The question says to solve sin (z) = cosh (2) by equating the real and imaginary parts.

These are the two equations to solve.

From this, I am just supposed to include when

then

Printable View

- Dec 22nd 2010, 03:52 PMdwsmithsin (z) = cosh (2)
The question says to solve sin (z) = cosh (2) by equating the real and imaginary parts.

These are the two equations to solve.

From this, I am just supposed to include when

then - Dec 22nd 2010, 07:14 PMProve It
From Equation 2, you have gotten two cases, OR .

Case 1: .

Substitute into Equation 1 to give

.

But since for all , that means we can only accept even values of , and therefore, that restricts to be .

When that is the case, .

So the first solution is .

Case 2: .

Substitute into Equation 1 to give

according to Mathematica.

So the second solution is . - Dec 22nd 2010, 07:56 PMdwsmith
- Dec 22nd 2010, 09:51 PMChris L T521