## 227. The exponential values of the sine and cosine.

From the formula $\exp (\xi + i\eta) = \exp \xi(\cos\eta + i\sin\eta),$ we can deduce a number of extremely important subsidiary formulae. Taking $$\xi = 0$$, we obtain $$\exp (i\eta) = \cos\eta + i\sin\eta$$; and, changing the sign of $$\eta$$, $$\exp (-i\eta) = \cos\eta – i\sin\eta$$. Hence \begin{aligned} {3} \cos\eta &= &&\tfrac{1}{2} &&\{\exp (i\eta) + \exp (-i\eta)\},\\ \sin\eta &= -&&\tfrac{1}{2}i&&\{\exp (i\eta) – \exp (-i\eta)\}.\end{aligned} We can of course deduce expressions for any of the trigonometrical ratios of $$\eta$$ in terms of $$\exp (i\eta)$$.

## 228. Definition of $$\sin\zeta$$ and $$\cos\zeta$$ for all values of $$\zeta$$.

We saw in the last section that, when $$\zeta$$ is real, $\begin{equation*} \cos\zeta =\quad \tfrac{1}{2} \{\exp (i\zeta) + \exp (-i\zeta)\},\tag{1a}\end{equation*}$ $\begin{equation*} \sin\zeta = -\tfrac{1}{2}i\{\exp (i\zeta) – \exp (-i\zeta)\}. \tag{1b} \end{equation*}$

The left-hand sides of these equations are defined, by the ordinary geometrical definitions adopted in elementary Trigonometry, only for real values of $$\zeta$$. The right-hand sides have, on the other hand, been defined for all values of $$\zeta$$, real or complex. We are therefore naturally led to adopt the formulae (1) as the definitions of $$\cos \zeta$$ and $$\sin \zeta$$ for all values of $$\zeta$$. These definitions agree, in virtue of the results of § 227, with the elementary definitions for real values of $$\zeta$$.

Having defined $$\cos \zeta$$ and $$\sin \zeta$$, we define the other trigonometrical ratios by the equations $\begin{equation*} \tan \zeta = \frac{\sin \zeta}{\cos \zeta},\quad \cot \zeta = \frac{\cos \zeta}{\sin \zeta},\quad \sec \zeta = \frac{1}{\cos \zeta},\quad \csc \zeta = \frac{1}{\sin \zeta}. \tag{2} \end{equation*}$ It is evident that $$\cos \zeta$$ and $$\sec \zeta$$ are even functions of $$\zeta$$, and $$\sin \zeta$$, $$\tan \zeta$$, $$\cot \zeta$$, and $$\csc \zeta$$ odd functions. Also, if $$\exp (i\zeta) = t$$, we have $\begin{gathered} \begin{equation*} \cos \zeta = \tfrac{1}{2} \{t + (1/t)\},\quad \sin \zeta = -\tfrac{1}{2}i \{t – (1/t)\},\\ \cos^{2} \zeta + \sin^{2} \zeta = \tfrac{1}{4}[\{t + (1/t)\}^{2} – \{t – (1/t)\}^{2}] = 1. \tag{3} \end{equation*}\end{gathered}$

We can moreover express the trigonometrical functions of $$\zeta + \zeta’$$ in terms of those of $$\zeta$$ and $$\zeta’$$ by precisely the same formulae as those which hold in elementary trigonometry. For if $$\exp (i\zeta) = t$$, $$\exp (i\zeta’) = t’$$, we have \begin{aligned} \cos (\zeta + \zeta’) &= \tfrac{1}{2} \left(tt’ + \frac{1}{tt’}\right) \\ &= \tfrac{1}{4} \left\{ \left(t + \frac{1}{t}\right) \left(t’ + \frac{1}{t’}\right) + \left(t – \frac{1}{t}\right) \left(t’ – \frac{1}{t’}\right)\right\}\\ &= \cos\zeta \cos\zeta’ – \sin\zeta \sin\zeta’; \begin{equation*} \tag{4} \end{equation*}\end{aligned} and similarly we can prove that $\begin{equation*} \sin (\zeta + \zeta’) = \sin\zeta \cos\zeta’ + \cos\zeta \sin\zeta’. \tag{5} \end{equation*}$ In particular $\begin{equation*} \cos(\zeta + \tfrac{1}{2}\pi) = -\sin\zeta,\quad \sin(\zeta + \tfrac{1}{2}\pi) = \cos\zeta. \tag{6} \end{equation*}$

All the ordinary formulae of elementary Trigonometry are algebraical corollaries of the equations (2)–(6); and so all such relations hold also for the generalised trigonometrical functions defined in this section.

