Definition 5.4.1. $\mathbf{\sin } z and \cos z$.
\begin{equation*}
\sin z=\sum_{n=0}^{\infty}(-1)^n\frac{z^{2n+1}}{(2n+1) !} \text{ and }
\cos z = \sum\limits_{n=0}^{\infty}(-1)^n\frac{z^{2n}}{(2n) !}
\end{equation*}
Section 5.4 Trigonometric and Hyperbolic Functions
Based on the success we had in using power series to define the complex exponential, we have reason to believe that this approach will also be fruitful for other elementary functions. The power series expansions for the real-valued sine and cosine functions are
\begin{equation*}
\sin x=\sum\limits_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!} \text{ and } \cos x = \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}\text{.}
\end{equation*}
so it is natural to make the following definitions.
With these definitions in place, we can now easily create the other complex trigonometric functions, provided the denominators in the following expressions are not zero.
Definition 5.4.2. Other Trigonometric functions.
\begin{equation*}
\tan z=\frac{\sin z}{\cos z}, \;\; \cot z=\frac{\cos z}{\sin z}, \;\; \sec z=\frac{1}{\cos z},
\text{ and } \csc z=\frac{1}{\sin z}
\end{equation*}
The series for the complex sine and cosine agree with the real sine and cosine when \(z\) is real, so the remaining complex trigonometric functions likewise agree with their real counterparts. What additional properties are common? For starters, we have
Theorem 5.4.3.
\(\sin z\) and \(\cos z\) are entire functions, with \(\frac{d}{dz}\sin z=\cos z\) and \(\frac{d}{dz}\cos z=-\sin z\text{.}\)
Proof.
The ratio test shows that the radius of convergence for both functions is infinity, so they are entire by Theorem 4.4.7, part (i). Part (iii) of that theorem gives
\begin{align*}
\frac{d}{dz}\sin z \amp = \frac{d}{dz}\left[ \sum\limits_{n=0}^{\infty}(-1)^n\frac{z^{2n+1}}{(2n+1) !}\right]\\
\amp = \sum\limits_{n=0}^{\infty}(-1)^n\frac{(2n+1) z^{2n}}{(2n+1) !} \qquad { {\text{ (Why does the index } n \text{ stay at 0 here?) } }}\\
\amp = \sum\limits_{n=0}^{\infty}(-1)^n\frac{z^{2n}}{(2n) !}\\
\amp = \cos z\text{.}
\end{align*}
We leave the proof that \(\frac{d}{dz}\cos z=-\sin z\) as an exercise.
We now list several additional properties, providing proofs for some and leaving others as exercises.
- For all complex numbers \(z\text{,}\)\begin{align*} \sin(-z) \amp = -\sin z,\\ \cos(-z) \amp = \cos z\text{ , and }\\ \sin^2 z+\cos^2z \amp = 1\text{.} \end{align*}The verification that \(\sin(-z) =-\sin z\) and \(\cos (-z) =\cos z\) comes from substituting \(-z\) for \(z\) in Definition 5.4.1. We leave verification of the identity \(\sin^2 z+\cos^2 z=1\) as an exercise (with hints).
- For all complex numbers \(z\) for which the expressions are defined,\begin{align*} \frac{d}{dz}\tan z \amp = \sec^2 z,\\ \frac{d}{dz}\cot z \amp = -\csc^2 z,\\ \frac{d}{dz}\sec z \amp = \sec z\tan z, \text{ and }\\ \frac{d}{dz}\csc z \amp = -\csc z\cot z\text{.} \end{align*}The proof that \(\frac{d}{dz}\tan z=\sec^2 z\) uses the identity \(\sin^2 z + \cos^2z=1\text{:}\)\begin{align*} \frac{d}{dz}\tan z \amp = \frac{d}{dz}\left(\frac{\sin z}{\cos z}\right) =\frac{\cos z\frac{d}{dz}\sin z-\sin z\frac{d}{dz}\cos z}{\cos^2 z}\\ \amp = \frac{\cos^2 z+\sin^2z}{\cos^2z}=\frac{1}{\cos^2z}\\ \amp = \sec^2 z\text{.} \end{align*}We leave the proofs of the other derivative formulas as exercises.
