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**Term Test 2 / Re: TT2B Problem 1**

« **on:**November 24, 2018, 10:38:02 PM »

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Cauchy's integral formula is $$ f(z) = \frac{1}{2𝜋i} \int_{Γ}\frac{f(z)}{z-a} dz$$

given formula is $$ \int_{Γ}\frac{dz}{z^2-6z+25} = \int_{Γ}\frac{dz}{(z-(3+4i))(z-(3-4i))}$$

For (a), as 3+4i is inside and 3-4i is outside

Let $$f(z)=\frac{1}{z-(3-4i)}$$

by Cauchy theorem $$ \int_{Γ}\frac{f(z) dz}{(z-(3+4i))}=2𝜋i f(3+4i) = \frac{2𝜋i}{8i} = \frac{𝜋}{4}$$

For (b), as 3-4i is inside and 3+4i is outside

Let $$f(z)=\frac{1}{z-(3+4i)}$$

by Cauchy theorem $$ \int_{Γ}\frac{f(z) dz}{(z-(3-4i))}=2𝜋i f(3-4i) = \frac{2𝜋i}{-8i} = \frac{-𝜋}{4}$$

For (c), as both points 3±4i are inside, we can use residue theorem

$$\int_{Γ}\frac{dz}{(z-(3+4i))(z-(3-4i))}$$

$$=2𝜋i(Res(f, 3+-4i) + Res(f, 3+4i))$$

$$=2𝜋i(\frac{1}{8i}+\frac{-1}{8i})=0$$

given formula is $$ \int_{Γ}\frac{dz}{z^2-6z+25} = \int_{Γ}\frac{dz}{(z-(3+4i))(z-(3-4i))}$$

For (a), as 3+4i is inside and 3-4i is outside

Let $$f(z)=\frac{1}{z-(3-4i)}$$

by Cauchy theorem $$ \int_{Γ}\frac{f(z) dz}{(z-(3+4i))}=2𝜋i f(3+4i) = \frac{2𝜋i}{8i} = \frac{𝜋}{4}$$

For (b), as 3-4i is inside and 3+4i is outside

Let $$f(z)=\frac{1}{z-(3+4i)}$$

by Cauchy theorem $$ \int_{Γ}\frac{f(z) dz}{(z-(3-4i))}=2𝜋i f(3-4i) = \frac{2𝜋i}{-8i} = \frac{-𝜋}{4}$$

For (c), as both points 3±4i are inside, we can use residue theorem

$$\int_{Γ}\frac{dz}{(z-(3+4i))(z-(3-4i))}$$

$$=2𝜋i(Res(f, 3+-4i) + Res(f, 3+4i))$$

$$=2𝜋i(\frac{1}{8i}+\frac{-1}{8i})=0$$

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