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Question Number 131094 by Chhing last updated on 01/Feb/21

Calculate

1 / I = \oint_{c^{+}} \frac{zdz}{{(z - 1)}^{2} (z^{2} - 2 z + 1 - 2 i)}, C = {z / ∣ z ∣= 2}

2 / J = \oint_{c^{+}} \frac{ch (z) dz}{z (e^{z} - 1)}, C = {z / ∣ z - 3 i ∣= 4}

3 / K = \oint_{c^{+}} \frac{\sin (z) dz}{z^{3} {(z + 1)}^{2}}, C = {z / ∣ z ∣= 2}

Commented by mathmax by abdo last updated on 02/Feb/21

sir what do you mean by C^{+} ? define it \dots

Answered by mathmax by abdo last updated on 01/Feb/21

1) I = \int_{C^{+}} \frac{zdz}{{(z - 1)}^{2} (z^{2} - 2 z + 1 - 2 i)} let φ (z) = \frac{z}{{(z - 1)}^{2} (z^{2} - 2 z + 1 - 2 i)}

z^{2} - 2 z + 1 - 2 i = 0 \Rightarrow Δ^{^{'}} = 1 - (1 - 2 i) = 2 i \Rightarrow z_{1} = 1 + \sqrt{2 i} = 1 + \sqrt{2} e^{\frac{i π}{4}}

z_{2} = 1 - \sqrt{2} e^{\frac{i π}{4}} we have z_{1} = 1 + \sqrt{2} {\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}} = 1 + 1 + i = 2 + i \Rightarrow

∣ z_{1} ∣= \sqrt{4 + 1} = \sqrt{5} > \sqrt{2} and z_{2} = 1 - \sqrt{2} {\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}} = 1 - 1 + i = i \Rightarrow∣ z_{1} ∣< 2

z_{2} \in C^{+} also z = 1 is double pole for φ \Rightarrow

I = \int_{C^{+}} φ (z) dz = 2 i π {Res (φ, 1) + Res (φ, i)}

Res (φ, 1) = \lim_{z \to 1} \frac{1}{(2 - 1)!} {{(z - 1)}^{2} φ (z)}^{(1)}

= \lim_{z \to 1} {(\frac{z}{z^{2} - 2 z + 1 - 2 i})}^{(1)}

= \lim_{z \to 1} (\frac{z^{2} - 2 z + 1 - 2 i - z (2 z - 2)}{{(z^{2} - 2 z + 1 - 2 i)}^{2}})

= \frac{1 - 2 + 1 - 2 i - o}{{(1 - 2 + 1 - 2 i)}^{2}} = \frac{- 2 i}{{(- 2 i)}^{2}} = \frac{- 2 i}{- 4} = \frac{i}{2}

Res (φ, i) = \lim_{z \to i} (z - i) φ (z) = \lim_{z \to i} (z - i) \frac{z}{{(z - 1)}^{2} (z - i) (z - 2 - i)}

= \lim_{z \to i} \frac{z}{{(z - 1)}^{2} (z - 2 - i)} = \frac{i}{{(i - 1)}^{2} (- 2)} = \frac{- i}{2 (- 2 i)} = \frac{1}{4} \Rightarrow

I = 2 i π {\frac{i}{2} + \frac{1}{4}} = - π + \frac{i π}{2}