To establish additional properties, expressing \(\cos z\) and \(\sin z\) in the Cartesian form \(u+iv\) will be useful. (Additionally, the applications in Chapters 10 and 11 will use these formulas.) We begin by observing that the argument given to prove part (iii) in Theorem 5.1.2 easily generalizes to the complex case with the aid of Definition 5.4.1. That is,
\begin{equation}
e^{iz}=\cos z+i\sin z\text{,}\tag{5.4.1}
\end{equation}
for all \(z\text{,}\) whether \(z\) is real or complex. Hence
\begin{equation}
e^{-iz}=\cos(-z) +i\sin(-z) =\cos z-i\sin z\text{.}\tag{5.4.2}
\end{equation}
\begin{align}
\sin z \amp = \frac{1}{2i}(e^{iz}-e^{-iz})\tag{5.4.3}\\
\amp = \frac{1}{2i}\left(e^{i(x+iy)}-e^{-i(x+iy)}\right)\notag\\
\amp = \frac{1}{2i}\left(e^{-y+ix}-e^{y-ix}\right)\notag\\
\amp = \frac{1}{2i}\left[ e^{-y}(\cos x+i\sin x) -e^y(\cos x-i\sin x) \right]\notag\\
\amp = \sin x\left(\frac{e^y+e^{-y}}{2}\right) +i\cos x\left(\frac{e^y-e^{-y}}{2}\right)\notag\\
\amp = \sin x\cosh y+i\cos x\sinh y\text{,}\tag{5.4.4}
\end{align}
where \(\cosh y=\frac{{e^y+e^{-y}}}{2}\) and \(\sinh y=\frac{{e^y-e^{-y}}}{2 }\text{,}\) respectively, are the hyperbolic cosine and hyperbolic sine functions that you studied in calculus.
Similarly,
\begin{align}
\cos z \amp = \frac{1}{2}(e^{iz}+e^{-iz})\tag{5.4.5}\\
\amp = \frac{1}{2}\left(e^{i(x+iy)}+e^{-i(x+iy)}\right)\notag\\
\amp = \frac{1}{2}\left(e^{-y+ix}+e^{y-ix}\right)\notag\\
\amp = \frac{1}{2}\left[e^{-y}(\cos x+i\sin x) +e^y(\cos x-i\sin x) \right]\notag\\
\amp = \cos x\left(\frac{e^y+e^{-y}}{2}\right) -i\sin x\left(\frac{ e^y-e^{-y}}{2}\right)\notag\\
\amp = \cos x\cosh y-i\sin x\sinh y\text{.}\tag{5.4.6}
\end{align}
Equipped with Identities (5.4.3)–(5.4.6), we can now establish many other properties of the trigonometric functions. We begin with some periodic results.
- For all complex numbers \(z=x+iy\text{,}\)\begin{align*} \sin(z+2\pi) \amp = \sin z,\\ \cos(z+2\pi) \amp = \cos z.\\ \sin(z+\pi) \amp = -\sin z,\\ \cos(z+\pi) \amp = -\cos z,\\ \tan(z+\pi) \amp = \tan z, \text{ and }\\ \cot(z+\pi) \amp = \cot z\text{.} \end{align*}Clearly, \(\sin(z+2\pi ) =\sin[(x+2\pi) +iy]\text{.}\) By Identity (5.4.4) this expression is\begin{align*} \sin(x+2\pi)\cosh y+i\cos(x+2\pi)\sinh y \amp = \sin x\cosh y+i\cos x\sinh y\\ \amp = \sin z\text{.} \end{align*}Again, the proofs for the other periodic results are left as exercises.
