constant ρ constant θ

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1 Answers to Problem Set Number 3 for MIT (Fall 999) Contents Roolfo R. Rosales Λ Boris Schlittgen y Zhaohui Zhang ẓ October 9, 999 Problems from the book by Saff an Snier. 2. Problem 04 in section Problem 07 in section Problem 6 in section Problem 09 in section Problem 2 in section Problem 08 in section Problem 4 in section Other problems Problem 3. in List of Figures.3. Level curves for the real an imaginary parts of Log(z) Stream lines for the flow given by w = z Λ MIT, Department of Mathematics, room 2-337, Cambrige, MA y MIT, Department of Mathematics, room 2-490, Cambrige, MA z MIT, Department of Mathematics, room 2-229, Cambrige, MA 0239.

2 8.04 MIT, Fall 999 (Rosales, Schlittgen an Zhang). Answers to Problem Set # 3. 2 Problems from the book by Saff an Snier.. Problem 04 in section 3.2. Let us write z = x + iy. Then Log(e z ) = Log(e x e iy ) = ln(e x ) + Arg(e x e iy )=x+i(y+2k 0 ß)where k 0 is an integer chosen such that ß <y+2k 0 ß»ß. Thus, we see that Log e z = z if an only if k 0 =0,which happens if an only if ß <y»ß..2 Problem 07 in section 3.2. Let us write z in polar form: z = re i. Then the polar form of the Cauchy-Riemann @r = an : Writing log z = ln(r)+i( +2kß), we have u = ln(r) an v = +2kß. r = =0 =0: So we see that the Cauchy-Riemann equations are inee satise, an thus log(z) is analytic. To n its erivative, consier approaching the limit raially, with angle = 0 xe. Then z log(z) = (e i 0 r) (ln(r)+i( +2kß)) = e i 0 (ln(r)+i( +2kß)) = r e i 0 r = z :.3 Problem 6 in section 3.2. Write z = ρe i, where ß <»ß. Then w = Log(z) = ln ρ + i. So the level curves for the real part of Log(z) are curves with ρ =constant, i.e.: circles centere at the origin. The level curves for the imaginary part of Log(z) are curves with constant argument, i.e. rays starting from the origin. See gure.3.. To see that the level curves are orthogonal at each point, we compute r(re(w)) r(im(w))=(=ρ)e ρ (=ρ)e =0: Alternatively: the level curves for the real an imaginary parts of Log(z) are the same as the coorinate lines for the (orthogonal an curvilinear) polar coorinate system.

3 8.04 MIT, Fall 999 (Rosales, Schlittgen an Zhang). Answers to Problem Set # 3. 3 constant ρ constant θ Figure.3.: Level curves for the real an imaginary parts of Log(z)..4 Problem 09 in section 3.3. Let z = cos(w) = 2 (eiw + e iw )= 2 ( + ), where = e iw. Multiplying both sies by 2 an then using the quaratic formula to solve for, we n: e iw = = z +(z 2 ) =2. Taking logs of both sies in this last equation an iviing by i, wen that w = cos (z) = ilog z +(z 2 ) =2 : (.4.) Now we ifferentiate this expression to n z cos (z) = i z+(z 2 ) =2 ψ z + (z 2 ) =2! = i : (.4.2) (z 2 =2 ) Remark.4. We coul now argue that, in the last expression in equation (.4.2) i(z 2 ) =2 =( z 2 ) =2 ; (.4.3) an thus write z cos (z) = : (.4.4) ( z 2 =2 )

4 8.04 MIT, Fall 999 (Rosales, Schlittgen an Zhang). Answers to Problem Set # 3. 4 We coul equally make the argument that an so conclue that i(z 2 ) =2 = ( z 2 ) =2 ; (.4.5) z cos (z) = ; (.4.6) ( z 2 =2 ) which is formula (3.3.) (page 92) in the book. However, now we seem to have arrive at two (apparently) ifferent answers equations (.4.4) an (.4.6) for the same question! So what is going on here? The answer to this conunrum lies in the multiple value nature of the functions involve, an it also teaches us that we have to be very careful when ealing with multiple value functions: Both (.4.3) an (.4.5) are true only in the multiple value sense, meaning that the set of values that the right han sies can take is equal to the set of values that the left han sies (respectively) can take. Since the values for square roots come in pairs with opposite signs, it is quite clear that these two equations are actually the same thing. Equations (.4.4) an (.4.6) are vali in precisely the same way. However, equation (.4.2) is vali in a somewhat stronger sense, as we explain next. Consier some arbitrary point z 0 6= ± in the complex plane. In some neighborhoo of it we can then ene (z 2 ) =2 as a single value function (i.e.: we pick a branch). Just so we o not get confuse in the argument that follows, we will give give a name to this branch of(z 2 ) =2 say: G(z). Thus G(z) is now some nice, single value, analytic function ene in some neighborhoo of z 0, which happens to have the property that G 2 = z 2. Let us also choose a branch for the logarithm, ene in a neighborhoo of z 0 + G(z 0 ). Then we can write (using equation (.4.)): w = arcos(z) = ilog(z + G) ; (.4.7) where LoG is the name of the branch for the log we just selecte an arcos is the name for the branch of cos that equation (.4.7) enes. The important point that istinguishes equation (.4.2) from equations (.4.4) an (.4.6) arises now, for we can substitute into it the various branches we have selecte to obtain a true equation, vali in a single value sense. That is: z arcos(z) = i G(z) : (.4.8)

