By Leif Mejlbro

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E. e. equal to the constant term of the polynomial. We see that the check is OK. 2 If one does not see immediately that 7 = 42 − 32 , then we may try to solve the equation (a + ib)2 = 7 + 24i. This gives us the equations a2 − b2 = 7 og 2ab = 24, hence 2 a2 + b2 = a2 − b2 2 + (2ab)2 = 72 + 242 = 49 + 576 = 625 = 252 , and thus a2 + b2 = 25, so when we combine it with a2 − b2 = 7 we obtain a2 = 16 and b2 = 9. 3 Solve the equation 1 2 z+ 1 2 =a for a ∈ C. Prove that the equation has precisely one solution in the open unit disc |z| < 1, if and only if a does not belong to the real interval [−1, 1].

Concerning polynomials, a good strategy is to identify the degrees of the pair (x, y), which occur. We see that we have the degrees 4 and 3, and since z 4 = (x + iy)4 = x4 + 4ix3 y − 6x2 y 2 − 4ixy 3 + y 4 = x4 − 6x2 y 2 + y 4 + i 4x3 y − 4xy 3 and z 3 = (x + iy)3 = x3 + 3ix2 y − 3xy 2 − iy 3 = x3 − 3xy 2 + i 3x2 y − y 3 , it follows that f (z) = 2x4 − 12x2 y 2 + 2y 4 − 3x3 + 9xy 2 + i 8x3 y − 8xy 3 − 9x2 y + 3y 3 = 2 x4 − 6x2 y 2 + y 4 + i 4x3 y − 4xy 3 − 3 x3 − 3xy 2 + i 3x2 y − y 3 = 2z 4 − 3z 3 , thus f (z) = 2z 4 − 3z 3 = 2z 3 z − 3 2 , and the roots are z = 0 (of multiplicity 3) and the simple root z = 3 .

2 2 (d) π π · exp −i 3 π π = π cos − i sin 3 3 =π· √ 1 3 −i 2 2 √ π π 3 = −i . com 33 Complex Funktions c-1 Polar form of complex numbers (d) exp i π 2 π π + i sin 2 2 = cos = i. 5 Assume ez = ew . Prove that there exists a k ∈ Z, such that z = w + 2π k i. e. same module) and (assuming that the modulus is = 0) if their arguments agree modulo 2π. If we put z = x + iy and w = u + iv into the exponential function, then ez = ex · ei y ew = eu · ei v . and The module is ex = eu = 0, hence x = u, and concerning the arguments we get y ≡ v (mod 2π), hence y = v + 2k π for some k ∈ Z.