If-z-3-what-is-the-maximum-and-minimum-value-of-z-1-i-3-

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 \dots .

Answered by 1549442205PVT last updated on 05/Sep/20

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}