5 8.04 MIT, Fall 999 (Rosales, Schlittgen an Zhang). Answers to Problem Set # 3. 5 This is easy to see, since all the operations in (.4.2) are consistent with the enitions of the various branches above. But we cannot o this with either (.4.4) or (.4.6), simply because these formulas involve yet another multiple value function (namely: ( z 2 ) =2 ) for which no branch has been selecte. So, short an sweet: once branches are ene for (.4.), equation (.4.2) can be use to calculate the erivative, without any longer having to worry about possible multiple values. Remark.4.2 We can also write (this is equivalent to (.4.), as can be seen using (.4.5)) w = cos (z) = ilog z + i( z 2 ) =2 : (.4.9) Then equation (.4.6) has the same relationship to this formula that (.4.2) has to (.4.). That is: once branches are ene for (.4.9), equation (.4.6) can be use to calculate the erivative, without any longer having to worry about possible multiple values..5 Problem 2 in section 3.3. Write z = tan w = i(e iw e iw )=(e iw + e iw ) an solve for w. First multiply both sies of the equation by ie iw (e iw + e iw ) to n iz(e 2iw +)=e 2iw. Hence: e 2iw =(+iz)=( iz). Now take the log of both sies an ivie by 2i to get: w = tan (z) = log +iz 2i iz We ifferentiate now this expression using the chain rule: z tan z = i 2 ψ i z i + z.6 Problem 08 in section 4.. = i 2 log i + z : i z (i z) (i + z)( ) (i z) 2! = +z 2 : The contour can be split naturally into two pieces, each one of which is easy to parametrize. Let us rst parametrize these two curves separately an then patch them together in a secon step. The rst curve can be parametrize as: z(t) =( 2+2i)+( 2i)t, for 0» t». The secon curve is escribe by: z(t) = exp(iß( t)), for 0» t». Patching these together, we have :z(t)= 8 >< >: ( 2+2i)+(2 4i)t; 0» t» 2 ; e 2iß( t) ; 2» t» :

6 8.04 MIT, Fall 999 (Rosales, Schlittgen an Zhang). Answers to Problem Set # 3. 6 We can now also parametrize the contour with the opposite orientation: :z(t)= 8 >< >: e 2iß(+t) ;» t» 2 ; ( 2+2i) (2 4i)t; 2» t»0:.7 Problem 4 in section 4.. Let I Now let t = f(s) so that t = f s s = I = Z c z (t) t Z c s f s z 2 (s) Z = c z (f(s)) s : s, since (by assumption) f=s > 0. Then: f Z s s = t=f() z Z t=f(c) t t = b z a t t :

7 8.04 MIT, Fall 999 (Rosales, Schlittgen an Zhang). Answers to Problem Set # Other problems. 2. Problem 3. in 999. Statement: Consier the complex potential for a flui given by w = Az 3, where A > 0 is a real number: (i) Fin the potential f, the stream-function ψ an the velocity el (u; v). (ii) Sketch the streamlines an the velocity el in the complex plane. (iii) Can you use this to n an incompressible, irrotational flow inawege (for some angle)? What is the angle of the wege you can o with this solution? Can you think ofawayof getting solutions for other angles? Solution: Let us set A = for convenience (you shoul be able to see that none of the arguments below will change in any essential way if the size of A is change). ffl i) Let = x + iy. Then w = z 3 =(x+iy) 3 =(x 3 3xy 2 )+i(3x 2 y y 3 ). Thus we can take: f(x; y) = Re(w) = x 3 3xy 2 (potential), ψ(x; y) = Im(w) = 3x 2 y y 3 (stream function), u = rf = 3(x 2 y 2 ) i 6xy j (u =(u; v) =velocity el) : Notice that u iv =3z 2 = erivative of the complex potential w =3z 2. This is always true; can you see why? ffl ii) See gure 2... ffl iii) We must impose the requirement that no flui can escape across the bounary, that is (on the bounary): u n = rf n =0,where n is the normal vector to the bounary. Since rf rψ = 0everywhere, the bounary must be a stream-line: ψ = constant. Then rψ is orthogonal to the bounary (on the bounary). Now clearly, y = 0 is one possible bounary, since ψ(x; 0) = 0. In orer to obtain a wege, we nee another bounary of the form y = ffx, for some constant ff. Since ψ(x; ffx) =ff(3 ff 2 )x 3, for ψ to be constant along the line (i.e. inepenent of x), we nee ff = 0; ± p 3. The case ff = 0 gives us the bounary we alreay have. The case ff = p 3 gives us a bounary at an

8 8.04 MIT, Fall 999 (Rosales, Schlittgen an Zhang). Answers to Problem Set # 3. 8 angle = ß=3 an the case ff = p 3 gives us a bounary at an angle = ß=3. Thus we can use the complex potential w = z 3 to escribe an incompressible, irrotational, two imensional flow in a wege of angle ß=3. See gure Im(z) Re(z) Figure 2..: Stream lines for the flow given by w = z 3. The flow is along the stream lines, in the irection inicate by the arrows. The magnitue of the flow spee is 3r 2, where r = jzj. In orer to obtain flow in weges of ifferent angles, consier w = z = ρ e i. Then the stream function is given by ψ = Im(w) =ρ sin( ) an two level curves where ψ = constant are clearly given by = 0 an = ß= (since ψ vanishes on these curves). THE END.