Solutions for Math 411 Assignment #10 1

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1 Solutions for Math 4 Assignment # AA. Compute the following integrals: a) + sin θ dθ cos x b) + x dx 4 Solution of a). Let z = e iθ. By the substitution = z + z ), sin θ = i z z ) and dθ = iz dz and Residue Theorem, we obtain + sin θ z = dθ = = Indeed, z = z + z dz 4i + z z ) z z + 4iz )z dz = i Res z + 4iz )z + i Res 3 )i z 3 )i)z )i)z = i z + i + 4iz z )i)z =. π + sin θ dθ = cosπ θ) π + sinπ θ) dθ π = π + sin θ dθ = + sin θ dθ Or simply, dθ =. + sin θ + sin θ dθ = d sin θ + sin θ = ln + sin θ) =. 3 )i xichen/math48f/hwasol.pdf

2 Solution of b). Since cos x/ + x 4 ) is even, cos x + x dx = cos x 4 + x dx = 4 Re e ix ) + x dx 4 Let us consider the complex integral ΓR e iz + z 4 dz = R R e iz + z dz + 4 e iz + z 4 dz where = { z = R, Imz) } is the semicircle oriented counterclockwise. For z = x + yi, y e iz = e y. Therefore, It follows that e CR iz + z dz 4 πr R 4 e CR iz dz =. + z4 e ΓR iz + z dz = e iz 4 + z dz. 4 On the other hand, by Residue Theorem, ΓR e iz + z 4 dz = i p <R,Imp)> Res p e iz + z 4 The polynomial + z 4 has four distinct roots + z 4 = z c )z c )z c )z c 3 ) for c n = exp πi 4 + nπi 4 ) = + i)in. So at each c = c n, e iz / + z 4 ) has a pole of order one with residue ) e iz e iz z c) Res c = = e ic z c) + z 4 z c + z 4 z c + z 4 ) because c 4 =. = eic 4c = 3 ceic 4 )

3 Clearly, Imc ) > and Imc ) >. Therefore, for R >, e ΓR iz c e ic dz = i + c e ic ) + z4 4 4 c e c = i + ic e c ) = π 4 4 c e c + ie c ) = π ) ) )) + i exp + i + i exp i ) )) ) = cos + sin exp. Therefore, e iz + z 4 dz = cos x + x 4 dx = ) )) ) cos + sin exp ) )) ) cos + sin exp. 4 3 AA. Prove the following generalization of Rouché s Theorem: Let fz) and gz) be two analytic functions on a simply connected open set D C with finitely many zeros in D and let γ be a continuous closed curve in D. Suppose that a fz) + b gz) + a fz) + b gz) > a + a )fz) + b + b )gz) for some real constants a, a, b, b and all z γ. Then p Z f νγ, p) mult p f = q Z g νγ, q) mult q g where νγ, z ) is the winding number of γ at a point z, Z f and Z g are the zero sets of f and g in D and mult p f and mult q g are the multiplicities of fz) and gz) at p Z f and q Z g, respectively. Hint: Show that the curve f/g) γ is contained in C\[, ).

4 4 Proof. Clearly, the inequality guarantees that fz) and gz) on γ. By Argument Principle, νγ, p) mult p f νγ, q) mult q g = p Z f q Z g i for hz) = fz)/gz). For z γ, we have γ = νf/g) γ, ) = νh γ, ) fz)/gz)) fz)/gz) dz a fz) + b gz) + a fz) + b gz) > a + a )fz) + b + b )gz) a hz) + b + a hz) + b > a + a )hz) + b + b ). If hz ) [, ) for some z γ, a hz ) + b + a hz ) + b = a + a )hz ) + b + b ) since a, a, b, b. Contradiction. Therefore, hz) [, ) for all z γ. Namely, h γ C\[, ), where C\[, ) C is simply connected. Consequently, νh γ, ) = νγ, p) mult p f = νγ, q) mult q g. p Z f q Z g AA.3 Show that the polynomial z 4 + z z + has one root in each quadrant. Proof. Since x 4 + x x + = x 4 + x + ) + x x + ) > for all x R, the polynomial fz) = z 4 + z z + has no roots on the real axis. And since fiy) = y 4 y + ) yi for all y R, fz) has no roots on the imaginary axis. Since fz) R[z], the roots of fz) appear in complex conjugate pairs. So it has two roots above the real axis and two roots below the real axis. To show it has one root in each quadrant, it suffices to show that it has exactly one root in the first quadrant.

