On the Function arc sin (x+iy).

Arthur Cayley

[From the Quarterly Journal of Pure and Applied Mathematics, vol. XV. (1878), pp. 171—174.]

The determination of the function in question, the arc to a given imaginary sine, is considered in Cauchy’s Exercises d’Analyse, &c., t. III. (1844), p. 382; but it appears, by two hydrodynamical papers by Mr. Ferrers and Mr. Lamb, Quarterly Mathematical Journal, t. XIII. (1874), p. 115, and t. XIV. (1875), p. 40, that the question is connected with the theory of confocal conics.

Taking c = \sqrt{a^{2} − b^{2}} a positive real quantity which may ultimately be put = 1, the question is to find the real quantities ξ, η, such that

ξ + iη = arc sin \frac{1}{c}(x + iy),

or say

x + iy = c sin(ξ + iη),

so that

x = c sin ξ cos iη, iy = c cos ξ sin iη.

It is convenient to remark that if a value of ξ +iη be ξ′ +iη′, then the general value is 2mπ + ξ′ + iη′ or (2m + 1)π − (ξ′ + iη′); hence, η may be made positive or negative at pleasure; cos iη is in each case positive, but \frac{1}{i} sin iη has the same sign as η; hence cos ξ has the same sign as x, but sin ξ has the same sign as y or the reverse sign, according as η is positive or negative; for any given values of x and y, we obtain, as will appear, determinate positive values of sin^{2} ξ and cos^{2} ξ; and the square roots of these must therefore be taken so as to give to sin ξ, cos ξ their proper signs respectively.

Suppose that λ, μ are the elliptic coordinates of the point (x, y); viz. that we have

\begin{aligned}\frac{x^{2}}{a^{2} + λ} + \frac{y^{2}}{b^{2} + λ} & = 1 \\ \frac{x^{2}}{a^{2} + μ} + \frac{y^{2}}{b^{2} + μ} & = 1 \end{aligned}

where a^{2} + λ, b^{2} + λ, and a^{2} + μ are positive, but b^{2} + μ is negative. Calling ρ, σ the distances of the point x, y from the points (c, 0) and (−c,), that is, assuming

\begin{aligned}ρ & = \sqrt{(x − c)^{2} + y^{2}}, \\ σ & = \sqrt{(x + c)^{2} + y^{2}}, \end{aligned}

then we have

\begin{aligned}\sqrt{a^{2} + λ} & = \frac{1}{2}(σ + ρ), \\ \sqrt{a^{2} + μ} & = \frac{1}{2}(σ − ρ), \end{aligned} whence also \begin{aligned}\sqrt{b^{2} + λ} & = \frac{1}{2} \sqrt{(σ + ρ)^{2} − 4c^{2}}, \\ \sqrt{b^{2} + μ} & = \frac{1}{2} \sqrt{(σ − ρ)^{2} − 4c^{2}}, \end{aligned}

which equations determine λ, μ as functions of x, y. Now we have

\begin{aligned}ρσ & = \sqrt{(x^{2} + y^{2} − c^{2})^{2} − 4c^{2}x^{2}} = \sqrt{(x^{2} − y^{2} − c^{2})^{2} + 4x^{2}y^{2}}, \\ ρ^{2} + σ^{2} & = 2(x^{2} + y^{2} + c^{2}); \end{aligned}

substituting herein for x, y their values

c sin ξ cos iη, −ci cos ξ sin iη,

we find

x^{2} − y^{2} − c^{2} = c^{2}{sin^{2} ξ cos^{2} iη + cos^{2} ξ sin^{2} iη − (sin^{2} ξ + cos^{2} ξ)(sin^{2} iη + cos^{2} iη)}
= −c^{2}(sin^{2} ξ sin^{2} iη + cos^{2} ξ cos^{2} iη),

whence

(x^{2} − y^{2} − c^{2})^{2} + 4x^{2}y^{2} = c^{4}(cos^{2} ξ cos^{2} iη + sin^{2} ξ sin^{2} iη)^{2} − 4c^{4} sin^{2} ξ cos^{2} ξ sin^{2} iη cos^{2} iη
= c^{4}(cos^{2} ξ cos^{2} iη − sin^{2} ξ sin^{2} iη)^{2}.

Hence

2ρσ = 2c^{2}(cos^{2} ξ cos^{2} iη − sin^{2} ξ sin^{2} iη),

and

ρ^{2} + σ^{2} = 2c^{2}(sin^{2} ξ cos^{2} iη − cos^{2} ξ sin^{2} iη + 1);

hence

(ρ + σ)^{2} = 2c^{2}(cos^{2} iη − sin^{2} iη + 1), = 4c^{2} cos^{2} iη,
(ρ − σ)^{2} = 2c^{2}(sin^{2} ξ − cos^{2} ξ + 1), = 4c^{2} sin^{2} ξ.

Consequently

\begin{aligned}a^{2} + λ & = c^{2} cos^{2} iη, \\ a^{2} + μ & = c^{2} sin^{2} ξ, \end{aligned} and thence \begin{aligned}b^{2} + λ & = −c^{2} sin^{2} iη, \\ b^{2} + μ & = −c^{2} cos^{2} ξ, \end{aligned}

values which verify as they should do the equations

\begin{aligned}\frac{x^{2}}{a^{2} + λ} + \frac{y^{2}}{b^{2} + λ} & = 1, \\ \frac{x^{2}}{a^{2} + μ} + \frac{y^{2}}{b^{2} + μ} & = 1, \end{aligned}

viz. these become

\frac{x^{2}}{c^{2} cos^{2} iη} + \frac{y^{2}}{−c^{2} sin^{2} iη} = sin^{2} ξ + cos^{2} ξ, = 1,
\frac{x^{2}}{c^{2} sin^{2} ξ} + \frac{y^{2}}{−c^{2} cos^{2} ξ} = cos^{2} iη + sin^{2} iη, = 1.

The same equations, or as we may also write them,

λ = −a^{2} sin^{2} iη − b^{2} cos^{2} iη,
μ = −a^{2} cos^{2} ξ − b^{2} sin^{2} ξ,

determine η as a function of λ, and ξ as a function of μ; λ, μ being by what precedes, given functions of x, y.

Or more simply, starting from the last-mentioned values of λ, μ, and substituting these in the expressions

x^{2} = \frac{(a^{2} + λ)(a^{2} + μ)}{a^{2} − b^{2}}, y^{2} = \frac{(b^{+}λ)(b^{2} + μ}{b^{2} − a^{2}},

we find

x^{2} = c^{2} sin^{2} ξ cos^{2} iη, y^{2} = −c^{2} cos^{2} ξ sin^{2} iη,

or say

x = c sin ξ cos iη, iy = c cos ξ sin iη,

whence

x + iy = c sin(ξ + iη),

the original relation between x, y and ξ, η.