By P.R. Graves-Morris, E.B. Saff, R.S. Varga
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Extra resources for Rational Approximation and Interpolation
Example text
Z z Solution. Let z = cos x + i sin x, x ∈ [0, 2π). Then 1= z z |z 2 + z 2 | = | cos 2x + i sin 2x + cos 2x − i sin 2x| = 2| cos 2x|, + = z z |z|2 whence cos 2x = 1 1 or cos 2x = − . 2 2 If cos 2x = 12 , then x1 = π , 6 x2 = 5π , 6 x3 = 7π , 6 x4 = 11π . 6 If cos 2x = − 12 , then x5 = π , 3 x6 = 2π , 3 x7 = 4π 5π , x8 = . 3 3 38 2 Complex Numbers in Trigonometric Form Hence there are eight solutions: zk = cos xk + i sin xk , k = 1, 2, . . , 8. 3 Operations with Complex Numbers in Polar Representation 1.
A) x2000 1 2 1 2 35. Factorize (in linear polynomials) the following polynomials: (a) x4 + 16; (b) x3 − 27; (c) x3 + 8; (d) x4 + x2 + 1. 36. Find all quadratic equations with real coefficients that have one of the following roots: (a) (2 + i)(3 − i); (b) 5+i ; (c) i51 + 2i80 + 3i45 + 4i38 . 2−i 37. (Hlawka’s inequality) Prove that the inequality |z1 + z2 | + |z2 + z3 | + |z3 + z1 | ≤ |z1 | + |z2 | + |z3 | + |z1 + z2 + z3 | holds for all complex numbers z1 , z2 , z3 . 38. Suppose that complex numbers xi , yi , i = 1, 2, .
Y 2 = −63 − 16i has the solution y1,2 = ± 65−63 2 +i 65+63 2 = ±(1 − 8i). It follows that z1,2 = 4 − 4i ± (1 − 8i). Hence z1 = 5 − 12i and z2 = 3 + 4i. Problem 2. Let p and q be complex numbers with q = 0. Prove that if the roots of the quadratic equation x2 + px + q 2 = 0 have the same absolute value, then pq is a real number. (1999 Romanian Mathematical Olympiad, Final Round) 18 1 Complex Numbers in Algebraic Form Solution. Let x1 and x2 be the roots of the equation and let r = |x1 | = |x2 |.