πi Z = 1 + z 1 z z = Z 1 Z + 1 πi 0 X (ii) In order to hae the branch ct of g (α1,α )(z) emanating from z = 1 going in the negatie real direction (as it does for w = f(z)), we can take α 1 = π. Similarl, we want the branch ct starting at z = 1 to lie along the positie real ais; we can take also α = π (the argment of the logarithm is 1 z, so the branch ct gets rotated as in Problem 6.14). Let s check that the second of the conditions in the statement holds for this choice. Between the two branch points, z = 1 and z = 1, both f(z) and g ( π, π) (z) are holomorphic. As z moes from z = 1 to z = 1, Z = (1 z)/(1 + z) moes along the real ais from Z = 0 to Z =. Bt the principal branch of the logarithm takes real ales on the positie real ais, and so we hae f() = Log 1 +, 1 < < 1. 1

15 where Log is jst the real logarithm. On the other hand, both 1 + z and 1 z are positie real for z on that segment, and so Log(1 + z) and Log(1 z) are also real on it, and which coincides with f() aboe. g ( π, π) () = Log(1 + ) Log(1 ), 1 < < Find the image of the (open) nit disc D nder the transformation w = f(z) in the last problem. To compte the image of the nit circle z = 1 nder the transformation Z = (1 + z)/(1 z), it is enogh to take three points on said circle. We can take, for eample, z = 1, z = 1 and z = i. The first goes to Z = 0, the second to Z =, and the third to (1 + i)/(1 i) = i. Ths, the image of z = 1 in the Z-plane is the imaginar ais, and the nit disc z < 1 goes to the right-hand plane Re Z > 0 (z = 0 is sent to Z = 1). On the other hand, w = Log Z maps the right half-plane in the Z-plane onto the strip π/ < Im w < π/. Hence the latter is also the image of the nit disc in the z-plane nder the composed transformation. 1 1 πi πi w = Log 1 + z 1 z πi/ Z = 1 + z 1 z Y πi/ w = Log Z πi 0 X