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?? That #1 is nothing but basic high school trig:

   Cos(2z) = 2Cos²(z)-1

   Cos(3z) = Cos(z+2z) = Cos(z)Cos(2x)-Sin(z)Sin(2z) = ...

Your #2 doesn't seem to make sense. Are you trying to write the hyperbolic sine and cosine in terms of the circular sine and cosine? What is your definition of the hyperbolics? I prefer:

   Sinh(z) = [e^z - e^(-z)] / 2   and   Cosh(z) = [e^z + e^(-z)] / 2

To connect that with the circular sine and cosine, do you know Euler's Formula (http://en.wikipedia.org/wiki/Euler%27s_formula ):

   e^(iz) = Cos(z) + i×Sin(z)

To get the required "i" in the exponent of "e" in the Sinh and Cosh definitions, replace "z" with "-i(iz)":

   Sinh(z) = [e^(i(-iz) - e^(i(iz))] / 2 = [Cos(iz) + Sin(-iz) - ...

Remember that Cos(-X) = Cos(X) and Sin(-X) = -Sin(X).

Your #3 is simple (especially if you use #2) provided you are using the exponential defintion of Sinh(z) and Cosh(z) and that you know Euler's formula. What's left is basic high school trigonometry: Special values of Sine and Cosine.

http://en.wikipedia.org/wiki/Trigonometric_functio...

That #4 is more interesting!

Remember that you are trying to find all solutions to:   Sin(z)=i.   and   Cos(z)=i

My suggestion is to replace "z" with "iw" so that you can use #2 to replace Sin with Sinh {and Cos with Cosh}, which is an expression in e^w. In fact, the result is just a quadratic equation in e^w {remember that e^(-w)=1/e^w, multiply through to clear the denominator}. Basic high school. Just remember that when you solve e^w=A in complex analysis, you get w=Ln(A)+2πiN for any integer N {since e^(2πi)=1}. Of course, both answers for #4 consist of a pair of arithmetic series of period 2π, just like ordinary high school trig with real numbers.