Question

Use Cauchy Integral Formula to evaluate

Integral on Curve [ e ^ (z+1) ] / [ (z-i) ( z^2 + (i-1)z-i)^3 ]

Note : 1))please the figure is close curve I cannot paint it by writing but these points (3i,i,-i,-2i,-1) inside the figure if you need it when find the singularity inside the curve
2))Also I need all singularity ( then sure only if inside curve i need all one with the working of cauchy integral , then plus it to find the total integral

Answer 1

Answer on Question #50258 – Math – Complex Analysis

Use Cauchy Integral Formula to evaluate

Integral on Curve [ e ^ (z+1) ] / [ {(z-i) ( z^2 + (i-1)z-i) }^3 ]

Note : 1)) please the figure is close curve I cannot paint it by writing but these points (3i,i,-i,-2i,-1) inside the figure if you need it when find the singularity inside the curve

2)) Also I need all singularity ( then sure only if inside curve i need all one with the working of cauchy integral , then plus it to find the total integral

Solution

$$\oint_{C} \frac{e^{z+1}}{(z-i)(z^2+(i-1)z-i)^3} dz,$$

where $C$ – closed curve.

Let us find singularities:

Numerator hasn't finite zeros. Denominator certainly has some, so let's find them.

Consider

$$z - i = 0$$ $$z_1 = i$$

Consider

$$(z^2 + (i-1)z - i)^3 = 0$$ $$z^2 + (i-1)z - i = 0$$

Due to the Vieta's formulas:

$$z_2 + z_3 = -\frac{i-1}{1} = 1 - i$$ $$z_2 z_3 = -\frac{i}{1} = -i$$

It's obvious now that

$$z_2 = 1$$ $$z_3 = -i$$

As we said, numerator hasn't finite zeros, also $z_1 \neq z_2 \neq z_3$, thus, zeros 1 and -i of denominator are nothing else, but poles of order 3, zero I of denominator is simple pole (pole of order 1)

$$\frac{n!}{2\pi i} \oint_{C} \frac{f(z)}{(z-a)^{n+1}} dz = f^{(n)}(a)$$

is called Cauchy's integral formula,

hence

$$\oint_{C} \frac{f(z)}{(z - a)^{n+1}} dz = \frac{2\pi i}{n!} f^{(n)}(a)$$

If region enclosed with a curve contains more than one singularity, we split up area/curve into parts, which contain each only one singularity. Let the singularity 1 is not included in the region.

Trigonometric form of complex number (angle between $-\pi$ and $\pi$):

$$i + i = 2i = 2\left(\cos\left(\frac{\pi}{2}\right) + \operatorname{isin}\left(\frac{\pi}{2}\right)\right),$$ $$i - 1 = \sqrt{2}\left(-\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}\right) = \sqrt{2}\left(\cos\left(\frac{3\pi}{4}\right) + \operatorname{isin}\left(\frac{3\pi}{4}\right)\right)$$ $$\begin{aligned} (i - 1)^3 &= 2^{3/2}\left(\cos\left(\frac{3\pi \cdot 3}{4}\right) + \operatorname{isin}\left(\frac{3\pi \cdot 3}{4}\right)\right) = 2\sqrt{2}\left(\cos\left(\frac{9\pi}{4}\right) + \operatorname{isin}\left(\frac{9\pi}{4}\right)\right) \\ &= 2\sqrt{2}\left(\cos\left(\frac{\pi}{4}\right) + \operatorname{isin}\left(\frac{\pi}{4}\right)\right) = 2\sqrt{2} \cdot \frac{1 + i}{\sqrt{2}} = 2 + 2i \end{aligned}$$