- If \(z_1\) and \(z_2\) are any complex numbers, then\begin{align*} \sin(z_1+z_2) \amp = \sin z_1\cos z_2+\cos z_1\sin z_2 \text{ and }\\ \cos(z_1+z_2) \amp = \cos z_1\cos z_2-\sin z_1\sin z_2 \text{ so }\\ \sin 2z \amp = 2\sin z\cos z.\\ \cos 2z \amp = \cos^2 z-\sin^2z\text{ , and }\\ \sin\left(\frac{\pi}{2}+z\right) \amp = \sin\left(\frac{\pi}{2}-z\right) = \cos z\text{.} \end{align*}We demonstrate that \(\cos(z_1+z_2) =\cos z_1\cos z_2-\sin z_1\sin z_2\) by making use of Identities (5.4.3)–(5.4.6):\begin{align*} \cos z_1\cos z_2 \amp = \frac{1}{4}\left[ e^{i(z_1+z_2)}+e^{i(z_1-z_2)}+e^{i(z_2-z_1)}+e^{-i(z_1+z_2)}\right], \text{ and }\\ -\sin z_1\sin z_2 \amp = \frac{1}{4}\left[ e^{i(z_1+z_2)}-e^{i(z_1-z_2)}-e^{i(z_2-z_1)}+e^{-i(z_1+z_2)}\right]\text{.} \end{align*}Adding these expressions gives\begin{equation*} \cos z_1\cos z_2-\sin z_1\sin z_2=\frac{1}{2}\left[e^{i(z_1+z_2)}+e^{-i(z_1+z_2)}\right] =\cos (z_1+z_2)\text{,} \end{equation*}which is what we wanted. A solution to the equation \(f(z) =0\) is called a zero of the given function \(f\text{.}\) As we now show, the zeros of the sine and cosine function are exactly where you might expect them to be.
- We have \(\sin z=0\) iff \(z=n\pi\text{,}\) where \(n\) is any integer, and \(\cos z=0\) iff \(z=(n+\frac{1}{2}) \pi\text{,}\) where \(n\) is any integer. We show the result for \(\cos z\) and leave the result for \(\sin z\) as an exercise. When we use Identity (5.4.6), \(\cos z=0\) iff\begin{equation*} 0=\cos x\cosh y-i\sin x\sinh y\text{.} \end{equation*}Equating the real and imaginary parts of this equation gives\begin{equation*} 0=\cos x\cosh y, \text{ and } 0 = \sin x\sinh y\text{.} \end{equation*}The real-valued function \(\cosh y\) is never zero, so the equation \(0=\cos x\cosh y\) implies that \(0=\cos x\text{,}\) from which we obtain \(x=(n+\frac{1}{ 2}) \pi\) for any integer \(n\text{.}\) Using the values for \(z=x+iy=(n+ \frac{1}{2}) \pi +iy\) in the equation \(0=\sin x\sinh y\) yields\begin{equation*} 0=\sin \left[(n+\frac{1}{2}) \pi \right] \sinh y=(-1)^n\sinh y\text{,} \end{equation*}which implies that \(y=0\text{,}\) so the only zeros for \(\cos z\) are the values \(z=(n+\frac{1}{2}) \pi\) for any integer \(n\text{.}\)
What does the mapping \(w=\sin z\) look like? We can get a graph of the mapping \(w=\sin z=\sin(x+iy) =\sin x\cosh y+i\cos x\sinh y\) by using parametric methods. Let’s consider the vertical line segments in the \(z\) plane obtained by successfully setting \(x=-\frac{\pi}{2}+\frac{k\pi}{12}\) for \(k=0,\,1,\ldots ,12\text{,}\) and for each \(x\) value and letting \(y\) vary continuously, \(-3 \le y \le 3\text{.}\) In the exercises we ask you to show that the images of these vertical segments are hyperbolas in the \(uv\) plane, as Figure 5.4.4 illustrates. In Chapter 10, we give a more detailed analysis of the mapping \(w=\sin z\text{.}\)
Figure 5.4.4 suggests one big difference between the real and complex sine functions. The real sine has the property that \(|\sin x| \le 1\) for all real \(x\text{.}\) In Figure 5.4.4, however, the modulus of the complex sine appears to be unbounded, which is indeed the case. Using Identity (5.4.4) gives
\begin{align*}
|\sin z|^2 \amp = |\sin z\cosh y+i\cos x\sinh y|^2\\
\amp = \sin^2 x\cosh^2y+\cos^2x\sinh^2y\\
\amp = \sin^2 x(\cosh^2y-\sinh^2y) +\sinh^2y(\cos ^2 x+\sin^2x)\text{.}
\end{align*}
The identities \(\cosh^2 y-\sinh^2y=1\) and \(\cos^2x+\sin^2x=1\) then yield
\begin{equation}
|\sin z|^2 =\sin^2x+\sinh^2y\text{.}\tag{5.4.7}
\end{equation}
A similar derivation produces
\begin{equation}
|\cos z|^2 =\cos^2x+\sinh^2y\text{.}\tag{5.4.8}
\end{equation}
\begin{equation*}
\lim_{y \to \infty}|\sin(x_0+iy)|^2 =\sin^2x_0+\lim_{y \to \infty}\sinh^2 y=\infty
\end{equation*}
As advertised, we have shown that \(\sin z\) is not a bounded function; it is also evident from Identity (5.4.8) that \(\cos z\) is unbounded.
