Have you tried going through the Cauchy-Riemann conditions with this function? If not, you should give it a go!
Why did Cauchy - RIemann invented this.. arrghh... help
Show that sin(z) satisfies the CAUCHY - RIEMANN conditions for analyticity for all values of z. Does [1/sin(z)] satisfy similar conditions? Calculate the derivative of [1/sin(z)] at z=0, . Comment on each answer.
f(z) = sin (z)
= sin (x + iy)
= sin x cosh y + i cos x sinh y
thus,
u(x,y)=sin x cosh y ....... v(x,y)= cos x sinh y
du/dx = cos x ............ dv/dx = -sin x
du/dy = -sinh y ........... dv/dy = cosh y
therefore I will compare it with the Cauchy Riemann formula.
Am I right to say it doesn't satisfies the condition??
Your calculation of the partial derivatives is incorrect. When calculating a partial derivative, all other variables are held constant, so are treated as constants. So with , when evaluating , the is treated as constant, and constant multiples in derivatives stay where they are... So it should be . Fix the rest.
Yes your right. I did not get 0 when I do it manually. By using my calculator my answer will get 0.
btw am i right to say that 1/sin (z) in (a+jb) format is
sin(x) cosh(y)/ sin^2 (x) cosh^2 (y) + cos^2 (x) sinh^2 (y) - j cos(x) sinh(y)/ sin^2 (x) cosh^2 (y) + cos^2 (x) sinh^2 (y) ??
Correct me if I'm wrong
It's nasty to do it manually..
If by theorem, 1/sin(z) will satisfy Cauchy - Riemann conditions for all values of z except z = k pi + pi/2, ( k=0, +- 1, +- 2 ...), where the denominator of the function equals to zero. ??
but I try it out, it seems it doesn't satisfy the equation. hmm...