By Euler formula, $e^{i+1} = e \cdot e^i = e(\cos(1) + \operatorname{isin}(1))$

For $z = i$ we have $f(z) = \frac{e^{z+1}}{(z+i)^3(z-1)^3}$, $n = 1, a = i$,

$$\begin{aligned} &\oint_{C_1} \frac{\frac{e^{z+1}}{(z+i)^3(z-1)^3}}{z - i} dz = 2\pi i f(i) = 2\pi i \frac{e^{i+1}}{(i+i)^3(i-1)^3} \\ &= 2\pi i \frac{e(\cos(1) + \operatorname{isin}(1))}{2^3\left(\cos\left(\frac{3\pi}{2}\right) + \operatorname{isin}\left(\frac{3\pi}{2}\right)\right) \cdot 2^{\frac{3}{2}} \left(\cos\left(\frac{3\pi}{4}\right) + \operatorname{isin}\left(\frac{3\pi}{4}\right)\right)} \\ &= 2\pi i \frac{e(\cos(1) + \operatorname{isin}(1)) \left(\cos\left(-\frac{3\pi}{2} - \frac{3\pi}{4}\right) + \operatorname{isin}\left(-\frac{3\pi}{2} - \frac{3\pi}{4}\right)\right)}{2^3 2^{\frac{3}{2}}} \\ &= 2\pi i \frac{e(\cos(1) + \operatorname{isin}(1)) \left(\cos\left(-\frac{9\pi}{4}\right) + \operatorname{isin}\left(-\frac{9\pi}{4}\right)\right)}{2^3 2^{\frac{3}{2}}} \\ &= 2\pi i \frac{e(\cos(1) + \operatorname{isin}(1)) \left(\cos\left(-\frac{\pi}{4}\right) + \operatorname{isin}\left(-\frac{\pi}{4}\right)\right)}{2^3 2^{\frac{3}{2}}} \\ &= 2\pi i \frac{e(\cos(1) + \operatorname{isin}(1)) (1 - i)}{2^3 2^{\frac{3}{2}} 2^{\frac{1}{2}}} = \pi e^{1+i} \frac{1 + i}{16} \end{aligned}$$

For $z = -i$ we have $f(z) = \frac{e^{z+1}}{(z-i)(z-1)^3}$, $n = 2, a = -i$

$$\oint_{C_2} \frac{\frac{e^{z+1}}{(z-i)(z-1)^3}}{(z+i)^3} dz = \frac{2\pi i}{2} f^*(-i) = \pi i f^*(-i) = \pi e^{1-i} \frac{19i - 27}{16}.$$

Method 1.

$$\begin{aligned} f^*(z) = \left(\frac{e^{z+1}}{(z-i)(z-1)^3}\right)' &= \frac{(e^{z+1})^*(z-i)(z-1)^3 - e^{z+1}((z-i)(z-1)^3)^*(z-i)^2(z-1)^6}{e^{z+1} \frac{(z-i)(z-1)^3 - 3(z-1)^2(z-i) - (z-1)^3}{(z-i)^2(z-1)^6}} \\ &= e^{z+1} \frac{(z-i)(z-1) - 3(z-i) - (z-1)}{(z-i)^2(z-1)^4} \\ &= e^{z+1} \frac{z^2 - z - iz + i - 3z + 3i - z + 1}{(z-i)^2(z-1)^4} = e^{z+1} \frac{z^2 - (5 + i)z + 4i + 1}{(z-i)^2(z-1)^4} \end{aligned}$$ $$\begin{aligned} f^*(z) = \left(e^{z+1} \frac{z^2 - (5 + i)z + 4i + 1}{(z-i)^2(z-1)^4}\right)' \\ = (e^{z+1})' \frac{z^2 - (5 + i)z + 4i + 1}{(z-i)^2(z-1)^4} + e^{z+1} \left(\frac{z^2 - (5 + i)z + 4i + 1}{(z-i)^2(z-1)^4}\right)' \\ = e^{z+1} \left(\frac{z^2 - (5 + i)z + 4i + 1}{(z-i)^2(z-1)^4}\right. \\ + \frac{(2z - 5 - i)(z-i)^2(z-1)^4 - (z^2 - (5 + i)z + 4i + 1)[2(z - i)(z-1)^4 + 4(z - i)^2(z-1)^3]}{(z - i)^4(z-1)^8} \\ = e^{z+1} \left(\frac{z^2 - (5 + i)z + 4i + 1}{(z-i)^2(z-1)^4}\right. \\ + \frac{(2z - 5 - i)(z - i)(z - 1) - (z^2 - (5 + i)z + 4i + 1)[2(z - 1) + 4(z - i)]}{(z - i)^4(z-1)^8} \end{aligned}$$

Method 2.

$$\begin{aligned} \frac{1}{(z-i)(z-1)^3} &= \frac{A}{z-i} + \frac{B}{(z-1)^3} + \frac{C}{(z-1)^2} + \frac{D}{z-1} \\ &= \frac{A(z-1)^3 + B(z-i) + C(z-1)(z-i) + D(z-i)(z-1)^2}{(z-i)(z-1)^3} \end{aligned}$$

Equate the left-hand and the right-hand sides of the previous equalities.

