MA 412 Complex Analysis Final Exam

Transcription

1 MA 4 Complex Analysis Final Exam Summer II Session, August 9, 00.. Find all the values of ( 8i) /3. Sketch the solutions. Answer: We start by writing 8i in polar form and then we ll compute the cubic root: ( 8i) /3 (8e iπ/ ) /3 8 /3 exp ( i( π 6 + πk 3 )), Hence z 0 exp( iπ/6), z exp(iπ/) i, z exp(i7π/6).. Suppose v is harmonic conjugated to u, and u is harmonic conjugated to v. Show that u and v must be constant functions. Answer By definition, v is harmonic conjugated to u if u v 0 and u x v y, u y v x hold. On the other hand, as u is harmonic conjugated to v, we also have v x u y and v y u x. Then, and u y v x v x implies v x 0 and u y 0 u x v y v y implies v y 0 and u x 0. Hence, u(x, y) and v(x, y) are constant functions. 3. Show that f(z) z is differentiable at the point z 0 0 but not at any other point.

2 z i z z 0 Figure : Roots of ( 8i) /3. Answer: We could show that f is not differentiable at any z 0 C 0 by computing different limits for different trajectories. But we ll use the necessary conditions of differentiability (i.e. the Cauchy-Riemann equations) to do so. First, we write f(z) u(x, y)+iv(x, y), so u(x, y) x +y and v(x, y) 0. Clearly, the first-order partials of u and v are continuous everywhere. But u x x equals v y 0, and u y y equals v x 0 if and only if x y 0. Then f can only be differentiable at z Construct a branch of f(z) log(z + 4) that is analytic at z 5 and takes on the value 7πi there. Sketch the branch cut in the complex plane. Answer: Let α 6π. Then define the branch of log as where 6π < θ 8π. Thus log α (z + 4) ln z iθ log α ( 5 + 4) log α ( ) ln + i(π + 6π) that is log α ( 5 + 4) 7πi. 5. Find C exp(z) z n dz

3 7π 4 α 6π Figure : The branch cut of log 6π. where C : z(t) e πit, 0 t and n is any non-negative integer. Answer: Let f(z) exp(z). Since z 0 0 lies inside the given contour and f is analytic in the interior of C, then by Cauchy s integral formula for derivatives exp(z) C z n dz πif(z 0) (n )! πi (n )! for n,,.... For n 0, the integrand reduces to f(z) which is an entire function. Hence, by Cauchy-Goursat theorem, we conclude exp(z) z n dz 0 n 0. C 6. Find two Laurent series expansions for f(z) z 3 z 4 that involves powers of z. Use the regions 0 < z < and z >. Answer: The given function has two singularities at z 0 0 and z. We ll proceed by expanding f in the regions, D 0 : 0 < z < and D : z >. 3

4 At D 0, for 0 < z <. Then f(z) z 3 z z 3 z n, f(z) z n 3 z 3 + z + z + z n, valid at D 0. At D, z > implies / z <. Then we must write f(z) as z 3 z z 4. z Hence, valid at D. f(z) z 4 z n z n, n4 7. Show.. + z ( ) n z n, z <, z ( ) n (z ) n, z <. Answers: (a) Using the Taylor series for /( z) zn, valid at z <, we obtain valid also at z <. + z ( z) ( ) n z n 4

5 (b) In this case, we ll use the result from question (a): valid for z <. z + (z ) ( ) n (z ) n, 8. Use Cauchy s Residue Theorem to evaluate the next integral around the positively oriented contour C : z 3 + z dz. C Answer Let p(z) and q(z) + z, so f(z) p(z)/q(z) represents the integrand given. Notice that q(z) has two zeros at z 0 i and z i, and both zeros are simple since q (z) z 0 if and only if z 0. Then f has two simple poles at z 0 and z, as p, being a constant function, has no zeros at all. By Cauchy s residue theorem, we obtain ( C + z dz πi Res zz0 + z + Res ) zz + z ( πi i + ) i 0 after evaluating q (z) at z 0 and z. 9. State and prove Liouville s theorem. pt. Answer: Let f be an entire function. If f is bounded, then f reduces to a constant function. Assume there exists M > 0 such that f M in the whole complex plane. Fix any point z 0 C. For any R > 0, Cauchy s inequality states f (z 0 ) M R when we restric f to the circular contour C R : z z 0 R. As R tends to, the absolute value of f (z 0 ) becomes zero. That implies u x v x 0 and u y v y 0, 5

6 at (x 0, y 0 ). Since z 0 was arbitrary, it follows that u and v are constant functions in the whole complex plane, and f reduces to a constant. 0. Compute 0 dx x 4 +. Let p(x) and q(x) x 4 +. Clearly, q has complex roots, so q(x) 0 for all x R. Since the degree of q is larger than, we can apply the theorem of indefinite integrals. We first compute the singularities of the integrand in complex form. The zeros of q(z) z 4 + are given by the fourth roots of, ( z k exp i( π 4 + πk ) 4 ) for k 0,,, 3. Only the roots z 0 ( + i)/, z ( + i)/ lie in the upper half plane, so we shall only compute the residues for these two singularities. Also notice that, since q (z) 4z 3 0 if and only if z 0, we only have simple zeros. It follows that p(z) Res zzk q(z) 4zk 3. Hence p(x) ( ) q(x) dx iπ e 3iπ 4 + e 9iπ 4 iπ ( ( + i) + ) ( i) iπ ( i + i) π Since the give integrand is an even function, we conclude 0 dx x 4 + dx x 4 + π. A final question What is the relationship between Complex Analysis and the animated gif of a solar flare found in the course webpage? 6