Question

Using Cauchy integral, calculate this....

Answer 1

\[\int\limits_{C} \frac{ e^{3z} dz }{ (z - \ln 2)^{4} }\]
C is the square which all angle at coordinate (\(\pm 1 \), \(\pm i \)) !

Answer 2

have idea @niksva ??

Answer 3

i m having no clue right now
gimme some time to think to think about it

Answer 4

ok

Answer 5

do u know cauchy integral general formula for a derivative of an analytic function?

Answer 6

@gerryliyana

Answer 7

little bit

Answer 8

the general expression is
\[f ^{n}(a) = \frac{ n! }{ 2*\pi*i }\int\limits_{C}^{} \ \frac{ f(z) dz }{ (z-a)^{n+1} }\]

Answer 9

is this ok? @gerryliyana ?

Answer 10

ok.., and then ?? what should i do ??

Answer 11

from the problem
it can be concluded that
\[f(z) = e ^{3z} \] , a = ln(2) and n=3
now we are going to find
\[f ^{3}(z) \] first then we will put z = ln2

Answer 12

is this ok?

Answer 13

\[f^{3} (\ln2) = \frac{ 3! }{ 2\pi i } \int\limits \frac{ e^{3x} dz }{ (z-\ln 2)4 }\]

Is this what you mean? how about C is the square which all angle at coordinate (±1, ±i) ??

Answer 14

@niksva

Answer 15

yes
u will be able to find out the integral after this step

Answer 16

i didn't get it :(..., i'm little bit confused

Answer 17

i m getting 72*pi*i

Answer 18

where are u getting confused?

Answer 19

how to get it ??

Answer 20

is the confusion lies in the formula? or in the calculation?

Answer 21

do i replace z become x+iy ??

Answer 22

no need to replace z with x+iy
first we will find \[f ^{3}(z)\]

Answer 23

f(z) = e^3z
f '(z) = 3e^3z
f ''(z) = 9 e^3z
f '''(z) = 27 e^3z
is this right?

Answer 24

ok..., then f'''(ln2) = 27 e^(3 ln 2) ???

Answer 25

yeah

Answer 26

are you sure about that ??

Answer 27

yeah i m sure

Answer 28

how about C is the square which all angle at coordinate (±1, ±i) ??

Answer 29

According to me
C is a square having its co-ordinates at (1,i) , (-1,-i) , (1 , -i) and (-1,i)

Answer 30

ok., thank you @niksva :) it's clearly now

Answer 31

Cauchy's differential formula yields:$$f^{(n)}(a)=\frac{n!}{2\pi i}\oint_\gamma\frac{f(z)}{(z-a)^{n+1}}\mathrm{d}z$$for any closed curve enclosing parameterized by \(\gamma\) that encloses \(a\).
Now, recognize that our integral is of a familiar form:$$\begin{align*}I&=\oint_C\frac{e^{3z}}{(z-\log2)^4}\mathrm{d}z\\&=\frac{2\pi i}{3!}\times\frac{3!}{2\pi i}\oint_C\frac{e^{3z}}{(z-\log2)^{3+1}}\mathrm{d}z\\&=\frac{2\pi i}{3!}f^{'''}(\log 2)\text{ where }f(z)=e^{3z}\\&=\frac{\pi i}3\times27e^{3\log 2}\\&=9\pi i\times2^3\\&=72\pi i\end{align*}$$

Answer 32

@oldrin.bataku : then for this questions, the answer is 72\(\pi i \) ??

Answer 33

yes

Answer 34

ok thank you