Geometry with Trigonometry

Sec. 10.3 Rotations and axial symmetries 151

which has the matrix form

(

x′−x 0

y′−y 0

)

=

(

cosα −sinα

sinα cosα

)(

x−x 0

y−y 0

)

Proof.

(i) For|z′−z 0 |=|(z−z 0 )cisα|=|z−z 0 ||cisα|=|z−z 0 |.

(ii) For by 10.2.2

z−z 0 =|z−z 0 |cisθ,z′−z 0 =|z′−z 0 |cisθ′,

and so|z′−z 0 |cisθ′=|z−z 0 |cisθcisα. Hence cisθ′=cisθcisα=cis(θ+α)and
soθ′=θ+αby 9.2.2.
(iii) Nowx′−x 0 +ı(y′−y 0 )=(cosα+ısinα)[x−x 0 +ı(y−y 0 )]and so

x′−x 0 =cosα.(x−x 0 )−sinα.(y−y 0 ),

y′−y 0 =sinα.(x−x 0 )+cosα.(y−y 0 ).

COMMENT. The rotationrα;Z 0 is characterised by (i) and (ii), as the steps can be
traced backwards. Why a frame of referenceF′is prominent in this characterisation
stems from the need to identify the anglesα,θ,θ′.

10.3.2Formulaforanaxialsymmetry ....................

O I

J H 1

H 2

H 4 H 3

Z 0 I 0

J 0 Z^1

Z

sl(Z)

α

l

Figure 10.3.

The form of equation of a line noted in 10.2.2 can be used in the formula in 6.6.1(iii)
for an axial symmetry. However, for practice with complex-valued coordinates we
deduce the result independently.