Question

Find the value(s) of constant B such that :

integral on curve for [ (1)- (3z) + (2 B {z^4}) +( z^6) + (3 {z^7}) + ( 11 {z ^ 100} ] /[ z^5] dz =
integral on curve for [ e ^ {Bz} + 2 z ] / [ z^2 ] dz
where C is the unit circle oriented counterclockwise

Answer 1

Answer on Question #50266 - Math - Complex Analysis

Find the value(s) of constant B such that :

integral on curve for $[(1)\text{-} (3z) + (2B(z^{\wedge}4)) + (z^{\wedge}6) + (3(z^{\wedge}7)) + (11(z^{\wedge}100)] / [z^{\wedge}5]\mathrm{d}z =$

integral on curve for $\left[\mathrm{e}^{\wedge}\{\mathrm{Bz}\} +2\mathrm{z}\right] / \left[\mathrm{z}^{\wedge}2\right]\mathrm{d}z$

where C is the unit circle oriented counterclockwise

Solution

We use Cauchy's integral formula for closed Jordan regions

$$\frac {1}{2 \pi i} \oint_ {T} \frac {f (z)}{z - z _ {0}} d z = \left\{ \begin{array}{l l} f (z _ {0}), & \text {if } z _ {0} \text { inside the closed contour} \\ 0, & \text {if } z _ {0} \text { outside the closed contour} \end{array} \right.$$

and its consequence (Cauchy's differentiation formula)

$$f ^ {(n)} (z _ {0}) = \frac {n !}{2 \pi i} \oint_ {T} \frac {f (z)}{(z - z _ {0}) ^ {n + 1}} d z,$$

hence

$$\oint_ {T} \frac {f (z)}{(z - z _ {0}) ^ {n + 1}} d z = \frac {2 \pi i f ^ {(n)} \left(z _ {0}\right)}{n !}$$

In the given problem $z_0 = 0$ is inside the closed region, curve $T$ is the unit circle.

In the first integral $\oint_{|z| = 1}\frac{1 - 3z + 2Bz^4 + z^6 + 3z^7 + 11z^{100}}{z^5} dz$ choose

$$f (z) = 1 - 3 z + 2 B z ^ {4} + z ^ {6} + 3 z ^ {7} + 1 1 z ^ {1 0 0}, n = 4,$$

In the second integral $\oint_{|z| = 1}\frac{e^{Bz} + 2z}{z^2} dz$ take $g(z) = e^{Bz} + 2z, n = 1$ .

To differentiate expressions, we need the following formula:

$$\left(z ^ {k}\right) ^ {(n)} = \left\{ \begin{array}{l l} \frac {k !}{(k - n) !} z ^ {k - n}, & \text {if } k \geq n, \\ 0, & \text {if } k < n \end{array} \right.$$

Next, compute

$$\begin{array}{l} f ^ {(4)} (z) = (1 - 3 z + 2 B z ^ {4} + z ^ {6} + 3 z ^ {7} + 1 1 z ^ {1 0 0}) ^ {(4)} = | \text {linearity of the nth derivative} | = \\ = (z ^ {0}) ^ {(4)} - 3 (z ^ {1}) ^ {(4)} + 2 B (z ^ {4}) ^ {(4)} + (z ^ {6}) ^ {(4)} + 3 (z ^ {7}) ^ {(4)} + 1 1 (z ^ {1 0 0}) ^ {(4)} = 0 - 0 + 2 B \frac {4 !}{(4 - 4) !} z ^ {4 - 4} + \\ + \frac {6 !}{(6 - 4) !} z ^ {6 - 4} + 3 \frac {7 !}{(7 - 4) !} z ^ {7 - 4} + 1 1 \frac {1 0 0 !}{(1 0 0 - 4) !} z ^ {1 0 0 - 4} = 2 B \cdot 4! + \frac {6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 !}{2 !} z ^ {2} + 3 \frac {7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 !}{3 !} z ^ {3} + \\ + 1 1 \frac {1 0 0 \cdot 9 9 \cdot 9 8 \cdot 9 7 \cdot 9 6 !}{9 6 !} z ^ {9 6}, \\ \end{array}$$ $$f ^ {(4)} (0) = 2 B \cdot 4! = 4 \cdot 3 \cdot 2 \cdot 2 B = 4 8 B,$$ $$g ^ {\prime} (z) = \left(e ^ {B z} + 2 z\right) ^ {\cdot} = \left(e ^ {B z}\right) ^ {\cdot} + 2 z ^ {\cdot} = \left(e ^ {B z}\right) ^ {\cdot} + 2 = B e ^ {B z} + 2$$ $$g ^ {\prime} (0) = B + 2$$

Using formula (1), rewrite equality

$$\oint_{|z|=1} \frac{1 - 3z + 2Bz^4 + z^6 + 3z^7 + 11z^{100}}{z^5} dz = \oint_{|z|=1} \frac{e^{Bz} + 2z}{z^2} dz$$

as

$$\frac{2\pi i f^{(4)}(0)}{4!} = \frac{2\pi i g'(0)}{1!},$$

which gives

$$\frac{f^{(4)}(0)}{4 \cdot 3 \cdot 2!} = \frac{g'(0)}{1!},$$

so

$$f^{(4)}(0) = 4 \cdot 3 \cdot 2 \cdot g'(0),$$

which is equivalent to

$$48B = 24(B + 2),$$

i.e.

$$48B - 24B = 48,$$

hence

$$24B = 48$$

and finally obtain

$$B = 2.$$

Answer: $B = 2$

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