## 229. The generalised hyperbolic functions.

In Ex. LXXXVII. 19, we defined $$\cosh \zeta$$ and $$\sinh \zeta$$, for real values of $$\zeta$$, by the equations $\begin{equation*} \cosh\zeta = \tfrac{1}{2} \{\exp \zeta + \exp (-\zeta)\},\quad \sinh\zeta = \tfrac{1}{2} \{\exp \zeta – \exp (-\zeta)\}. \tag{1} \end{equation*}$

We can now extend this definition to complex values of the variable; i.e. we can agree that the equations (1) are to define $$\cosh \zeta$$ and $$\sinh \zeta$$ for all values of $$\zeta$$ real or complex. The reader will easily verify the following relations: $\cos i\zeta = \cosh \zeta,\quad \sin i\zeta = i\sinh \zeta,\quad \cosh i\zeta = \cos \zeta,\quad \sinh i\zeta = i\sin \zeta.$

We have seen that any elementary trigonometrical formula, such as the formula $$\cos 2\zeta = \cos^{2} \zeta – \sin^{2} \zeta$$, remains true when $$\zeta$$ is allowed to assume complex values. It remains true therefore if we write $$\cos i\zeta$$ for $$\cos \zeta$$, $$\sin i\zeta$$ for $$\sin \zeta$$ and $$\cos 2i\zeta$$ for $$\cos 2\zeta$$; or, in other words, if we write $$\cosh \zeta$$ for $$\cos \zeta$$, $$i\sinh \zeta$$ for $$\sin \zeta$$, and $$\cosh 2\zeta$$ for $$\cos 2\zeta$$. Hence $\cosh 2\zeta = \cosh^{2} \zeta + \sinh^{2} \zeta.$ The same process of transformation may be applied to any trigonometrical identity. It is of course this fact which explains the correspondence noted in Ex. LXXXVII. 21 between the formulae for the hyperbolic and those for the ordinary trigonometrical functions.

## 230. Formulae for $$\cos(\xi + i\eta)$$, $$\sin(\xi + i\eta)$$, etc.

It follows from the addition formulae that \begin{aligned} {4} \cos (\xi + i\eta) &= \cos\xi \cos i\eta &&- \sin\xi \sin i\eta &&= \cos\xi \cosh \eta &&- i\sin\xi \sinh \eta,\\ \sin (\xi + i\eta) &= \sin\xi \cos i\eta &&+ \cos\xi \sin i\eta &&= \sin\xi \cosh \eta &&+ i\cos\xi \sinh \eta.\end{aligned} These formulae are true for all values of $$\xi$$ and $$\eta$$. The interesting case is that in which $$\xi$$ and $$\eta$$ are real. They then give expressions for the real and imaginary parts of the cosine and sine of a complex number.

Example XCV
1. Determine the values of $$\zeta$$ for which $$\cos\zeta$$ and $$\sin\zeta$$ are (i) real (ii) purely imaginary. [For example $$\cos\zeta$$ is real when $$\eta = 0$$ or when $$\xi$$ is any multiple of $$\pi$$.]

2. \begin{aligned} {2} |\cos (\xi + i\eta)| &= \sqrt{\cos^{2} \xi + \sinh^{2} \eta} &&= \sqrt{\tfrac{1}{2} (\cosh 2\eta + \cos 2\xi)}, \\ |\sin (\xi + i\eta)| &= \sqrt{\sin^{2} \xi + \sinh^{2} \eta} &&= \sqrt{\tfrac{1}{2} (\cosh 2\eta – \cos 2\xi)}.\end{aligned}

[Use (e.g.) the equation $$|\cos(\xi + i\eta)| = \sqrt{\cos(\xi + i\eta) \cos(\xi – i\eta)}$$.]

3. $$\tan (\xi + i \eta) = \dfrac{\sin 2\xi + i\sinh 2\eta}{\cosh 2\eta + \cos 2\xi}$$, $$\cot (\xi + i \eta) = \dfrac{\sin 2\xi – i\sinh 2\eta}{\cosh 2\eta – \cos 2\xi}$$.

[For example $\tan (\xi + i\eta) = \frac{\sin (\xi + i\eta) \cos (\xi – i\eta)} {\cos (\xi + i\eta) \cos (\xi – i\eta)} = \frac{\sin 2\xi + \sin 2i\eta}{\cos 2\xi + \cos 2i\eta},$ which leads at once to the result given.]

