# z_1_5 - We apply the Mathematica command Solve(z...

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Extra Problems Sheet 1 Stephen Taylor April 28, 2005 5. Consider the equation ( z + 1) 5 = z 5 . (a) Without attempting to solve the equation, show geometrically that all four solutions must lie on the vertical line, Re z = - 1 2 . We ﬁrst note that if one expands the left side of the equation, the leading term is z 5 which cancels with the like term on the right side. The result is a fourth degree polynomial, with by the fundamental theorem has four roots. To show geometrically all such roots lie on the line Re z = - 1 2 we take the modulus of the equation: ( z + 1) 5 = z 5 ⇐⇒ | ( z + 1) 5 | = | z 5 | ⇐⇒ | ( z + 1) | 5 = | z | 5 ⇐⇒ | z + 1 | = | z | But this is analogous to (3 c ) and can be interpreted as a hyperbola with zero curvature with foci at (0 , 0) , ( - 1 , 0), the line that is the perpendicular bisector of the segment adjoining the aforementioned foci, or the line parameterized by ( - 1 / 2 , t ) where t R . We therefore conclude that all solutions to the desired equation must lie on the claimed line. (b) and (c) Solve the equation

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Unformatted text preview: We apply the Mathematica command Solve[(z+1)^5==z^5,z] to ﬁnd the four solutions of the equation to be z 1 =-1 2-i 10 q 5(5-2 √ 5) z 2 =-1 2 + i 10 q 5(5-2 √ 5) z 3 =-1 2-i 10 q 5(5 + 2 √ 5) z 4 =-1 2 + i 10 q 5(5 + 2 √ 5) which veriﬁes our geometric argument. 1 We also note in the absence of a symbolic manipulator, we may solve the above equation in the following manner where we apply the identity sin( θ ) = e iθ-e-iθ 2 i We start by dividing both sides by z 5 ( z + 1) 5 = z 5 ⇐⇒ (1 + z-1 ) 5 = 1 ⇐⇒ 1 + z-1 = 1 1 / 5 ⇐⇒ 1 + z-1 = e i 2 πk 5 Solving for z , we ﬁnd z = 1 e i 2 πk 5-1 = 1 e i πk 5 ( e i πk 5-e-i πk 5 ) = e-i πk 5 2 i sin( πk 5 ) = cos( πk 5 )-i sin( πk 5 ) 2 i sin( πk 5 ) which reduces to z =-1 2-i 2 cot ± 2 πk 5 ² Letting k run from 1 to 4 we obtain the above roots. ± 2...
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z_1_5 - We apply the Mathematica command Solve(z...

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