The periodic character of the trigonometric functions makes apparent that any point in their ranges is actually the image of infinitely many points.
Example 5.4.5.
Find the values of \(z\) for which \(\cos z=\cosh 2\text{.}\)
Solution.
Starting with Identity (5.4.6), we write
\begin{equation*}
\cos z=\cos x\cosh y-i\sin x\sinh y=\cosh 2\text{.}
\end{equation*}
If we equate real and imaginary parts, then we get
\begin{equation*}
\cos x\cosh y=\cosh 2\text{ and } \sin x\sinh y=0\text{.}
\end{equation*}
The equation \(\sin x\sinh y=0\) implies either that \(x=\pi n\text{,}\) where \(n\) is an integer, or that \(y=0\text{.}\) Using \(y=0\) in the equation \(\cos x\cosh y=\cosh 2\) leads to the impossible situation \(\cos x=\frac{\cosh 2}{\cosh 0}=\cosh 2>1\text{.}\) Therefore \(x=\pi n\text{,}\) where \(n\) is an integer. Since \(\cosh y \ge 1\) for all values of \(y\text{,}\) the term \(\cos x\) in the equation \(\cos x\cosh y=\cosh 2\) must also be positive. For this reason we eliminate the odd values of \(n\) and get \(x=2\pi k\text{,}\) where \(k\) is an integer.
Finally, we solve the equation \(\cos 2\pi k\cosh y=\cosh y=\cosh 2\) and use the fact that \(\cosh y\) is an even function to conclude that \(y=\pm 2\text{.}\) Therefore the solutions to the equation \(\cos z=\cosh 2\) are \(z=2\pi k \pm 2i\text{,}\) where \(k\) is an integer.
The hyperbolic functions also have practical use in putting the tangent function into the Cartesian form \(u+iv\text{.}\) Using Definition 5.4.2, and Equations (5.4.4) and (5.4.6), we have
\begin{equation*}
\tan z=\tan(x+iy) =\frac{\sin(x+iy)}{\cos(x+iy)}=\frac{\sin x\cosh y+i\cos x\sinh y}{\cos x\cosh y-i\sin x\sinh y}
\end{equation*}
If we multiply each term on the right by the conjugate of the denominator, the simplified result is
\begin{equation}
\tan z=\frac{\cos x\sin x+i\cosh y\sinh y}{\cos^2 x\cosh^2y+\sin^2 x\sinh^2y}\text{.}\tag{5.4.9}
\end{equation}
We leave it as an exercise to show that the identities \(\cosh^2 y-\sinh^2 y=1\) and \(\sinh 2y=2\cosh y\sinh y\) can be used in simplifying Equation (5.4.9) to get
\begin{equation}
\tan z=\frac{\sin 2x}{\cos 2x+\cosh 2y}+i\frac{\sinh 2y}{\cos 2x+\cosh 2y}\text{.}\tag{5.4.10}
\end{equation}
As with \(\sin z\text{,}\) we obtain a graph of the mapping \(w=\tan z\) parametrically. Consider the vertical line segments in the \(z\) plane obtained by successively setting \(x=-\frac{\pi}{4}+\frac{k\pi}{16}\) for \(k=0,\,1,\ldots ,8\) and for each \(z\) value letting \(y\) vary continuously, \(-3 \le y \le 3\text{.}\) In the exercises we ask you to show that the images of these vertical segments are circular arcs in the \(uv\) plane, as Figure 5.4.6 shows. In Section 9.4 we give a more detailed investigation of the mapping \(w=\tan z\text{.}\)
How should we define the complex hyperbolic functions? We begin with
Definition 5.4.7. $\mathbf{\cosh} z$ and $\sinh z$.
\begin{equation*}
\cosh z=\frac{1}{2}(e^z+e^{-z}) \text{ and } \sinh z= \frac{1}{2}(e^z-e^{-z})
\end{equation*}
With these definitions in place, we can now easily create the other complex hyperbolic trigonometric functions, provided the denominators in the following expressions are not zero.
Definition 5.4.8. Complex hyperbolic functions.
\begin{equation*}
\tanh z =\frac{\sinh z}{\cosh z}, \; \coth z=\frac{\cosh z}{\sinh z}, \; \mathrm{sech}\,z=\frac{1}{\cosh z}
\text{ and } \mathrm{csch}\,z=\frac{1}{\sinh z}
\end{equation*}
As the series for the complex hyperbolic sine and cosine agree with the real hyperbolic sine and cosine when \(z\) is real, the remaining complex hyperbolic trigonometric functions likewise agree with their real counterparts. Many other properties are also shared. We state several results without proof, as they follow from the definitions we gave using standard operations, such as the quotient rule for derivatives. We ask you to establish some of these identities in the exercises.
The derivatives of the hyperbolic functions follow the same rules as in calculus:
\begin{align*}
\frac{d}{dz}\cosh z \amp = \sinh z \text{ and } \frac{d}{dz}\sinh z=\cosh z.\\
\frac{d}{dz}\tanh z \amp = \mathrm{sech}^2 z \text{ and } \frac{d}{dz} \coth z=-\mathrm{csch}^2 z.\\
\frac{d}{dz}\mathrm{sech}\,z \amp = -\mathrm{sech}\,z\tanh z \text{ and } \frac{d}{dz}\mathrm{csch}\,z=-\mathrm{csch}\,z\coth z\text{.}
\end{align*}
The hyperbolic cosine and hyperbolic sine can be expressed as
\begin{align*}
\cosh z \amp = \cosh x\cos y+i\sinh x\sin y, \text{ and }\\
\sinh z \amp = \sinh x\cos y+i\cosh x\sin y\text{.}
\end{align*}
The complex trigonometric and hyperbolic functions are all defined in terms of the exponential function, so we can easily show them to be related by
\begin{align*}
{3} \amp \cosh(iz) = \cos z \amp \amp \text{ and } \amp \amp \sinh(iz) = i\sin z;\\
\amp \cos(iz) = \cosh z \amp \amp \text{ and } \amp \amp \sin (iz) = i\sinh z\text{.}
\end{align*}
Some of the important identities involving the hyperbolic functions are
\begin{align*}
\cosh^2 z-\sinh^2z \amp = 1.\\
\sinh(z_1+z_2) \amp = \sinh z_1\cosh z_2+\cosh z_1\sinh z_2,\\
\cosh(z_1+z_2) \amp = \cosh z_1\cosh z_2+\sinh z_1\sinh z_2,\\
\cosh(z+2\pi i) \amp = \cosh z,\\
\sinh(z+2\pi i) \amp = \sinh z,\\
\cosh(-z) \amp = \cosh z, \text{ and }\\
\sinh(-z) \amp = -\sinh z\text{.}
\end{align*}
We conclude this section with an example from electronics. In electric circuits, the voltage drop, \(E_{R}\text{,}\) across a resistance \(R\) obeys Ohm’s law,
\begin{equation*}
E_{R}=IR\text{,}
\end{equation*}
where \(I\) is the current flowing through the resistor. Additionally, the current and voltage drop across an inductor, \(L\text{,}\) obey the equation
\begin{equation*}
E_{L}=L\frac{dI}{dt}
\end{equation*}
The current and voltage across a capacitor, \(C\text{,}\) are related by