4. \begin{aligned} \sec (\xi + i \eta) &= \frac{\cos\xi \cosh\eta + i\sin\xi \sinh\eta} {\frac{1}{2} (\cosh 2\eta + \cos 2\xi)}, \\ \csc (\xi + i \eta) &= \frac{\sin\xi \cosh\eta – i\cos\xi \sinh\eta} {\frac{1}{2} (\cosh 2\eta – \cos 2\xi)}.\end{aligned}

5. If $$|\cos (\xi + i\eta)| = 1$$ then $$\sin^{2} \xi = \sinh^{2} \eta$$, and if $$|\sin (\xi + i\eta)| = 1$$ then $$\cos^{2} \xi = \sinh^{2} \eta$$.

6. If $$|\cos (\xi + i\eta)| = 1$$, then $\sin \{\am \cos (\xi + i\eta)\} = \pm\sin^{2} \xi = \pm\sinh^{2} \eta.$

7. Prove that $$\log \cos (\xi + i\eta) = A + iB$$, where $A = \tfrac{1}{2} \log \{\tfrac{1}{2} (\cosh 2\eta + \cos 2\xi)\}$ and $$B$$ is any angle such that $\frac{\cos B}{\cos\xi \cosh\eta} = -\frac{\sin B}{\sin\xi \sinh\eta} = \frac{1}{\sqrt{\frac{1}{2} (\cosh 2\eta + \cos 2\xi)}}.$ Find a similar formula for $$\log \sin (\xi + i\eta)$$.

8. Solution of the equation cos $\zeta = a$, where $a$ is real. Putting $$\zeta = \xi + i\eta$$, and equating real and imaginary parts, we obtain $\cos\xi \cosh\eta = a,\quad \sin\xi \sinh\eta = 0.$ Hence either $$\eta = 0$$ or $$\xi$$ is a multiple of $$\pi$$. If (i) $$\eta = 0$$ then $$\cos\xi = a$$, which is impossible unless $$-1 \leq a \leq 1$$. This hypothesis leads to the solution $\zeta = 2k\pi \pm \arccos a,$ where $$\arccos a$$ lies between $$0$$ and $$\frac{1}{2}\pi$$. If (ii) $$\xi = m\pi$$ then $$\cosh\eta = (-1)^{m}a$$, so that either $$a \geq 1$$ and $$m$$ is even, or $$a \leq -1$$ and $$m$$ is odd. If $$a = \pm 1$$ then $$\eta = 0$$, and we are led back to our first case. If $$|a| > 1$$ then $$\cosh\eta = |a|$$, and we are led to the solutions \begin{aligned} {4} \zeta &=& 2k &\pi pm i\log \{ &&a + \sqrt{a^{2} – 1}\}\quad &&(a > 1), \\ \zeta &=&(2k + 1) &\pi \pm i\log \{-&&a + \sqrt{a^{2} – 1}\}\quad &&(a < -1).\end{aligned} For example, the general solution of $$\cos\zeta = -\frac{5}{3}$$ is $$\zeta = (2k + 1)\pi pm i\log 3$$.

9. Solve $$\sin\zeta = \alpha$$, where $$\alpha$$ is real.

10. Solution of $$\cos\zeta = \alpha + i\beta$$, where $$\beta \neq 0$$. We may suppose $$\beta > 0$$, since the results when $$\beta < 0$$ may be deduced by merely changing the sign of $$i$$. In this case $\begin{equation*} \cos\xi \cosh\eta = \alpha,\quad \sin\xi \sinh\eta = -\beta, \tag{1} \end{equation*}$ and $(\alpha/\cosh\eta)^{2} + (\beta/\sinh\eta)^{2} = 1.$

If we put $$\cosh^{2} \eta = x$$ we find that $x^{2} – (1 + \alpha^{2} + \beta^{2})x + \alpha^{2} = 0$ or $$x = (A_{1} \pm A_{2})^{2}$$, where $A_{1} = \tfrac{1}{2}\sqrt{(\alpha + 1)^{2} + \beta^{2}},\quad A_{2} = \tfrac{1}{2}\sqrt{(\alpha – 1)^{2} + \beta^{2}}.$ Suppose $$\alpha > 0$$. Then $$A_{1} > A_{2} > 0$$ and $$\cosh\eta = A_{1} \mp A_{2}$$. Also $\cos\xi = \alpha/(\cosh\eta) = A_{1} \mp A_{2},$ and since $$\cosh\eta > \cos\xi$$ we must take $\cosh\eta = A_{1} + A_{2},\quad \cos\xi = A_{1} – A_{2}.$ The general solutions of these equations are $\begin{equation*} \xi = 2k\pi \pm \arccos M,\quad \eta = \pm\log \{L + \sqrt{L^{2} – 1}\}, \tag{2} \end{equation*}$ where $$L = A_{1} + A_{2}$$, $$M = A_{1} – A_{2}$$, and $$\arccos M$$ lies between $$0$$ and $$\frac{1}{2}\pi$$.

