Menu Close

z-2iz-3-3i-geometrical-representation-is-

Tinku Tara 0 Comments
Pin




Question Number 146041 by alcohol last updated on 10/Jul/21
z′ = 2iz + (3−3i) geometrical representation is?
$${z}'\:=\:\mathrm{2}{iz}\:+\:\left(\mathrm{3}−\mathrm{3}{i}\right) \\ $$$${geometrical}\:{representation}\:{is}? \\ $$
Answered by ajfour last updated on 10/Jul/21
(dx/dt)+i(dy/dt)=2i(x+iy)+3(1−i) ⇒ (dx/dt)=3−2y (dy/dt)=2x−3 (dx/(3−2y))=(dy/(2x−3)) ∫(2x−3)dx+∫(2y−3)dy=c x^2 −3x+y^2 −3y=c (x−(3/2))^2 +(y−(3/2))^2 =c+(9/2)
$$\frac{{dx}}{{dt}}+{i}\frac{{dy}}{{dt}}=\mathrm{2}{i}\left({x}+{iy}\right)+\mathrm{3}\left(\mathrm{1}−{i}\right) \\ $$$$\Rightarrow\:\frac{{dx}}{{dt}}=\mathrm{3}−\mathrm{2}{y} \\ $$$$\:\:\:\:\frac{{dy}}{{dt}}=\mathrm{2}{x}−\mathrm{3} \\ $$$$\frac{{dx}}{\mathrm{3}−\mathrm{2}{y}}=\frac{{dy}}{\mathrm{2}{x}−\mathrm{3}} \\ $$$$\int\left(\mathrm{2}{x}−\mathrm{3}\right){dx}+\int\left(\mathrm{2}{y}−\mathrm{3}\right){dy}={c} \\ $$$${x}^{\mathrm{2}} −\mathrm{3}{x}+{y}^{\mathrm{2}} −\mathrm{3}{y}={c} \\ $$$$\left({x}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({y}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} ={c}+\frac{\mathrm{9}}{\mathrm{2}} \\ $$
Commented by alcohol last updated on 10/Jul/21
z′ is the image of z
$${z}'\:{is}\:{the}\:{image}\:{of}\:{z} \\ $$$$ \\ $$
Commented by mr W last updated on 10/Jul/21
how is your image of z defined?
$${how}\:{is}\:{your}\:{image}\:{of}\:{z}\:{defined}? \\ $$
Commented by alcohol last updated on 10/Jul/21
it is a transformation (similitude) and am asked to give the geometrical interpretion
$${it}\:{is}\:{a}\:{transformation} \\ $$$$\left({similitude}\right)\:{and}\:{am}\:{asked}\:{to}\:{give}\:{the} \\ $$$${geometrical}\:{interpretion} \\ $$
Answered by Ar Brandon last updated on 10/Jul/21
Soient A(a) et B(a) deux points distinct et A′(a′), B′(b′) leurs images par s, alors on a; a′=2ia+(3−3i), b′=2ib+(3−3i) a′−b′=2i(a−b)⇒∣a′−b′∣=2∣a−b∣. Par suite A′B′=2AB s est une similitude de rapport 2.
$$\mathrm{Soient}\:\mathrm{A}\left(\mathrm{a}\right)\:\mathrm{et}\:\mathrm{B}\left(\mathrm{a}\right)\:\mathrm{deux}\:\mathrm{points}\:\mathrm{distinct}\:\mathrm{et}\:\mathrm{A}'\left(\mathrm{a}'\right),\:\mathrm{B}'\left(\mathrm{b}'\right)\: \\ $$$$\mathrm{leurs}\:\mathrm{images}\:\mathrm{par}\:\mathrm{s},\:\mathrm{alors}\:\mathrm{on}\:\mathrm{a}; \\ $$$$\mathrm{a}'=\mathrm{2ia}+\left(\mathrm{3}−\mathrm{3i}\right),\:\mathrm{b}'=\mathrm{2ib}+\left(\mathrm{3}−\mathrm{3i}\right) \\ $$$$\mathrm{a}'−\mathrm{b}'=\mathrm{2i}\left(\mathrm{a}−\mathrm{b}\right)\Rightarrow\mid\mathrm{a}'−\mathrm{b}'\mid=\mathrm{2}\mid\mathrm{a}−\mathrm{b}\mid.\:\mathrm{Par}\:\mathrm{suite}\:\mathrm{A}'\mathrm{B}'=\mathrm{2AB} \\ $$$$\mathrm{s}\:\mathrm{est}\:\mathrm{une}\:\mathrm{similitude}\:\mathrm{de}\:\mathrm{rapport}\:\mathrm{2}. \\ $$
Answered by physicstutes last updated on 11/Jul/21
let the image of a be a′ and the image of b be b′ then, a′ = 2ia + (3−3i).....(i) b′ = 2ib + (3−3i)......(ii) (ii)−(i) ⇒ b′−a′ = 2i(b−a) ⇒ ∣b′−a′∣ = 2∣b−a∣ So we say your transformation call it f is a similitude of radius 2, since similitudes are enlargments . Then Geometrically you will inteprete that f is an enlargment scale factor 2 followed by a translation about the vector ((3),((−3)) )
$$\mathrm{let}\:\mathrm{the}\:\mathrm{image}\:\mathrm{of}\:{a}\:\mathrm{be}\:{a}'\:\mathrm{and}\:\mathrm{the}\:\mathrm{image}\:\mathrm{of}\:{b}\:\mathrm{be}\:\:{b}' \\ $$$$\mathrm{then},\:\:{a}'\:=\:\mathrm{2}{ia}\:+\:\left(\mathrm{3}−\mathrm{3}{i}\right)…..\left({i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{b}'\:=\:\mathrm{2}{ib}\:+\:\left(\mathrm{3}−\mathrm{3}{i}\right)……\left({ii}\right) \\ $$$$\left({ii}\right)−\left({i}\right)\:\Rightarrow\:\:{b}'−{a}'\:=\:\mathrm{2}{i}\left({b}−{a}\right) \\ $$$$\Rightarrow\:\mid{b}'−{a}'\mid\:=\:\mathrm{2}\mid{b}−{a}\mid \\ $$$$\mathrm{So}\:\mathrm{we}\:\mathrm{say}\:\mathrm{your}\:\mathrm{transformation}\:\mathrm{call}\:\mathrm{it}\:{f}\:\mathrm{is}\:\mathrm{a}\:\mathrm{similitude}\:\mathrm{of}\:\mathrm{radius} \\ $$$$\mathrm{2},\:\mathrm{since}\:\mathrm{similitudes}\:\mathrm{are}\:\mathrm{enlargments}\:.\:\mathrm{Then}\:\mathrm{Geometrically} \\ $$$$\mathrm{you}\:\mathrm{will}\:\mathrm{inteprete}\:\mathrm{that}\:{f}\:\mathrm{is}\:\mathrm{an}\:\mathrm{enlargment}\:\mathrm{scale}\:\mathrm{factor}\:\mathrm{2} \\ $$$$\mathrm{followed}\:\mathrm{by}\:\mathrm{a}\:\mathrm{translation}\:\mathrm{about}\:\mathrm{the}\:\mathrm{vector}\:\begin{pmatrix}{\mathrm{3}}\\{−\mathrm{3}}\end{pmatrix} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *