Question

how to prove tan z is analytic using cauchy-riemann conditions

Answer 1

So what version of the C-R equations do you have? Is it expression in terms of x, y, u, v?

w = u + iv = f(z) = f(x + iy)

Answer 2

i'll do some steps...i am basically stuck at the integration part..idk...
so it goes like this
tan z
z=x+iy
tan (x+iy)=(tan x+i tanh y)/1-itanx tanh y

Answer 3

Right, and from that you can derive the real and imaginary parts. Then show those function u and v satisfy the C-R equations.

Answer 4

ya but i doubt its right
i need some one to check them

Answer 5

tan iy=tanh y

Answer 6

Yes, right; bring out the i.

You know it's right if the resulting functions do satisfy the C-R equations.

Answer 7

ya as i told u i am stuck in the differentiation part

Answer 8

\[u=(\tan x+\tan hx \tan ^{2}y)/1+\tan ^{2}hx \tan ^{2}y\]

Answer 9

this is what i got for real part but i dont knw how to proceed frm here

Answer 10

I think the differentiation part is the easy part.
tan(x+iy) can be simplified into:
\[\tan(x+iy)={\sin(2x) \over \cos(2x)+\cosh(2y)}+{i\sinh(2y) \over \cos(2x)+\cosh(2y)}.\]
Then the real part is \(\large \large u(x,y)=\large{\frac{\sin(2x)}{\cos(2x)+\cosh(2y)}}.\)

And the imaginary part is \(\large v(x,y)=\large{\frac{\sinh(2y)}{\cos(2x)+\cosh(2y)}}.\)

Given that, we can find the partial derivatives as:

\[u_x={2\cos(2x)(\cos(2x)+\cosh(2y))+2\sin^2(2x)\over (\cos(2x)+\cosh(2y))^2}\]
\[={2+\cos(2x)\cosh(2y) \over (\cos(2x)+\cosh(2y))^2}\]

\[v_y={2\cosh(2y)(\cos(2x)+\cosh(2y))-2\sinh^2(2x) \over (\cos(2x)+\cosh(2y))^2}\]
\[={2+\cosh(2y)\cos(2x) \over (\cos(2x)+\cosh(2y))^2}\]
Thus \(u_x=v_y\).

Answer 11

In a similar manner you can show the other C-R equation \(u_y=-v_x\).

Answer 12

let me try this way...thanz a lot

Answer 13

ya bt i ddnt understand dis!

Answer 14

cn u help me ?
i gt stuck with tan(x+iy)=tanx+itanhy /(1-itanxtanhy)

Answer 15

what should i do next?