By C. W. Celia, A. T. F. Nice, K. F. Elliott
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Hence (z")" and (z")" are equal. Consider the polynomial P(z) given by 34 Advanced mathematics 3 where n E 7!.. , Provided that the coefficients ao, a lo [P(z)]* P(z*) . + P(z) = P(x = a + ib, iy) = [P(z)]* P(z*) = When P(z) an are real , then = an(zn)* + an -I (zn-I)* + + a1z* + a o = an(z*)" + an -I (Z*)" -I + . + a. z" + a o = Let Then • • •, = a, b where a - ib P(x - iy) =a- E IR. ib. o. = 0, both a and b must be zero, so that P(z*) = P(z) = 0 Hence P(z*) = <:> o. This proves that, when ex is a complex root of the equation P(z) = 0 in which the coefficients are real, a* will also be a root of this equation.
C! (d) On CO , let z = it, where 0 ~ t ~ 1. Then w = l iCit) = -i/t. As t decreases from 1 to 0, w moves along the negative v-axis away from the origin. Note that the centre of the square, given by z = (1 + i)/2, is mapped to the point given by w = 1 - i. 10 1 A region is defined in the z-plane by 1 ~ [z] ~ 2, 0 ~ arg z ~ n/4. Find the region to which this is mapped by the transformation w = l iz. 2 Show that the points given by z = 1, 1 + i and i lie on the circle 12z - 1 - i] = J 2. Show also that the transformation w = l /z maps these points to points on a straight line.
2i = R(cos ex + i sin ex). R = 12 + 2il = 2J2, ex = arg (2 + 2i) = 45° 2 + 2i = 2J2(cos 45° + i sin 45°). = The positive cube root of R is J 2 and ex/3 = 15°. Hence one cube root is J 2(cos 15° + i sin 15°). In Fig. 2 the point P represents 2 + 2i and the point A represents the cube Let Then 2 root with argument 15°. r p 2 B A 2 x - I Fig. 2 Cube roots of 2 + 2i The other two cube roots, represented by Band C, will have the same modulus J 2, and their arguments will be (15" + 120°) and (15° - 120°) respectively .