227–230. The trigonometrical and hyperbolic functions

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

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

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

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

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 $$\sin \{\text{am}\cos (\xi +i\eta )\}=\pm {\sin }^2\xi =\pm {\sinh }^2\eta .$$

7. Prove that $\log \cos (\xi +i\eta )=A+iB$, where $$A=\frac{1}{2}\log \{\frac{1}{2}(\cosh 2\eta +\cos 2\xi )\}$$ 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 $$\cos \xi \cosh \eta =a,\sin \xi \sinh \eta =0.$$ Hence either $\eta =0$ or $\xi$ is a multiple of $\pi$. If (i) $\eta =0$ then $\cos \xi =a$, which is impossible unless $-1\le a\le 1$. 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 $a\ge 1$ and $m$ is even, or $a\le -1$ 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 $$\begin{array}{rlrlrlrl}4\zeta & =& 2k& \pi pmi\log \{& & a+\sqrt{a^2–1}\}& & (a>1),\\ \zeta & =& (2k+1)& \pi \pm i\log \{-& & a+\sqrt{a^2–1}\}& & (a<-1).\end{array}$$ For example, the general solution of $\cos \zeta =-\frac{5}{3}$ is $\zeta =(2k+1)\pi pmi\log 3$.

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

10. Solution of $\cos \zeta =\alpha +i\beta$, where $\beta \ne 0$. We may suppose $\beta >0$, since the results when $\beta <0$ may be deduced by merely changing the sign of $i$. In this case $$\begin{array}{}\text{(1)}& \cos \xi \cosh \eta =\alpha ,\sin \xi \sinh \eta =-\beta ,\end{array}$$ 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 $$A_1=\frac{1}{2}\sqrt{(\alpha +1{)}^2+{\beta }^2},A_2=\frac{1}{2}\sqrt{(\alpha –1{)}^2+{\beta }^2}.$$ 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 $$\cosh \eta =A_1+A_2,\cos \xi =A_1–A_2.$$ The general solutions of these equations are $$\begin{array}{}\text{(2)}& \xi =2k\pi \pm \arccos M,\eta =\pm \log \{L+\sqrt{L^2–1}\},\end{array}$$ 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 $$\begin{array}{}\text{(3)}& \cos \xi \cosh \eta =\alpha ,\sin \xi \sinh \eta =\beta ,\end{array}$$ 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 $\beta \ne 0$, is $$\zeta =k\pi +\frac{1}{2}\theta +\frac{1}{4}i\log \left\{\frac{{\alpha }^2+(1+\beta {)}^2}{{\alpha }^2+(1–\beta {)}^2}\right\},$$ 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 $$\begin{array}{rl}\surd [\frac{1}{2}\{1+\cos 4\pi c+\cos (2\pi \xi \sqrt{2})& +\cosh (2\pi \xi \sqrt{2})\\ & -4\cos 2\pi c\cos (\pi \xi \sqrt{2})\cosh (\pi \xi \sqrt{2})\}]\end{array}.$$

17. Prove that $$\begin{array}{c}|\exp \exp (\xi +i\eta )|=\exp (\exp \xi \cos \eta ),\\ \begin{array}{rl}\mathrm{R}\{\cos \cos (\xi +i\eta )\}& =\cos (\cos \xi \cosh \eta )\cosh (\sin \xi \sinh \eta ),\\ \mathrm{I}\{\sin \sin (\xi +i\eta )\}& =\cos (\sin \xi \cosh \eta )\sinh (\cos \xi \sinh \eta ).\end{array}\end{array}$$

18. Prove that $|\exp \zeta |$ tends to $\mathrm{\infty }$ 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 $\mathrm{\infty }$ 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.