If $z = 1$ then $B(1 - i) = 1$, hence $B = \frac{1}{1-i} = \frac{1+i}{(1-i)(1+i)} = \frac{1+i}{1-i^2} = \frac{1+i}{1+1} = \frac{1+i}{2}$

If $z = i$ then $A(i - 1)^3 = 1$, hence

$$A = \frac{1}{(i-1)^3} = \frac{(i+1)^3}{(i^2 - 1)^3} = \frac{i^3 + 3i^2 + 3i + 1}{(-2)^3} = \frac{i^2 \cdot i + 3 \cdot (-1) + 3i + 1}{-8} = \frac{3i - i + 1 - 3}{-8} = \frac{2i - 2}{-8} = \frac{1 - i}{4}$$

Consider $\frac{1}{(z-i)(z-1)^3} - \frac{A}{z-i} - \frac{B}{(z-1)^3} = \frac{C}{(z-1)^2} + \frac{D}{z-1}$,

$$\frac{1}{(z-i)(z-1)^3} - \frac{1 - i}{4(z-i)} - \frac{1 + i}{2(z-1)^3} = \frac{C}{(z-1)^2} + \frac{D}{z-1}$$ $$\frac{4 - (1 - i)(z - 1)^3 - 2(1 + i)(z - i)}{4(z - i)(z - 1)^3} = \frac{4C(z - i)(z - 1) + 4D(z - i)(z - 1)^2}{4(z - i)(z - 1)^3},$$

hence

$$\begin{array}{l} 4 - (1 - i) (z ^ {3} - 3 z ^ {2} + 3 z - 1) - 2 (1 + i) (z - i) = 4 C (z ^ {2} - (1 + i) z + i) + 4 D (z - i) (z ^ {2} - 2 z + 1), \\ (i - 1) z ^ {3} + (3 - 3 i) z ^ {2} + ((3 i - 3) - 2 (1 + i)) z + 1 - i + 4 + 2 i + 2 i ^ {2} = 4 C z ^ {2} - 4 C (1 + i) z + 4 C i + 4 D (z ^ {3} - 2 z ^ {2} + z - i z ^ {2} + 2 i z - i), \\ (i - 1) z ^ {3} + (3 - 3 i) z ^ {2} + (i - 5) z + 3 + i = 4 D z ^ {3} + (4 C - 8 D - 4 D i) z ^ {2} + (- 4 C (1 + i) + 4 D + 8 D i) z + 4 C i - 4 D i, \end{array}$$

hence $4D = i - 1$ , $4C - 8D - 4Di = 3 - 3i$ , $-4C(1 + i) + 4D + 8Di = i - 5$ , $4Ci - 4Di = 3 + i$

Solve for $D = \frac{i - 1}{4}$ , substitute into other expressions: $4C - 2i + 2 - i(i - 1) = 3 - 3i$ , $-4C(1 + i) + i - 1 + i(2i - 2) = i - 5$ , $4Ci - i(i - 1) = i + 3$ .

Simplify these expressions: $4C - i + 3 = 3 - 3i$ , $-4C(1 + i) - i - 3 = i - 5$ , $4Ci + i + 1 = i + 3$ , so $C = -\frac{i}{2}$ , check $2i(1 + i) - i - 3 = i - 5$ , $-2i^2 + i + 1 = i + 3$ , which also hold true.

Thus,

$$\begin{array}{l} A = \frac {1 - i}{4}, B = \frac {1 + i}{2}, C = - \frac {i}{2}, D = \frac {i - 1}{4} \\ \frac {1}{(z - i) (z - 1) ^ {3}} = \frac {A}{z - i} + \frac {B}{(z - 1) ^ {3}} + \frac {C}{(z - 1) ^ {2}} + \frac {D}{z - 1} = \\ = \frac {1 + i}{2 (z - 1) ^ {3}} - \frac {i}{2 (z - 1) ^ {2}} + \frac {i - 1}{4 (z - 1)} + \frac {1 - i}{4 (z - i)} \\ \end{array}$$

