Question

Test the analyticity of the function w=sin z hence show that the derivative of sin z = cos z.

Answer 1

to show function is analytic , you have to show that it satisfies CR equations and have continuous partial derivatives. derivative of sin(z) is cos(z) follows from following definition.
\[ \lim_{\Delta z \to 0}\frac{\sin(z+\Delta z) - \sin(z)}{\Delta z} = \cos z\]

Answer 2

What of analyticity? If I could just get a clear working,I can be able to tackle other problems.

Answer 3

sin(z) = sin(x+iy) <-- can you show that it satisfies CR equations?
just like i did in your other question

Answer 4

sin(z) = (e^(iz) - e^(iz))/(2i)
the simplest logic would be to use the definition like above. you know exponential function is entire function and the sum of entire function is also entire. so sin(z) is also an entire function i.e. holomorphic in entire domain.
holomorphic function are analytic.

Answer 5

Thanks but we are not deep into analytic functions yet,I've tried the other but got stuck midway,please show me a clear working of the problem.

Answer 6

show me your working first ... it will help me to help you.

Answer 7

hey..there's a working we have tried but got stuck while testing the partial derivatives..we have a problem keying in our working for both problems.

Answer 8

if you have typed, it would make thins easier for me .. anyway let me write it in short.

Answer 9

we would appreciate.

Answer 10

\[ \sin(x+iy)= \sin(x)\cos(iy) + \sin(iy)\cos(x) = \sin(x)\cosh(y) + i \cos(x) \sinh(y) \\ = u(x,y) + v(x,y)\]
^^ test the above function for CR equations. I have used following formula
\[ \sin(ix) = i \sinh(x) \\
\cos(ix) = \cosh(x)\]

Answer 11

also it is not difficult to show that they are continuous on entire domain. because they are just product of two continuous function.
to find the derivative, just use the usual definition used in elementary calculus.

Answer 12

Thank you.

Answer 13

thank you ..i appreciate your help

Answer 14

you are welcome ...
also to find the derivative, you can directly do
\[ f'(z) = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial y}\]
which you should get u+iv form of cos(x+iy)

Answer 15

Close, @experimentX, but the Wirtinger derivative has a \(1/2\).
https://en.wikipedia.org/wiki/Wirtinger_derivative

Answer 16

that comes from chain rule ... dw/dx = dw/dz.dz/dx + dw/dzbar. dzbar/dx
do same for y part and solve for dw/dz, you should get 1/2