Linear Algebra Solutions 58

Linear Algebra Solutions 58 - and hence pw + pz = n. (iii)...

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Unformatted text preview: and hence pw + pz = n. (iii) Let p = b − a and n = |b|2 − |a|2 . Then |z − a| = |z − b| ⇔ |z − a|2 = |z − b|2 ⇔ (z − a)(z − a) = ⇔ z z − az − z a + aa = ⇔ (z − a)(z − a) = ⇔ (b − a)z + (b − a)z = ⇔ pz + pz = (z − b)(z − b) (z − b)(z − b) z z − bz − z b + bb |b|2 − |a|2 n. Suppose z lies on the circle z −a and let w be the reflection of z in the z −b line pz + pz = n. Then by part (ii) pw + pz = n. Taking conjugates gives pw + pz = n and hence z= n − pw p (a) Substituting for z in the circle equation, using (a) gives λ= n−pw p n−pw p −a −b = n − pw − pa . n − pw − pb However n − pa = |b|2 − |a|2 − (b − a)a = bb − aa − ba + aa = b(b − a) = bp. Similarly n − pb = ap. Consequently (b) simplifies to λ= which gives w−a w−b w−b bp − pw b−w = = , ap − pw a−w w−a 1 = λ. 61 (b) ...
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This note was uploaded on 12/19/2011 for the course MAS 3105 taught by Professor Dreibelbis during the Fall '10 term at UNF.

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