Complex Analysis Homework 1: Solutions

Transcription

1 Complex Analysis Fall 007 Homework 1: Solutions a) + i)4 + i) 8 ) )i i b) 8 + 6i) 64 6) )i i c) 1 + ) 1 + i i) 1 + i)1 i) 1 + i ) 5 ) i ) i ) ) i a) If z x + iy we have so that ) z + 1 Re z 5 z + 1 z 5 z + 1) z 5) z 5) z 5) z z 5z + z 5 4z z 10z + z) + 5 z 5z + z 5 4 z 10z + z) + 5 x + y ) 5x + x 5 + 5yi yi 4x + y ) 0x + 5 x + y ) x 5 4x + y ) 0x y 4x + y ) 0x + 5 i x + y ) x 5 4x + y ) 0x + 5 ) z + 1, Im z 5 7y 4x + y ) 0x

2 b) If z x + iy then z x + iy) x + x yi + xy i + y i x xy ) + x y y )i so that Rez x xy, Imz x y y a) 1 i) 1 1+i 1 i)1+i) 1+i i b) 1+i 1 i 1 + i)1 i) i) i) 1 1 ) ) i i 1... a) The equation z is equivalent to z 6 8. Since 8 8 and arg 8) π, the solutions to the latter equation are z k 6 π 8 cos 6 + πk ) π + i sin 6 + πk )) for k 0, 1,..., 5. b) The equation z 4 0 is equivalent to z 4 which, since 4 4 and arg4) 0, has the solutions z k ) )) πk πk 4 cos + i sin for k 0, 1, Recalling that conjugation preserves the arithmetic of C, we have ) 8 i) i) i)10 4 6i) DeMoivre s formula and the binomial theorem give cos 6x + i sin 6x cos x + i sin x) 6 Equating real and imaginary parts gives cos 6 x 15 cos 4 x sin x + 15 cos x sin 4 x sin 6 x) +i6 cos 5 x sin x 0 cos x sin x + 6 cos x sin 5 x). cos 6x cos 6 x 15 cos 4 x sin x + 15 cos x sin 4 x sin 6 x sin 6x 6 cos 5 x sin x 0 cos x sin x + 6 cos x sin 5 x.

3 1..8. Since preserves multiplication and division we have i) i 8 + 6i) 8 + 6i Let pz) a j z j j0 be a polynomial with a j R for all j. To say that the roots of pz) occur in complex pairs means that if z 0 C is a root, then so too is z 0. To see that this is the case, let z 0 C with pz 0 ) 0. Then, since conjugation preserves arithmetic and a j a j for all j, 0 0 pz 0 ) Notice first that z w 1 z w a j z j 0 j0 z w 1 z w a j z j 0 j0 w z 1 zw a j z j 0 pz 0 ) j0 w z 1 w z. That is, the expression in question is symmetric in z and w. We may therefore assume, without loss of generality, that w 1. The second, and perhaps more important observation, is that if w 1 then w 1 w, since 1 w w w. We therefore have z w 1 z w z w 1 zw 1 z w w z)w 1 wz w) w z which is what we wanted to show w z w w z a) Since z R, z z z and congruence mod π is an equivalence relation, if z 0 we have 0 arg z mod π) arg z z mod π) arg z + arg z mod π) 1 which shows that arg z arg z mod π). b) As above, we have arg z argzw/w) mod π) argz/w) + arg w mod π) which shows that argz/w) arg z arg w mod π).

4 c) Let z x + iy C. Then z x + y so that z 0 iff z 0 iff x + y 0. Since x, y R, this can happen iff x y 0 iff z Since cubing a complex number triples its argument and cubes its modulus, it is clear that z maps the disk {z C z < } onto the disk {w C w < 8}. Multiplication by i rotates this disk by π/4 radians back onto itself and adding 1 shifts its radius to the point 1 on the real axis. Since the right-most edge of the shifted disk passes through the point 9, we find that sup z < Reiz + 1) a) e i e cos 1) + i sin 1)) e cos 1 ie sin 1 b) cos + i) ei+i) + e i+i) e +i + e i e cos + i sin ) + e cos ) + i sin )) cos e + e ) i sin e e cos cosh i sin sinh a) If sin z + i)/4 then 4 sin z + i. The definition of sin z then gives ie iz e iz ) + i which, upon multiplication by e iz which is nonzero), is equivalent to or The quadratic formula yields ie iz ) + i + i)e iz ie iz ) + i)e iz + i 0. e iz + i ± + i) 16) + i ± 8 + 6i. At this point we need to compute the square root. Writing 8 + 6i rcos θ + i sin θ) r > 0) we find that r 10 and cos θ 8/10. Using the standard half-angle formulas, we get cosθ/) 1/ 10 and sinθ/) / 10, so that one of the square roots of 8 + 6i is rcosθ/) + i sinθ/)) 101/ 10 + i/ 10) 1 + i. 4

5 Therefore, our solutions satisfy e iz + i ± 8 + 6i 4 + 4i, i 1 + i, i. Let log z denote the branch of the logarithm with imaginary part in the interval [ π, pi). Applying this branch to the equation above we find that all of the solutions are given by iz log 1 + i) + nπi 1 iz log + 1 ) i + mπi for m, n Z. Since log 1+i) log +πi/4 and log1+i)/) log1/ )+πi/4 log + πi/4 the two equations above can be rewritten as z nπ + π 4 i log z mπ + π 4 + i log. Letting m, n vary through all possible integer values yields the complete set of solutions to the original equation. b) The procedure is exactly the same as that above. Plugging in the definition of sin z, the equation sin z 4 becomes the equation e iz ) 8ie iz 1 0, which the quadratic formula transforms to e iz 4 ± 15)i. Using the same branch of log z as above we get iz log4 ± 15)i) + nπi log4 ± π ) 15) + i + nπ where n Z is arbitrary. Dividing by i gives the final answer: for any n Z. z π + nπ i log 4 ± 15) If w e x+iy e x e iy then w e x and arg w y mod π). Hence, if y is fixed, e z lies on a ray emanating from the origin, making an angle of y with the real axis. As x, w e x indicating that e z moves out along this ray away from the origin indefinitely. On the other hand, as x, w e x 0 so that e z moves along the ray closer and closer to the origin. If x is fixed, then w is a constant, indicating that e z remains on the circle of radius e x centered at 0. Since arg w y mod π), as y the point w e z simply moves around this circle repeatedly counterclockwise and as y the circle is traced out infinitely often in the clockwise direction. 5

6 1..4. We start by noting that if z x + iy then cos z eiz + e iz e y cos x + i sin x) + e y cos x) + i sin x)) cos x cosh y i sin x sinh y. A horizontal line has the form z x+iy 0 where y 0 R is fixed and x R varies. Writing cos z u + iv we find that the real and imaginary parts of the image of this horizontal line satisfy u cos x cosh y 0 v sin x sinh y 0. For variable x, this is the standard clockwise parametrization of the ellipse u cosh + v y 0 sinh 1. 1 y 0 That is, horizontal lines are mapped onto ellipses. A vertical line is given by z x 0 + iy where x 0 R is fixed and y R is free to vary. In this case we have, again writing cos z u + iv, u cos x 0 cosh y v sin x 0 sinh y which as y R varies gives the standard parametrization of the hyperbola u v cos x 0 sin 1. x 0 That is, vertical lines are mapped onto hyperbolas. 1 Provided sinh y 0 0. It is left to the reader to describe the image in the case sinh y 0 0. Provided that cos x 0 and sin x 0 are both nonzero. It is left to the reader to determine the image when cos x 0 0 or sin x