yes. I think is more simplifying the equation.
I am trying to find the limit z = 0. When I see inside my text before they use l'hopital rule, they simplified the equation A to B. Thereafter they apply the l'hopital rule
Hahaha.. do this by hand?? oh my god! I would suffer from brain damage. ouh well. Ok a quick question. If I need to calculate the derivative of 1/sin(z) at z=0, +- pi/2, +- pi, +-3pi/2...
I used quotient rule to get -cos (z)/ sin^2 (z)
Thus, sub the values of z into the equation. I will...
It's nasty..
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 manually, it seems it doesn't satisfy the equation. hmm...
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
thanks for notifying. I mange to solve it. and yes it satifies the condition.
What if the equation is 1/sin (z)?
I try to conjugate the equation but end up with 0.
i guess i did it wrong.
How do I start with this equation?
Thanks.
Homework Statement
Show that sin(z) satisfies the condition. (Stated in the title)
Homework Equations
The Attempt at a Solution
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...
Will try it out. Actually, this is not homework. I am just trying out past year exam questions. Sad to say, solutions is not provided. So I will never know whether I am doing it right or wrong.
Hi Simon,
Is it possible to show me your method starting from $$z^6-1=\left [ \frac{64j(1-j)}{1+2j} \right ]^6$$ till the end equation $$z^6$$= ##a+jb##
please?