\begin{equation*}
E_{C}=\frac{1}{C}\int_{t_0}^tI(\tau ) d\tau
\end{equation*}
The voltages \(E_{L}, \, E_{R}\text{,}\) and \(E_{C}\) and the impressed voltage \(E(t)\) illustrated in Figure 5.4.9 satisfy the equation
\begin{equation}
E_{L}+E_{R}+E_{C}=E(t)\text{.}\tag{5.4.11}
\end{equation}
Suppose that the current \(I(t)\) in the circuit is given by
\begin{equation*}
I(t) = I_0\sin \omega t
\end{equation*}
Using this in the equations for \(E_{R}\) and \(E_{L}\) gives
\begin{align}
E_{R} \amp = RI_0\sin \omega t, \text{ and }\tag{5.4.12}\\
E_{L} \amp = \omega LI_0\cos \omega t\text{.}\tag{5.4.13}
\end{align}
We then set \(t_0=\frac{\pi}{2}\) in the equation for \(E_{C}\) to obtain
\begin{equation}
E_{C} = \frac{1}{C}\int\limits_\frac{\pi}{2}^tI(\tau ) d\tau = \frac{1}{C}\int\limits_\frac{\pi}{2}^tI_0\sin \omega td\tau = -\frac{1}{\omega C}I_0\cos \omega t\text{.}\tag{5.4.14}
\end{equation}
We rewrite the equation \(I(t) =I_0\sin \omega t\) as a “complex current,”
\begin{equation*}
I^*=I_0e^{i\omega t}
\end{equation*}
with the understanding that the actual physical current \(I\) is the imaginary part of \(I^*\text{.}\) Similarly, we rewrite Equations (5.4.12)–(5.4.14) as
\begin{align*}
{3} \amp E_{R}^* \amp \amp = RI_0e^{i\omega t} \amp \amp = RI^*.\\
\amp E_{L}^* \amp \amp = i\omega LI_0e^{i\omega t} \amp \amp = i\omega LI^*, \text{ and }\\
\amp E_{C}^* \amp \amp = \frac{1}{i\omega C}I_0e^{i\omega t} \amp \amp = \frac{1}{i\omega C} I^*\text{.}
\end{align*}
Substituting these terms leads to an extension of Equation (5.4.11),
\begin{equation}
E^*=E_{R}^*+E_{L}^*+E_{C}^*=\left[R+i\left(\omega L-\frac{1}{\omega C}\right)\right]I^*\text{.}\tag{5.4.15}
\end{equation}
The complex quantity \(Z\) defined by
\begin{equation*}
Z=R+i\left(\omega L-\frac{1}{\omega C}\right)
\end{equation*}
is called the complex impedance. Substituting this expression into Equation (5.4.15) gives
\begin{equation*}
E^*=ZI^*\text{.}
\end{equation*}
which is the complex extension of Ohm’s law.
Exercises Exercises
1.
Establish that \(\frac{d}{dz}\cos z=-\sin z\) for all \(z\text{.}\)
Solution.
\(\frac{d}{dz}\cos z = \frac{d}{dz}\left[\sum \limits_{n=0}^{\infty}(-1)^n\frac{z^{2n}}{(2n)!}\right] = \sum\limits_{n=1}^{\infty}(-1)^n\frac{(2n)z^{2n-1}}{(2n)!}\text{.}\) Explain why the index \(n\) begins at 1 in the last expression. The result follows from simplification and reindexing.
2.
Demonstrate that, for all \(z\text{,}\) \(\sin^2 z+\cos^2z=1\text{,}\) as follows.
(a)
Define the function \(g(z) =\sin^2 z+\cos^2z\text{.}\) Explain why \(g\) is entire.
(b)
Show that \(g\) is constant. \hint{Look at \(g\,'(z)\text{.}\)}
(c)
Use part (b) to establish that, for all \(z\text{,}\) \(\sin^2 z+\cos^2z=1\text{.}\)
3.
\hint{Use the facts that \(\cosh^2 y-\sinh^2y=1\) and \(\sinh 2y=2\cosh y\sinh y\text{.}\)}
Solution.
