Question

Write 3cos(x) + 2isin(x) in a form involving the (x+iy) factor... Help?! Thanks!

Answer 1

Your formula sent me on a wild goose
chase for a while, because I interpreted sin(2x) / cosh(2y)+cos(2x)
as:

sin(2x)
------- + cos(2x)
cosh(2y)

instead of:

sin(2x)
-----------------
cosh(2y)+ cos(2x)

I would expect to see this last expression written as

sin(2x)/(cosh(2y)+cos(2x))

Ahlfors' argument, as he admits, is heuristic, and is more of a
mnemonic, but he assures the reader that it can be proved.

In this case, u(0,0) = 0, and the key is that for an imaginary
argument iw (w real),

e^(iw) + e^(-iw)
cosh(iw) = ----------------
2

(you get this from the formula for cosh(x) when x is real, by just
replacing x by a complex value, in this case, a pure imaginary one).
Using Euler's formula:

e^(it) = cos(t) + i*sin(t)

cos(w) + i*sin(w) + cos(w) - i*sin(t)
cosh(iw) = -------------------------------------
2

= cos(w)

Also,

cosh(-iw) = cosh(i*(-w))

= cos(-w)

= cos(w)

Since i^4 = 1

1
- = -i
i

so

f(z) = 2*u(z/2,z/(2i)) - u(0,0)

sin(2*(z/2))
= 2 * -------------------------- - 0
cosh(2*(z/2i))+cos(2*(z/2)

sin(z)
= 2 * ------------------
cosh(-iz) + cos(z)

sin(z)
= 2 * ---------------
cos(z) + cos(z)

= tan(z)

Answer 2

w=3cos(x) + 2isin(x)= Ln|w| +iArg(w)

Answer 3

Woah Rohangrr! Impressive reply, but you lost me from the get go! .. How did you get sin(2x) / cosh(2y)+cos(2x) in the first place?

Answer 4

@feldy90 it was just to make easy for u!

Answer 5

nothing else!

Answer 6

see the work done

Answer 7

okay.. cheers!

Answer 8

feldy u must have give reward ft that

Answer 9

i did...

Answer 10

thanx!