C-z-1-2i-Az-2-2z-A-dz-where-C-is-a-unit-circle-with-radius-1-

Question Number 131743 by frc2crc last updated on 08/Feb/21

\int_{C :∣ z ∣= 1} \frac{- 2 i}{{A z}^{2} + 2 z + A} d z w h e r e C i s a u n i t c i r c l e w i t h r a d i u s 1

Answered by mathmax by abdo last updated on 08/Feb/21

let φ (z) = \frac{- 2 i}{{az}^{2} + 2 z + a} poles of φ ?

Δ^{^{'}} = 1 - a^{2} case 1 ∣ a ∣< 1 \Rightarrow z_{1} = \frac{- 1 + \sqrt{1 - a^{2}}}{a} and z_{2} = \frac{- 1 - \sqrt{1 - a^{2}}}{a}

(we suppose a > 0) ∣ z_{1} ∣ - 1 = \frac{1}{a} (1 - \sqrt{1 - a^{2}}) - 1 = \frac{1 - \sqrt{1 - a^{2}} - a}{a}

= \frac{1 - a - \sqrt{1 - a^{2}}}{a} < 0 because

{(1 - a)}^{2} - (1 - a^{2}) = a^{2} - 2 a + 1 - 1 + a^{2} = {2 a}^{2} - 2 a = 2 a (a - 1) < 0

∣ z_{2} ∣ - 1 = \frac{1 + \sqrt{1 - a^{2}}}{a} - 1 = \frac{1 - a + \sqrt{1 - a^{2}}}{a} > 0 (out of circle) \Rightarrow

\int_{∣ z ∣= 1} φ (z) dz = 2 i π Res (φ, z_{1}) = 2 i π \times \frac{- 2 i}{a (z_{1} - z_{2})}

= \frac{4 π}{a . \frac{2 \sqrt{1 - a^{2}}}{a}} = \frac{2 π}{\sqrt{1 - a^{2}}}

case 2 ∣ a ∣> 1 \Rightarrow z_{1} = \frac{- 1 + i \sqrt{a^{2} - 1}}{a} and z_{2} = \frac{- 1 - i \sqrt{a^{2} - 1}}{a}

∣ z_{1} ∣ = \frac{1}{a} \sqrt{1 + a^{2} - 1} = 1 and ∣ z_{2} ∣= 1 \Rightarrow

\int_{∣ z ∣= 1} φ (z) dz = 2 i π {Res (φ, z_{1}) + Res (φ, z_{2})}

= 2 i π {\frac{- 2 i}{a (z_{1} - z_{2})} + \frac{- 2 i}{a (z_{2} - z_{1})}} = 0