Article 29
Functions of Pure Imaginaries.

In the defining identities

coshu = 1 + 1 2!u2 + 1 4!u4 + , sinhu = 1 + 1 3!u3 + 1 5!u5 + ,

put for u the pure imaginary iy, then

coshiy = 1 1 2!y2 + 1 4!y4 = cosy,  (57) sinhiy = iy + 1 3!(iy)3 + 1 5!(iy)5 + = i y 1 3!y3 + 1 5!y5 = isiny,  (58)  and, by division, tanhiy = itany.  (59)

These formulas serve to interchange hyperbolic and circular functions. The hyperbolic cosine of a pure imaginary is real, and the hyperbolic sine and tangent are pure imaginaries.

The following table exhibits the variation of sinhu, coshu. tanhu, expu, as u takes a succession of pure imaginary values.






usinhucoshutanhuexpu





00101





1 4iπ.7i.712i.7(1 + i)





1 2iπi0ii





3 4iπ.7i .7 i.7(1 i)





iπ0 10 1





5 4iπ .7i .7i .7(1 + i)





3 2iπ i0i i





7 4iπ .7i.7 i .7(1 i)





2iπ0101





Prob. 81.
Prove the following identities: cos y = cosh iy = 1 2 exp iy + exp(iy) , sin y = 1 i sinh iy = 1 2i exp iy exp(iy) , cos y + isin y = cosh iy + sinh iy = exp iy, cos y isin y = cosh iy sinh iy = exp (iy), cos iy = cosh y,sin iy = isinh y.
Prob. 82.
Equating the respective real and imaginary parts on each side of the equation cos ny + isin ny = (cos y + isin y)n, express cos ny in powers of cos y, sin y; and hence derive the corresponding expression for cosh ny.
Prob. 83.
Show that, in the identities (57) and (58), y may be replaced by a general complex, and hence that

sinh(x ±iy) = ±isin(y ix), cosh(x ±iy) = cos(y ix), sin(x ±iy) = ±isinh(y ix), cos(x ±iy) = cosh(y ix).
Prob. 84.
From the product-series for sin x derive that for sinh x:

sin x = x 1 x2 π2 1 x2 22π2 1 x2 32π2 , sinh x = x 1 + x2 π2 1 + x2 22π2 1 + x2 32π2 .