If ∣z∣ = 3 , what is the maximum and minimum value of ∣z−1+i(√3) ∣ ?


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Question Number 111850 by bemath last updated on 05/Sep/20

$I f ∣ z ∣ = 3, w h a t i s t h e m a x i m u m$ $a n d m i n i m u m v a l u e o f ∣ z - 1 + i \sqrt{3} ∣ ?$
	Answered by ajfour last updated on 05/Sep/20

	$m a x ∣ z - 1 + i \sqrt{3} ∣= 5$ $m i n ∣ z - 1 + i \sqrt{3} ∣= 1$
	Commented by ajfour last updated on 05/Sep/20

	Answered by Her_Majesty last updated on 05/Sep/20

	$∣ z ∣= 3 \Rightarrow z = 3 \cos θ + 3 i \sin θ$ ${i t}^{'} s a c i r c l e w i t h r a d i u s 3$ $- 1 + i \sqrt{3} = 2 c o s \frac{2 π}{3} + 2 i \sin \frac{2 π}{3}$ $m a x ∣ z - 1 + i \sqrt{3} ∣= 5$ $m i n ∣ z - 1 + i \sqrt{3} ∣= 1$
	Answered by bemath last updated on 05/Sep/20

	$g r e a t s a n t u y b o t h . . . .$
	Answered by 1549442205PVT last updated on 05/Sep/20
	$Put z=a+bi⇒z−1+i(√3)=(a−1)+i(b+(√3)) ∣z∣=3⇔(√(a^2 +b^2 )) =3⇔a^2 +b^2 =9(1) ∣z−1+i(√3)∣=(√((a−1)^2 +(b+(√3))^2 )) We need find least and greaest value of the expression P=(√((a−1)^2 +(b+(√3))^2 )) which is equivalent to find least and greaest value of Q=(a−1)^2 +(b+(√3))^2 =13+2b(√3)−a(2) Apply the inequality ∣ax+by∣≤ (√((a^2 +b^2 )(x^2 +y^2 ))) we have ∣2b(√3)−a∣=∣(2(√3)).b+(−2).a∣≤ (√([(2(√3))^2 +(−2)^2 ](a^2 +b^2 ))) =(√(16.9))=12 ⇒−12≤2b(√3)−2a≤12 (3) From (2)and (3) we get 1≤Q≤25⇔1≤P≤5 P=1⇔Q=1⇔ { ((a=3/2)),((b=−3(√3)/2)) :} P=5⇔Q=25⇔ { ((a=−3/2)),((b=3(√3)/2)) :} Thus,P=∣z−1+i(√3) ∣ has the smallest value equal to 1 when (a,b)=((3/2),((−3(√3))/2)) i.e when z=(3/2)−((3i(√3))/2) and the greatest value equal to 5 when (a,b)=(((−3)/2),((3(√3))/2)) i.e z=((−3)/2)+((3i(√3))/2)$
	$Put z = a + bi \Rightarrow z - 1 + i \sqrt{3} = (a - 1) + i (b + \sqrt{3})$ $∣ z ∣= 3 \Leftrightarrow \sqrt{a^{2} + b^{2}} = 3 \Leftrightarrow a^{2} + b^{2} = 9 (1)$ $∣ z - 1 + i \sqrt{3} ∣= \sqrt{{(a - 1)}^{2} + {(b + \sqrt{3})}^{2}}$ $We need find least and greaest value$ $of the expression$ $P = \sqrt{{(a - 1)}^{2} + {(b + \sqrt{3})}^{2}} which is$ $equivalent to find least and greaest$ $value of Q = {(a - 1)}^{2} + {(b + \sqrt{3})}^{2}$ $= 13 + 2 b \sqrt{3} - a (2)$ $Apply the inequality ∣ ax + by ∣⩽$ $\sqrt{(a^{2} + b^{2}) (x^{2} + y^{2})} we have$ $∣ 2 b \sqrt{3} - a ∣=∣ (2 \sqrt{3}) . b + (- 2) . a ∣⩽$ $\sqrt{[{(2 \sqrt{3})}^{2} + {(- 2)}^{2}] (a^{2} + b^{2})} = \sqrt{16.9} = 12$ $\Rightarrow - 12 ⩽ 2 b \sqrt{3} - 2 a ⩽ 12 (3)$ $From (2) and (3) we get$ $1 ⩽ Q ⩽ 25 \Leftrightarrow 1 ⩽ P ⩽ 5$ $P = 1 \Leftrightarrow Q = 1 \Leftrightarrow {\begin{cases} a = 3 / 2 \\ b = - 3 \sqrt{3} / 2 \end{cases}$ $P = 5 \Leftrightarrow Q = 25 \Leftrightarrow {\begin{cases} a = - 3 / 2 \\ b = 3 \sqrt{3} / 2 \end{cases}$ $Thus, P =∣ z - 1 + i \sqrt{3} ∣ has the smallest$ $value equal to 1 when (a, b) = (\frac{3}{2}, \frac{- 3 \sqrt{3}}{2})$ $i . e when z = \frac{3}{2} - \frac{3 i \sqrt{3}}{2}$ $and the greatest value equal to 5$ $when (a, b) = (\frac{- 3}{2}, \frac{3 \sqrt{3}}{2}) i . e z = \frac{- 3}{2} + \frac{3 i \sqrt{3}}{2}$
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