Trigonometric form of complex number (angle between $-\pi$ and $\pi$ ):

$$- i - 1 = \sqrt {2} \left(- \frac {1}{\sqrt {2}} - i \frac {1}{\sqrt {2}}\right) = \sqrt {2} \left(\cos \left(- \frac {3 \pi}{4}\right) + i \sin \left(- \frac {3 \pi}{4}\right)\right),$$

hence

$$\begin{array}{l} (- i - 1) ^ {- 5} = 2 ^ {- \frac {5}{2}} \left(\cos \left(\frac {3 \pi \cdot 5}{4}\right) + i \sin \left(\frac {3 \pi \cdot 5}{4}\right)\right) = \frac {1}{4 \sqrt {2}} \left(\cos \left(\frac {1 5 \pi}{4}\right) + i \sin \left(\frac {1 5 \pi}{4}\right)\right) = \\ = \frac {1}{4 \sqrt {2}} \left(\cos \left(- \frac {\pi}{4}\right) + i \sin \left(- \frac {\pi}{4}\right)\right) = \frac {1}{4 \sqrt {2}} \left(\frac {1}{\sqrt {2}} - i \frac {1}{\sqrt {2}}\right) = \frac {1 - i}{1 6}, \\ \end{array}$$ $$\begin{array}{l} (- i - 1) ^ {- 4} = 2 ^ {- \frac {4}{2}} \left(\cos (3 \pi) + i \sin (3 \pi)\right) = \frac {1}{4} \left(\cos (\pi) + i \sin (\pi)\right) = - \frac {1}{4}, \\ (- i - 1) ^ {- 3} = 2 ^ {- \frac {3}{2}} \left(\cos \left(\frac {3 \pi \cdot 3}{4}\right) + i \sin \left(\frac {3 \pi \cdot 3}{4}\right)\right) = \frac {1}{2 \sqrt {2}} \left(\cos \left(\frac {9 \pi}{4}\right) + i \sin \left(\frac {9 \pi}{4}\right)\right) = \\ \frac {1}{2 \sqrt {2}} \left(\cos \left(\frac {\pi}{4}\right) + i \sin \left(\frac {\pi}{4}\right)\right) = \frac {1}{2 \sqrt {2}} \left(\frac {1}{\sqrt {2}} + i \frac {1}{\sqrt {2}}\right) = \frac {i + 1}{4}, \\ (- i - 1) ^ {- 2} = 2 ^ {- \frac {2}{2}} \left(\cos \left(\frac {3 \pi \cdot 2}{4}\right) + i \sin \left(\frac {3 \pi \cdot 2}{4}\right)\right) = \frac {1}{2} \left(\cos \left(\frac {6 \pi}{4}\right) + i \sin \left(\frac {6 \pi}{4}\right)\right) = \\ \end{array}$$ $$= \frac {1}{2} \left(\cos \left(\frac {3 \pi}{2}\right) + i \sin \left(\frac {3 \pi}{2}\right)\right) = \frac {- i}{2},$$ $$(- i - 1) ^ {- 1} = 2 ^ {- \frac {1}{2}} \left(\cos \left(\frac {3 \pi}{4}\right) + i \sin \left(\frac {3 \pi}{4}\right)\right) = \frac {1}{\sqrt {2}} \left(- \frac {1}{\sqrt {2}} + i \frac {1}{\sqrt {2}}\right) = \frac {i - 1}{2}.$$

Applying formulas $\left((ax + b)^{\beta}\right)^{(n)} = a^{n}\beta (\beta -1)\ldots (\beta -n + 1)(ax + b)^{\beta -n}$, obtain

$$((z - 1) ^ {- 3}) ^ {(2)} = - 3 (- 3 - 1) (z - 1) ^ {- 3 - 2} = \frac {1 2}{(z - 1) ^ {5}}$$ $$((z - 1) ^ {- 3}) ^ {(1)} = - 3 (z - 1) ^ {- 3 - 1} = - \frac {3}{(z - 1) ^ {4}}$$ $$((z - 1) ^ {- 2}) ^ {(2)} = - 2 (- 2 - 1) (z - 1) ^ {- 2 - 2} = \frac {6}{(z - 1) ^ {4}}$$ $$((z - 1) ^ {- 2}) ^ {(1)} = - 2 (z - 1) ^ {- 2 - 1} = - \frac {2}{(z - 1) ^ {3}}$$ $$((z - 1) ^ {- 1}) ^ {(2)} = - 1 (- 1 - 1) (z - 1) ^ {- 1 - 2} = \frac {2}{(z - 1) ^ {3}}$$ $$((z - 1) ^ {- 1}) ^ {(1)} = - (z - 1) ^ {- 1 - 1} = - \frac {1}{(z - 1) ^ {2}}$$ $$((z - i) ^ {- 1}) ^ {(2)} = - 1 (- 1 - 1) (z - i) ^ {- 1 - 2} = \frac {2}{(z - i) ^ {3}}$$ $$((z - i) ^ {- 1}) ^ {(1)} = - (z - i) ^ {- 1 - 1} = - \frac {1}{(z - i) ^ {2}}$$

