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Question Number 195848 by pete last updated on 11/Aug/23

Please how did ∣z−a∣=r became z= a + re^(iθ) ?

$$\mathrm{Please}\:\mathrm{how}\:\mathrm{did}\:\mid\mathrm{z}−\mathrm{a}\mid=\mathrm{r}\:\mathrm{became}\: \\ $$$$\mathrm{z}=\:\mathrm{a}\:+\:\mathrm{re}^{\mathrm{i}\theta} ? \\ $$

Answered by Frix last updated on 11/Aug/23

∣z−a∣=r ⇒ z−a=re^(iθ) This step is always true, even for z−a∈R because we can always find a θ: if z−a>0 ⇒ θ=0, if z−a<0 ⇒ θ=π z−a=re^(iθ) ⇔ z=a+re^(iθ)

$$\mid{z}−{a}\mid={r}\:\Rightarrow\:{z}−{a}={r}\mathrm{e}^{\mathrm{i}\theta} \\ $$$$\mathrm{This}\:\mathrm{step}\:\mathrm{is}\:\mathrm{always}\:\mathrm{true},\:\mathrm{even}\:\mathrm{for}\:{z}−{a}\in\mathbb{R} \\ $$$$\mathrm{because}\:\mathrm{we}\:\mathrm{can}\:\mathrm{always}\:\mathrm{find}\:\mathrm{a}\:\theta:\:\mathrm{if}\:{z}−{a}>\mathrm{0} \\ $$$$\Rightarrow\:\theta=\mathrm{0},\:\mathrm{if}\:{z}−{a}<\mathrm{0}\:\Rightarrow\:\theta=\pi \\ $$$${z}−{a}={r}\mathrm{e}^{\mathrm{i}\theta} \:\Leftrightarrow\:{z}={a}+{r}\mathrm{e}^{\mathrm{i}\theta} \\ $$

Answered by deleteduser1 last updated on 11/Aug/23

∣z−a∣=r is the equation for the circle centered at a with radius r So,we get z(points on the circle) by translating a circle with centre at the origin(0,0) and radius r to a re^(iθ) ⇒the locus of points generated by rotating a vector with magnitude r around the origin(this gives a circle) +a⇒translating re^(iθ) by a,the centre(origin) is also translated by a ⇒re^(iθ) +a gives a circle with centre at a[∣z−a∣=r]

$$\mid{z}−{a}\mid={r}\:{is}\:{the}\:{equation}\:{for}\:{the}\:{circle}\:{centered} \\ $$$${at}\:{a}\:{with}\:{radius}\:{r} \\ $$$${So},{we}\:{get}\:{z}\left({points}\:{on}\:{the}\:{circle}\right)\:{by}\:{translating} \\ $$$${a}\:{circle}\:{with}\:{centre}\:{at}\:{the}\:{origin}\left(\mathrm{0},\mathrm{0}\right)\:{and}\:{radius} \\ $$$${r}\:{to}\:{a} \\ $$$${re}^{{i}\theta} \:\Rightarrow{the}\:{locus}\:{of}\:{points}\:{generated}\:{by}\:{rotating} \\ $$$${a}\:{vector}\:{with}\:{magnitude}\:{r}\:{around}\:{the}\:{origin}\left({this}\right. \\ $$$$\left.{gives}\:{a}\:{circle}\right) \\ $$$$+{a}\Rightarrow{translating}\:{re}^{{i}\theta} \:\:{by}\:{a},{the}\:{centre}\left({origin}\right)\:{is} \\ $$$${also}\:{translated}\:{by}\:{a} \\ $$$$\Rightarrow{re}^{{i}\theta} +{a}\:{gives}\:{a}\:{circle}\:{with}\:{centre}\:{at}\:{a}\left[\mid{z}−{a}\mid={r}\right] \\ $$

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