Z-C-satisfies-the-condition-Z-3-Then-find-the-least-value-of-Z-1-Z-

Question Number 49272 by rahul 19 last updated on 05/Dec/18

Z ϵ C s a t i s f i e s t h e c o n d i t i o n ∣ Z ∣⩾ 3 .

T h e n f i n d t h e l e a s t v a l u e o f ∣ Z + \frac{1}{Z} ∣ ?

Answered by tanmay.chaudhury50@gmail.com last updated on 05/Dec/18

Commented by tanmay.chaudhury50@gmail.com last updated on 05/Dec/18

Commented by rahul 19 last updated on 05/Dec/18

thank you sir! ��

Answered by mr W last updated on 05/Dec/18

Z = r (\cos θ + i \sin θ)

∣ Z ∣= r ⩾ 3

Z + \frac{1}{Z} = r (\cos θ + i \sin θ) + \frac{1}{r (\cos θ + i \sin θ)}

Z + \frac{1}{Z} = r (\cos θ + i \sin θ) + \frac{\cos θ - i \sin θ}{r}

Z + \frac{1}{Z} = (r + \frac{1}{r}) \cos θ + i (r - \frac{1}{r}) \sin θ

Z + \frac{1}{Z} = \frac{1}{r} [(r^{2} + 1) \cos θ + i (r^{2} - 1) \sin θ]

∣ Z + \frac{1}{Z} ∣= \frac{1}{r} \sqrt{{(r^{2} + 1)}^{2} \cos^{2} θ + {(r^{2} - 1)}^{2} \sin^{2} θ}

∣ Z + \frac{1}{Z} ∣= \frac{1}{r} \sqrt{{(r^{2} + 1)}^{2} - [{(r^{2} + 1)}^{2} - {(r^{2} - 1)}^{2}] \sin^{2} θ}

∣ Z + \frac{1}{Z} ∣= \frac{1}{r} \sqrt{{(r^{2} + 1)}^{2} - 4 r^{2} \sin^{2} θ}

∣ Z + \frac{1}{Z} ∣= \sqrt{{(r + \frac{1}{r})}^{2} - {4 \sin}^{2} θ}

⩾ \sqrt{{(r + \frac{1}{r})}^{2} - 4}

= \sqrt{{(r - \frac{1}{r})}^{2}}

=∣ r - \frac{1}{r} ∣ (i n c r e a s i n g f u n c t i o n u p o n r = 1)

⩾∣ 3 - \frac{1}{3} ∣= \frac{8}{3} = 2.67

Commented by rahul 19 last updated on 05/Dec/18

thank you sir! ��