# Computational physics by Steven E Koonin By Steven E Koonin

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The length of the other side is b. As the angle θ between the two sides changes, determine the locus of the center of the parallelogram. 6 Polar Form of Complex Numbers B 43 C iθ be z θ a O A Fig. 6. The curve described by the center of a parallelogram. 15. Let the origin of the coordinates be at the left bottom corner of the parallelogram. So zA = a, zB = beiθ . Let the center of the parallelogram be z which is at the midpoint of the diagonal OC. Thus z= 1 1 1 (zA + zB ) = a + beiθ 2 2 2 or 1 1 z − a = beiθ 2 2 It follows that: 1 1 1 z − a = beiθ = b.

Note only in this particular case, − ln(2 + 3) = ln(2 − 3), since √ √ −1 √ √ 2− 3 1 √ = ln √ = ln(2 − 3). 317. 317i + 2πn, n = 0, ±1, ±2, . . In real variable domain, the maximum value of cosine is one. Therefore we expect cos−1 2 to be complex numbers. Also note that ± solutions may be expected since cos (−z) = cos (z) . 14. Show that tan−1 z = i [ln(i + z) − ln (i − z)]. 14. Let w = tan−1 z, so z = tan w = eiw − e−iw sin w = , cos w i (eiw + e−iw ) iz eiw + e−iw = eiw − e−iw , (iz − 1) eiw + (iz + 1) e−iw = 0.

5. Express −64 as a point in the complex plane −64 = 64eiπ+ik2π , k = 0, 1, 2, . . It follows that: z = (−64) 1/4 1/4 i(π+2kπ)/4 = (64) e , k = 0, 1, 2, 3, ⎧ √ 2 2(cos π4 + i sin π4 ) = 2 + 2i, ⎪ ⎪ ⎨ √ 3π 2√2(cos 3π 4 + i sin 4 ) = −2 + 2i, z= 5π ⎪ 2 2(cos 4 + i sin 5π ⎪ 4 ) = −2 − 2i, ⎩ √ 7π 2 2(cos 7π + i sin 4 4 ) = 2 − 2i, √ Note that the four roots are on a circle of radius 8 They are 90◦ apart. 6. 6. 3/2 k=0 k=1 k=2 k = 3. centered at the origin. √ π 2eiθ , θ = tan−1 (−1) = − . 4 √ i3θ+ik2π 3 (1 − i) = 2 2e , k = 0, 1, 2, .