Mid Term-1 : Solutions to practice problems

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1 Mid Term- : Solutions to practice problems 0 October, 06. Is the function fz = e z x iy holomorphic at z = 0? Give proper justification. Here we are using the notation z = x + iy. Solution: Method-. Use Cauchy-Riemann equations. Setting f = u + iv then by a simply calculation we get ux, y = e x x cos y + y sin y and vz, y = e x x sin y y cos y. It is then easy to see that u x 0, 0 = v y 0, 0 = u y 0, 0 = v x 0, 0 = 0, and in particular the Cauchy-Riemann equaitons are satisfied at z = 0. Also, it is clear that all the first partials are continuous, and hence the function is in fact holomorphic at z = 0. Note that the Cauchy Riemann equations will actually not be satisfied anywhere other than z = 0. Method-. You could also use the limit definition. Since f0 = 0, we need to establish if the following limit exists. e h+ik h ik lim. h,k 0 h + ik Since e h+ik = e h cos k + i sin k, e h+ik e h < if h is small. Claim. h ik h+ik h + k. To see this note that by completing squares h ik = h4 + k 4 h + ik h + k = h + k h k h + k h + k, and hence the claim is proved. So combining with our estimate on e z if h is small enough then eh+ik h ik h h + ik + k h,k 0 0. So by the squeeze theorem the limit is zero. In particular, the derivative exists and f 0 = 0.

2 . Find all possible values of i, i and + 4i /. Solution: = e ln /+nπi and so i = e nπ e i ln / = e nπ cos ln + sin ln. i = e iiπ+nπi = e π+n. For + 4i /, one could do this using the same way as above. OR alternatively, let + 4i / = a + ib. The squaring we get a system of equations a b = ab = 4. To solve this use substitutions b = /a and a = x, then the system reduces to which reduces to a quadratic equation x 4 x = x x 4 = 0, which of course has solutions x = 4,. Since a is real, x has to be positive. this gives us x = 4 or a = ± and b = ±. So the roots are + i and i.. Find a closed form formula for + cos θ + cos θ + + cos nθ. Solution: By Euler formula e ikθ = cos kθ + i sin kθ and e ikθ = cos kθ i sin kθ. Solving for cosine, cos kθ = eikθ + e ikθ. So by the geometric series formula namely + r + r + r n = r n+ / r, n cos kθ = e in+θ e iθ + e in+θ e iθ. k=0 You could either leave it in this form or simplify using Euler formula again. 4. Find the branch points including infinity for fz = z z.

3 Show that fz defines a single valued holomorphic function on C \ [0, ]. Solution: Using the double polar coordinates, let z = r e θ = + r e iθ. Then fz, being a product of the multivalued functions z / and z /, has a branch given by fz = r / e i θ + θ. Of course, we need to now investigate where all is this continuous. Let θ, θ [0, π. Then only possible points of discontinuity of θ and θ, and hence of fz are on the interval [0,. There are two cases : continuity on [0, ] or,. Case- [0, ]. As z approaches [0, ] from above, θ 0 and θ π draw a picture!. Then fz r / e πi/ = r / + i. On the other hand as z approaches [0, ] from below, θ π and θ π and hence fz r / e π/+πi/ = r / e 4πi/ r / and hence f is discontinuous on this interval. i, Case-,. As z approaches [, ] from above, θ 0 and θ 0, and hence fz r /. On the other hand, As z approaches [, ] from below, θ π and θ π, and hence fz r / e πi/+4πi/ = r /, and hence fz is continuous on,. Hence the branch defined by equation is continuous, and hence holomorphic, on C \ [0, ]. 5. Find the values of z for which the following series are convergent.. n= np z n. n= n!zn. z n. n=0 +z Solution: n= np z n. The limiting ratio is lim n + p z n+ = z lim + n p z n p = z. n 0 n n

4 Hence the series converges for all z <. n= n!zn. Again looking at the limiting ratio - lim n n +!zn+ = z lim n!z n n. n So the only point where the series converges is z = 0. z n. n=0 +z The limiting ratio is z, + z which we need to be less than one. This gives us z < z +. Geometrically these are the points that are closer from 0 than. Points that are equidistant from 0 and are points for which Rez = /. Hence the set of points for which the series converges is Rez > /. 6. Let a Show that there exists a R > 0 such that pz = z n + a n z n + + a z + a 0. z n pz < z n, whenever z > R. Hint. Triangle inequality both sides! Solution: By the triangle inequality, pz z n + a z + a 0 = z n + a n z + + a 0 z n z n if z > R and R is sufficiently big. For the other inequality, we use the other side of the triangle inequality. That is, But a n z pz z n a n z + + a 0 z n. + + a 0 a n z n + + a 0 z z n <, if z > R when R is as above. So when z > R, pz z n. This proves the inequality that we need for any R > max R, R. 4

5 b For n, if is any circle that contains all the roots of pz in the interior, then show that dz = 0. pz Hint. What happens as r is made bigger and bigger? Solution: If r is big enough such that contains all roots of pz in the interior, then /pz is holomorphic outside. Hence by Cauchy s theorem, if R > r and denotes the circle of radius R with positive orientation, then pz dz = pz dz. But if R is sufficiently big, by the estimate above /pz < / z n, and hence by triangle inequality pz dz pz dz z n dz R n len = 4π R R n 0, since n. But since R is arbitrary, this shows that dz = 0. pz c More generally, if P z and Qz are polynomials such that degqz degp z +, show that P z dz = 0, Qz for any circle such that all the roots of Qz lie in the interior of. Solution: The same argument as above works. Using the estimate in part we can prove that for R sufficiently big, if n = deg Q and m = deg P, pz dz C R n m, for some constant C > 0 independent of R. Since n > m +, the right hand again goes to zero, and then you can use the same argument as before. d Is the above statement true if the assumption on the degrees is dropped? Either prove it, or provide a counter-example. Solution: The statement is not true without the degree assumption, as can be seen by for instance taking P z = and Qz = z. 7. Let pz = z α z α z α n, 5

6 such that all the roots lie in the upper half plane i.e Imα j > 0 for all j. a Show that p z pz = n j= z α j. Solution: This follows from the product rule, p z = z α z α n +z α z α z α n + +z α z α n, and dividing by pz. b Show that if Imz 0, then for all j, Im > 0. z α j Solution: Imz α j = Imz Imα j < 0, and taking inverse changes the sign of the imaginary part prove this!. c Use this to show that any root α of p z also satisfies Imα > 0. Solution: Proceed by contradiction, that is assume that Imα 0. Then in part, since α is a root of p z, the left hand side is zero when you plug in z = α. But from part the imaginary part of the right hand side is strictly positive, and hence cannot be zero. This is a contradiction. d Challenge. Dont waste too much time on this before the exam! A convex hull of α, α n is the set Sα,, α n = {t α + + t n α n t + + t n = }. More generally, show that a root α of p z lies in Sα,, α n. Solution: Keep trying...maybe after the exam! 6