$$ \int _ { c } \frac { z - 1 } { ( z + 1 ) ^ { 2 } ( z - 2 ) } d z \quad c \cdot | z - i | = 2 $$

Question

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We are to evaluate

$$
\int_{c}\frac{z-1}{(z+1)^2(z-2)}\,dz,
$$

where $$ c $$ is the circle defined by

$$
|z-i|=2.
$$

**Step 1. Identify the singularities.**

The function

$$
f(z)=\frac{z-1}{(z+1)^2(z-2)}
$$

has singularities at $$ z=-1 $$ and $$ z=2 $$. The singularity at $$ z=-1 $$ is a double pole (order 2) and at $$ z=2 $$ is a simple pole (order 1).

**Step 2. Determine which singularities lie inside the contour.**

The circle is centered at $$ i $$ with radius 2.

- For $$ z=-1 $$, compute

$$
  |(-1)-i|=\sqrt{(-1)^2+(-1)^2}=\sqrt{1+1}=\sqrt{2}\approx1.414.
  $$

Since $$ \sqrt{2}<2 $$, the pole at $$ z=-1 $$ is inside $$ c $$.

- For $$ z=2 $$, compute

$$
  |2-i|=\sqrt{2^2+(-1)^2}=\sqrt{4+1}=\sqrt{5}\approx2.236.
  $$

Since $$ \sqrt{5}>2 $$, the pole at $$ z=2 $$ lies outside $$ c $$.

Only the pole at $$ z=-1 $$ contributes to the integral.

**Step 3. Compute the residue at $$ z=-1 $$.**

Since $$ z=-1 $$ is a double pole (order 2), the residue is given by

$$
\operatorname{Res}(f,-1)=\lim_{z\to-1}\frac{d}{dz}\left[(z+1)^2 f(z)\right].
$$

Notice that

$$
(z+1)^2 f(z)=\frac{z-1}{z-2}.
$$

Now, differentiate

$$
\frac{d}{dz}\left(\frac{z-1}{z-2}\right).
$$

Use the quotient rule. Let $$ u=z-1 $$ and $$ v=z-2 $$, so that $$ u'=1 $$ and $$ v'=1 $$. Then

$$
\frac{d}{dz}\left(\frac{z-1}{z-2}\right)=\frac{u'v-uv'}{v^2}=\frac{(1)(z-2)-(z-1)(1)}{(z-2)^2}=\frac{z-2-z+1}{(z-2)^2}=\frac{-1}{(z-2)^2}.
$$

Now, evaluate at $$ z=-1 $$:

$$
\operatorname{Res}(f,-1)=\frac{-1}{((-1)-2)^2}=\frac{-1}{(-3)^2}=\frac{-1}{9}.
$$

**Step 4. Apply the residue theorem.**

The residue theorem states

$$
\int_{c}f(z)\,dz=2\pi i \sum_{\text{inside } c}\operatorname{Res}(f,z),
$$

so we have

$$
\int_{c}\frac{z-1}{(z+1)^2(z-2)}\,dz = 2\pi i\left(-\frac{1}{9}\right) = -\frac{2\pi i}{9}.
$$

Thus, the answer is

$$
-\frac{2\pi i}{9}.
$$
The integral evaluates to $$ -\frac{2\pi i}{9} $$.

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