if-z-1-z-C-a-R-find-max-value-of-z-2-az-1-

Question Number 449 by 123456 last updated on 25/Jan/15

if ∣ z ∣= 1, z \in C, a \in R

find \max value of ∣ z^{2} + a z - 1 ∣

Answered by prakash jain last updated on 09/Jan/15

z = i

z^{2} = - 1

z^{2} + a z - 1 = - 2 + a i

∣ - 2 + a i ∣= \sqrt{4 + a^{2}}

Commented by prakash jain last updated on 09/Jan/15

z = x + i y

z^{2} = x^{2} - y^{2} + i 2 x y

y = \sqrt{1 - x^{2}}

z^{2} = 2 x^{2} - 1 + 2 i x \sqrt{1 - x^{2}}

a z = a x + i a \sqrt{1 - x^{2}}

z^{2} + a z - 1 = 2 x^{2} - 1 + 2 i x \sqrt{1 - x^{2}} + a x + i a \sqrt{1 - x^{2}} - 1

= (2 x^{2} + a x - 2) + i (2 x + a) \sqrt{1 - x^{2}}

∣ z^{2} + a z - 1 ∣^{2} = {(2 x^{2} + a x - 2)}^{2} + {(2 x + a)}^{2} (1 - x^{2})

4 x^{4} + a^{2} x^{2} + 4 + 4 {a x}^{3} - 8 x^{2} - 4 a x + (4 x^{2} + 4 a x + a^{2}) (1 - x^{2})

= 4 - 8 x^{2} + 4 x^{2} + a^{2}

= a^{2} + 4 - 4 x^{2}

Maximum value when x = 0

and is given by

∣ z^{2} + a z - 1 ∣= \sqrt{a^{2} + 4}