Roots of a Complex Equation
Date: 04/28/97 at 01:44:57 From: raylim Subject: Complex number The following question is from the British A Level exam. Can you solve it for me? I couldn't get the solution because I had no idea how to begin. You are given that cos t + isin t is the root of z^3 + az + b = 0 in which sin t is not equal to 0. If a and b are real and not equal to zero, show that a + b^2 = 1. Thanks, Raylim
Date: 04/28/97 at 16:25:59 From: Doctor Anthony Subject: Re: Complex number Raylim, If z = cos(t) + i.sin(t) is a root, then since the coefficients of the equation are real, it is also true that z = cos(t) - i.sin(t) is a root. It follows that [z-cos(t)-i.sin(t)] and [z-cos(t)+i.sin(t)] are factors: = (z-cos(t))^2 + sin^2(t) is a factor = z^2 - 2z.cos(t) + cos^2(t) + sin^2(t) is a factor = z^2 - 2z.cos(t) + 1 is a factor If we do long division of this expression into z^3 + az + b, it must go in exactly with a remainder of 0. In fact, the quotient is z + 2.cos(t) and the remainder is: Remainder = z[a-1+4cos^2(t)] + [b - 2.cos(t)] = 0 We can put each part of this expression separately equal to 0 since one part is multiplied by z (meaning that is also multiplied by i and so is imaginary) and the other part, b-2.cos(t), is real: b-2.cos(t) = 0 b = 2.cos(t) b^2 = 4.cos^2(t) Also: a-1 + 4.cos^2(t) = 0 a-1 + b^2 = 0 a + b^2 = 1 -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Date: 04/28/97 at 16:45:55 From: Doctor Wilkinson Subject: Re: Complex number Raylim, You are given that one of the roots is of the form cos t + i sin t, where sin t is different from 0. In particular, this is a complex root, and you are probably expected to know that the complex roots of an equation with real coefficients occur in conjugate pairs, so that a second root is cos t - i sin t. The product of these two roots is cos^ t + sin^ t, which is 1. Now we can forget about the trigonometry and call the complex roots u and v, where uv = 1. The remaining root must be real, and we'll call it w. Now we can write: z^3 + az + b = (z - u) (z - v)(z - w) Now multiply out the right side. This gives: a^3 - (u + v + w) z^2 + (uv + uw +vw) z - uvw This must be identically equal to the left side, so we must have: u + v + w = 0 uv + uw +vw = a uvw = -b Now we already now that uv = 1, so we can simplify these to: (u + v) + w = 0 1 + (u + v) w = a w = -b The first equation gives us u + v = -w. Substitute this into the second equation to get: 1 - w^2 = a But the third equation says w = -b, so: 1 - b^2 = a or a + b^2 = 1 -Doctor Wilkinson, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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