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


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

$Please how did ∣ z - a ∣= r became$ $z = a + {re}^{i θ} ?$
	Answered by Frix last updated on 11/Aug/23

	$∣ z - a ∣= r \Rightarrow z - a = r e^{i θ}$ $This step is always true, even for z - a \in R$ $because we can always find a θ : if z - a > 0$ $\Rightarrow θ = 0, if z - a < 0 \Rightarrow θ = π$ $z - a = r e^{i θ} \Leftrightarrow z = a + r e^{i θ}$
	Answered by deleteduser1 last updated on 11/Aug/23

	$∣ z - a ∣= r i s t h e e q u a t i o n f o r t h e c i r c l e c e n t e r e d$ $a t a w i t h r a d i u s r$ $S o, w e g e t z (p o i n t s o n t h e c i r c l e) b y t r a n s l a t i n g$ $a c i r c l e w i t h c e n t r e a t t h e o r i g i n (0, 0) a n d r a d i u s$ $r t o a$ ${r e}^{i θ} \Rightarrow t h e l o c u s o f p o i n t s g e n e r a t e d b y r o t a t i n g$ $a v e c t o r w i t h m a g n i t u d e r a r o u n d t h e o r i g i n (t h i s$ $g i v e s a c i r c l e)$ $+ a \Rightarrow t r a n s l a t i n g {r e}^{i θ} b y a, t h e c e n t r e (o r i g i n) i s$ $a l s o t r a n s l a t e d b y a$ $\Rightarrow {r e}^{i θ} + a g i v e s a c i r c l e w i t h c e n t r e a t a [∣ z - a ∣= r]$
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