The values of $$\eta$$ and $$\xi$$ thus found above include, however, the solutions of the equations $\begin{equation*} \cos\xi \cosh\eta = \alpha,\quad \sin\xi \sinh\eta = \beta, \tag{3} \end{equation*}$ as well as those of the equations , since we have only used the second of the latter equations after squaring it. To distinguish the two sets of solutions we observe that the sign of $$\sin\xi$$ is the same as the ambiguous sign in the first of the equations , and the sign of $$\sinh\eta$$ is the same as the ambiguous sign in the second. Since $$\beta > 0$$, these two signs must be different. Hence the general solution required is $\zeta = 2k\pi \pm [\arccos M – i\log \{L + \sqrt{L^{2} – 1}\}].$

11. Work out the cases in which $$\alpha < 0$$ and $$\alpha = 0$$ in the same way.

12. If $$\beta = 0$$ then $$L = \frac{1}{2}|\alpha + 1| + \frac{1}{2}|\alpha – 1|$$ and $$M = \frac{1}{2}|\alpha + 1| – \frac{1}{2}|\alpha – 1|$$. Verify that the results thus obtained agree with those of Ex. 8.

13. Show that if $$\alpha$$ and $$\beta$$ are positive then the general solution of $$\sin\zeta = \alpha + i\beta$$ is $\zeta = k\pi +(-1)^{k} [\arcsin M + i\log \{L + \sqrt{L^{2} – 1}\}],$ where $$\arcsin M$$ lies between $$0$$ and $$\frac{1}{2}\pi$$. Obtain the solution in the other possible cases.

14. Solve $$\tan\zeta = \alpha$$, where $$\alpha$$ is real. [All the roots are real.]

15. Show that the general solution of $$\tan \zeta = \alpha + i\beta$$, where $$\beta \neq 0$$, is $\zeta = k\pi + \tfrac{1}{2}\theta + \tfrac{1}{4} i\log\left\{ \frac{\alpha^{2} + (1 + \beta)^{2}} {\alpha^{2} + (1 – \beta)^{2}} \right\},$ where $$\theta$$ is the numerically least angle such that $\cos \theta : \sin \theta : 1 :: 1 – \alpha^{2} – \beta^{2} : 2\alpha : \sqrt{(1 – \alpha^{2} – \beta^{2})^{2} + 4\alpha^{2}}.$

16. If $$z = \xi\exp(\frac{1}{4}\pi i)$$, where $$\xi$$ is real, and $$c$$ is also real, then the modulus of $$\cos 2\pi z – \cos 2\pi c$$ is \begin{aligned}[b] \surd[\tfrac{1}{2}\{1 + \cos 4\pi c + \cos(2\pi\xi\sqrt{2}) &+ \cosh(2\pi\xi\sqrt{2}) \\ &- 4\cos 2\pi c \cos(\pi\xi\sqrt{2}) \cosh(\pi\xi\sqrt{2})\}] \end{aligned}.

17. Prove that \begin{gathered} |\exp \exp(\xi + i\eta)| = \exp(\exp\xi \cos\eta), \\ \begin{aligned} \operatorname{R} \{\cos\cos(\xi + i\eta)\} &= \cos(\cos\xi \cosh\eta) \cosh(\sin\xi \sinh\eta),\\ \operatorname{I} \{\sin\sin(\xi + i\eta)\} &= \cos(\sin\xi \cosh\eta) \sinh(\cos\xi \sinh\eta). \end{aligned}\end{gathered}

18. Prove that $$|\exp\zeta|$$ tends to $$\infty$$ if $$\zeta$$ moves away towards infinity along any straight line through the origin making an angle less than $$\frac{1}{2}\pi$$ with $$OX$$, and to $$0$$ if $$\zeta$$ moves away along a similar line making an angle greater than $$\frac{1}{2}\pi$$ with $$OX$$.

19. Prove that $$|\cos\zeta|$$ and $$|\sin\zeta|$$ tend to $$\infty$$ if $$\zeta$$ moves away towards infinity along any straight line through the origin other than either half of the real axis.

20. Prove that $$\tan\zeta$$ tends to $$-i$$ or to $$i$$ if $$\zeta$$ moves away to infinity along the straight line of Ex. 19, to $$-i$$ if the line lies above the real axis and to $$i$$ if it lies below.