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Unformatted text preview: t follows from (11) and (12) that x2 + y 2 = a2 + b2 , x2 − y 2 = a, 2xy = b. where the square root is non-negative. Together with (11), we obtain (13) x=
2 a+ √ a2 + b2 2 and y= 2 −a + √ a2 + b2 . 2 Note that the equations (13) generally yield two solutions for x and two solutions for y . However, note that by (12), the product xy has to have the same sign as b. It follows that √ √ √ a + a2 + b2 −a + a2 + b2 b a + bi = ± +i , 2 |b| 2
Chapter 1 : The Number System page 12 of 19 First Year Calculus c W W L Chen, 1982, 2005 where the square roots are non-negative. This is a rather cumbersome approach, and is not to be recommended for higher order roots. As we have shown earlier, it is more convenient to do multiplication of complex numbers in polar coordinates, so let us attempt to ﬁnd roots using polar coordinates. Suppose that c = R(cos α + i sin α), where c, α ∈ R and c > 0. Consider the equation z n = c, where n ∈ N is ﬁxed. Writing z = r(cos θ + i sin θ), where r, θ ∈ R and r > 0, we have, using de Moivre’s theorem, that z...
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