a,b ∈C : ab^− + b = 0 f : z′ = az^− + b such that f(M) = M′ 1. let z_A = z and z_(A′) = z′ and f(A) = A show that 2Re(b^− z) = bb^− (A is the set of invariant points and describes a line (△) ) 2. Deduce that (△) is a line with gradient u^( →) with affix z_u^→ = ib 3. show that (z_(MM ′) /z_u ) = ((bb^− − 2Re(bz^− ))/(ibb^− )) 4. show that 2Re(b^− z_0 ) = bb^_ where z_0 = ((z + z ′)/2) 5. Deduce that for M ∉ (△) , M is a perpendicular bisector of [MM ′]


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Question Number 209707 by alcohol last updated on 19/Jul/24

$a, b \in C : a \bar{b} + b = 0 f : z^{'} = a \bar{z} + b$ $s u c h t h a t f (M) = M^{'}$ $1 . l e t z_{A} = z a n d z_{A^{'}} = z^{'} a n d f (A) = A$ $s h o w t h a t 2 R e (\bar{b z}) = b \bar{b}$ $(A i s t h e s e t o f i n v a r i a n t p o i n t s a n d$ $d e s c r i b e s a l i n e (△))$ $2 . D e d u c e t h a t (△) i s a l i n e w i t h$ $g r a d i e n t \vec{u} w i t h a f f i x z_{\vec{u}} = i b$ $3 . s h o w t h a t \frac{z_{M M^{'}}}{z_{u}} = \frac{b \bar{b} - 2 R e (b \bar{z})}{i b \bar{b}}$ $4 . s h o w t h a t 2 R e ({\bar{b z}}_{0}) = b \overline{b} w h e r e$ $z_{0} = \frac{z + z^{'}}{2}$ $5 . D e d u c e t h a t f o r M \notin (△), M i s$ $a p e r p e n d i c u l a r b i s e c t o r o f [M M^{'}]$
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