f(z) = sin(2z). I have so far got to:

f(x + iy) = sin[2(x+iy)

f(x + iy) = sin(2x) cos(2iy) + cos(2x) sin(2iy).

I'm stuck on converting this into the form f(x + iy) = u(x,y) + iv(x,y).

Any help on that step? After that step I should be ok.

Printable View

- May 20th 2009, 06:55 PMkiwijoeyShow this function obeys the Cauchy Riemann equations
f(z) = sin(2z). I have so far got to:

f(x + iy) = sin[2(x+iy)

f(x + iy) = sin(2x) cos(2iy) + cos(2x) sin(2iy).

I'm stuck on converting this into the form f(x + iy) = u(x,y) + iv(x,y).

Any help on that step? After that step I should be ok. - May 20th 2009, 07:35 PMputnam120
I haven't tried to work it all out but I would suggest that you write $\displaystyle \sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$.