Besides, $(e^{z + 1})^{\cdot} = (e^{z + 1})^{\cdot \cdot} = e^{z + 1}$

Using Leibnitz formula, rewrite

$$\begin{array}{l} \left(\frac {e ^ {z + 1}}{(z - i) (z - 1) ^ {3}}\right) ^ {\cdot} = \frac {1 + i}{2} \left(\frac {e ^ {z + 1}}{(z - 1) ^ {3}}\right) ^ {\cdot} - \frac {i}{2} \left(\frac {e ^ {z + 1}}{(z - 1) ^ {2}}\right) ^ {\cdot} + \frac {i - 1}{4} \left(\frac {e ^ {z + 1}}{z - 1}\right) ^ {\cdot} + \frac {1 - i}{4} \left(\frac {e ^ {z + 1}}{z - i}\right) ^ {\cdot} = \\ = \frac {1 + i}{2} (e ^ {z + 1} (z - 1) ^ {- 3}) ^ {\cdot} - \frac {i}{2} (e ^ {z + 1} (z - 1) ^ {- 2}) ^ {\cdot} + \frac {i - 1}{4} (e ^ {z + 1} (z - 1) ^ {- 1}) ^ {\cdot} \\ + \frac {1 - i}{4} (e ^ {z + 1} (z - i) ^ {- 1}) ^ {\cdot} = \\ = \frac {1 + i}{2} \left[ (e ^ {z + 1}) ^ {(2)} (z - 1) ^ {- 3} + 2 (e ^ {z + 1}) ^ {(1)} ((z - 1) ^ {- 3}) ^ {(1)} + e ^ {z + 1} ((z - 1) ^ {- 3}) ^ {(2)} \right] \\ - \frac {i}{2} \left[ (e ^ {z + 1}) ^ {(2)} (z - 1) ^ {- 2} + 2 (e ^ {z + 1}) ^ {(1)} ((z - 1) ^ {- 2}) ^ {(1)} + e ^ {z + 1} ((z - 1) ^ {- 2}) ^ {(2)} \right] \\ + \frac {i - 1}{4} \left[ (e ^ {z + 1}) ^ {(2)} (z - 1) ^ {- 1} + 2 (e ^ {z + 1}) ^ {(1)} ((z - 1) ^ {- 1}) ^ {(1)} + e ^ {z + 1} ((z - 1) ^ {- 1}) ^ {(2)} \right] + \\ \frac {1 - i}{4} \left[ (e ^ {z + 1}) ^ {(2)} (z - i) ^ {- 1} + 2 (e ^ {z + 1}) ^ {(1)} ((z - i) ^ {- 1}) ^ {(1)} + e ^ {z + 1} ((z - i) ^ {- 1}) ^ {(2)} \right] = \\ \end{array}$$ $$= \frac {1 + i}{2} e ^ {z + 1} \left[ (z - 1) ^ {- 3} + 2 ((z - 1) ^ {- 3}) ^ {(1)} + ((z - 1) ^ {- 3}) ^ {(2)} \right]$$ $$\begin{array}{l} - \frac {i}{2} e ^ {z + 1} \left[ (z - 1) ^ {- 2} + 2 ((z - 1) ^ {- 2}) ^ {(1)} + ((z - 1) ^ {- 2}) ^ {(2)} \right] \\ + \frac {i - 1}{4} e ^ {z + 1} \left[ (z - 1) ^ {- 1} + 2 ((z - 1) ^ {- 1}) ^ {(1)} + ((z - 1) ^ {- 1}) ^ {(2)} \right] + \\ + \frac {1 - i}{4} e ^ {z + 1} \left[ (z - i) ^ {- 1} + 2 ((z - i) ^ {- 1}) ^ {(1)} + ((z - i) ^ {- 1}) ^ {(2)} \right] = \\ = \frac {1 + i}{2} e ^ {z + 1} \left[ \frac {1}{(z - 1) ^ {3}} - \frac {6}{(z - 1) ^ {4}} + \frac {12}{(z - 1) ^ {5}} \right] \\ - \frac {i}{2} e ^ {z + 1} \left[ \frac {1}{(z - 1) ^ {2}} - \frac {4}{(z - 1) ^ {3}} + \frac {6}{(z - 1) ^ {4}} \right] \\ + \frac {i - 1}{4} e ^ {z + 1} \left[ \frac {1}{z - 1} - \frac {2}{(z - 1) ^ {2}} + \frac {2}{(z - 1) ^ {3}} \right] + \\ + \frac {1 - i}{4} e ^ {z + 1} \left[ \frac {1}{z - i} - \frac {2}{(z - i) ^ {2}} + \frac {2}{(z - i) ^ {3}} \right] = \\ = \frac {1 + i}{2} e ^ {z + 1} \left[ \frac {z ^ {2} - 2 z + 1 - 6 z + 6 + 12}{(z - 1) ^ {5}} \right] \\ - \frac {i}{2} e ^ {z + 1} \left[ \frac {z ^ {2} - 2 z + 1 - 4 z + 4 + 6}{(z - 1) ^ {4}} \right] \\ + \frac {i - 1}{4} e ^ {z + 1} \left[ \frac {z ^ {2} - 2 z + 1 - 2 z + 2 + 2}{(z - 1) ^ {3}} \right] + \\ + \frac {1 - i}{4} e ^ {z + 1} \left[ \frac {z ^ {2} - 2 i z - 1 - 2 z + 2 i + 2}{z - i} \right] = \\ = e ^ {z + 1} \left(\frac {1 + i}{2} \left[ \frac {z ^ {2} - 8 z + 19}{(z - 1) ^ {5}} \right] - \frac {i}{2} \left[ \frac {z ^ {2} - 6 z + 11}{(z - 1) ^ {4}} \right] + \frac {i - 1}{4} \left[ \frac {z ^ {2} - 4 z + 5}{(z - 1) ^ {3}} \right] + \frac {1 - i}{4} \left[ \frac {z ^ {2} - (2 i + 2) z + 2 i + 1}{(z - i) ^ {3}} \right]\right), \end{array}$$ $$\begin{array}{l} \left(\frac {e ^ {z + 1}}{(z - i) (z - 1) ^ {3}}\right) ^ {\prime \prime} \bigg | _ {z = - i} \\ = e ^ {i + 1} \left(\frac {1 + i}{2} \left[ \frac {i ^ {2} - 8 i + 1 9}{(i - 1) ^ {5}} \right] - \frac {i}{2} \left[ \frac {i ^ {2} - 6 i + 1 1}{(i - 1) ^ {4}} \right] + \frac {i - 1}{4} \left[ \frac {i ^ {2} - 4 i + 5}{(i - 1) ^ {3}} \right]\right) \\ \left. + \frac {1 - i}{4} \left[ \frac {i ^ {2} - (2 i + 2) i + 2 i + 1}{(z - i) ^ {3}} \right]\right) \\ \end{array}$$

If region enclosed with a curve contains more than one singularity, we split up area/curve into parts, which contain each only one singularity. Let the singularity 1 is not included in the region.

Thus,

$$\oint_ {C} \frac {e ^ {z + 1}}{(z - i) (z ^ {2} + (i - 1) z - i) ^ {3}} d z =$$ $$\begin{array}{l} \oint_{C} \frac{e^{z+1}}{(z-i)(z+i)^3(z-1)^3} dz = \\ \oint_{C_1} \frac{\frac{e^{z+1}}{(z+i)^3(z-1)^3}}{z-i} dz + \oint_{C_2} \frac{\frac{e^{z+1}}{(z-i)(z-1)^3}}{(z+i)^3} dz = \\ = 2\pi i \left. \frac{e^{z+1}}{(z+i)^3(z-1)^3} \right|_{z=i} + \pi i \left( \frac{e^{z+1}}{(z-i)(z-1)^3} \right)'' \Big|_{z=-i} = \\ = \pi e^{1+i} \frac{1+i}{16} + \pi e^{1-i} \frac{19i - 27}{16} \end{array}$$

The region does not contain $z = 1$.

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