5 We let Γ R = X R + + Y R, where X R t) = tr for t [, ] t) = Re it for t [, π/] Y R t) = ir t) for t [, ] for some R >. By Argument Principle, it suffices to show that = f z) i Γ R fz) dz = f z) i X R fz) dz + f z) i fz) dz + f z) i Y R fz) dz = d arg w + f z) f X R i fz) dz + d arg w f Y R for R sufficiently large. For z X R, we have prove that fz) >. So f X R {Rew) > }. And since arg w has a branch in {Rew) > } with we have f X R d arg w = π < arg w < π arg fr) arg f)) = argr4 + R R + ) arg)) =. For z = yi Y R, we have Imfz)) = y and fz). So f Y R {Imw) }\{}. And since arg w has a branch in {Imw) }\{} with we have π arg w d arg w = arg f) arg fri)) f Y R = arg) argr4 R + ) Ri)) = )) + tan R R 4 R + ) R R 4 R + = tan 5

6 6 On, we have f z) i fz) dz i d arg w =. f Y R 4 z dz = i zf z) 4fz) zfz) Since degzf z) 4fz)) degzfz)), we have zf z) 4fz) zfz) M R for some constant M and z = R sufficiently large. Therefore, zf z) 4fz) ) ) πr M dz i zfz) R = M 4R i Consequently, i f z) fz) Γ R f z) fz) dz = dz = = + + i 4 z dz =. d arg w f X R f z) fz) dz i On the other hand, since f z) i Γ R fz) dz = νf Γ R, ) is an integer, f z) i Γ R fz) dz = f Y R d arg w for R is sufficiently large. So fz) has exactly one root in the first quadrant. By complex conjugation, it has exactly one root in the fourth quadrant. So it has two roots in total in the second and third quadrant. By complex conjugation again, it must have one root each in the seond and third quadrant as well. dz

7 AA.4 Suppose that fz) = z+a z +...+a n z n is one-to-one on the unit disk D = { z < }. Show that a n )/ and a n /n. Hint: Use the fact that f z) in D. Proof. Since fz) is one-to-one on D, f z) in D. That is, all roots of f z) are in { z }. Then f z) = + a z na n z n = na n z c )z c )...z c n ) = c z) c z)... c n z) where c j are the roots of f z) and c j. Then n a = j= n na n = ) n a n c j j= j= c j n a n = n c j n j= 7 c j n. AA.5 Let fz) be an analytic function on D = { < z < }. If f : D C is one-to-one, then fz) must have a removable singularity or a pole of order at. Hint: First you need to einate the possibility that fz) has an essential singularity at. Since f : D C is one-to-one, f : D fd) is biholomorphic proper. Choose K = { w w ε} fd) and f K) is compact in D. Therefore, f K) {r z < } for some r >. Then f{ < z < r}) K =. Use Casorati-Weierstrass to conclude that fz) does not have an essential singularity at. Next, show that fz) cannot have a pole of order m at. If it does, choose some < r < such that fz) in D r = { < z < r}; then apply Rouché s theorem or directly, Bifurcation Theorem) to show that the map /fz) : D r C is not one-to-one. Proof. We first show that fz) cannot have an essential singularity at. Since f : D C is one-to-one, f : D fd) is biholomorphic proper. We choose w fd) and

8 8 K = { w w ε} fd) for some ε >. Since K is compact, f K) D is compact. Therefore, for some r >. Then f K) {r z < } f{ < z < r}) K =. Since K has nonempty interior, f{ < z < r}) cannot be dense in C. By Casorati-Weierstrass, fz) cannot have an essential singularity at. Suppose that fz) has a pole at of order m. Since z fz) =, there exists < r < such that fz) > for all z D r = { < z < r}. Let gz) = /fz) : D r C. Then gz) is also one-to-one on D r. Since z gz) =, gz) has a removable singularity at. Indeed, gz) has a zero at of multiplicity m, i.e., g) = g ) =... = g m ) ) = and g m) ). By Bifurcation Theorem, there exist ε, δ > such that the map g : { < z < ε} g { < w < δ}) { < w < δ} is m-to-. Contradiction. So fz) has either a removable singularity or a pole of order at.