\(\tan z=\frac{2\cos x \sin x}{2(\cos^2x \cosh^2y + \sin^2x \sinh^2y)} + i\frac{2\cosh y \sinh y}{2(\cos^2x \cosh^2y + \sin^2x \sinh^2y)}\text{.}\) The numerators simplify to \(\sin 2x\) and \(\sinh 2y\text{,}\) respectively. Show that the denominator equals \(\cos 2x + \cosh 2y\) by using the identities \(\cos 2x = \cos^2x - \sin^2x\) and \(\cosh^2y - \sinh^2y = 1\text{.}\)
4.
Explain why the diagrams in Figures 5.4.6 and Figure 5.4.9 came out they way they did.
5.
Show that, for all \(z\) ,
(a)
\(\sin(\pi -z) =\sin z\text{.}\)
Solution.
This follows immediately from \(\sin(z_1+z_2) = \sin z_1 \cos z_2 + \cos z_1 \sin z_2\text{.}\)
(b)
\(\sin(\frac{\pi}{2}-z) =\cos z\text{.}\)
(c)
\(\sinh(z+i\pi ) =-\sinh z\text{.}\)
Solution.
This follows immediately from \(\sinh z = \sinh x \cos y + i\cosh x \sin y\text{,}\) where we replace \(z=x+iy\) with \(z=x+iy+i\pi = x+i(y+\pi)\text{.}\)
(d)
\(\tanh(z+i\pi ) =\tanh z\text{.}\)
(e)
\(\sin(iz) =i\sinh z\text{.}\)
Solution.
This follows immediately from \(\sin z = \sin x \cosh y + i\cos x \sinh y\text{,}\) where we replace \(z=x+iy\) with \(iz=-y+ix\text{.}\)
(f)
\(\cosh(iz) =\cos z\text{.}\)
6.
Express the following quantities in \(u+iv\) form.
(a)
\(\cos(1+i)\text{.}\)
(b)
\(\sin(\frac{\pi +4i}{4})\text{.}\)
(c)
\(\sin 2i\text{.}\)
(d)
\(\cos(-2+i)\text{.}\)
(e)
\(\tan(\frac{\pi +2i}{4})\text{.}\)
(f)
\(\tan(\frac{\pi +i}{2})\text{.}\)
(g)
\(\sinh(1+i\pi )\text{.}\)
(h)
\(\cosh \frac{i\pi}{2}\text{.}\)
(i)
\(\cosh(\frac{4-i\pi}{4})\text{.}\)
7.
Find the derivatives of the following and state where they are defined.
(a)
\(\sin(\frac{1}{z})\text{.}\)
Solution.
\(-\frac{1}{z^2}\cos(\frac{1}{z})\text{,}\) valid for \(z \ne 0\text{.}\)
(b)
\(z\tan z\text{.}\)
(c)
\(\sec z^2\text{.}\)
Solution.
\(2z\sec z^2\tan z^2\text{,}\) valid for \(z \ne (k+\frac{1}{2}) \pi\text{,}\) where \(k\) is an integer.
(d)
\(z\csc^2 z\text{.}\)
(e)
\(z\sinh z\text{.}\)
Solution.
\(z\cosh z + \sinh z\text{,}\) valid for all \(z\text{.}\)
(f)
\(\cosh z^2\text{.}\)
(g)
\(z\tan z\text{.}\)
8.
Show that
(a)
\(\sin \overline{z}=\overline{\sin z}\) holds for all \(z\text{.}\)
(b)
\(\sin \overline{z}\) is nowhere analytic.
(c)
\(\cosh \overline{z}=\overline{\cosh z}\) holds for all \(z\text{.}\)
(d)
\(\cosh \overline{z}\) is nowhere analytic.
9.
Show that
(a)
\(\lim\limits_{z \to 0}\frac{\cos z-1}{z}=0\text{.}\)
Solution.
Use the same methods as in Exercise 11a of Section 5.1.
(b)
\(\lim\limits_{y \to +\infty}\tan(x_0+iy) =i\text{,}\) where \(x_0\) is any fixed real number.
10.
Find all values of \(z\) for which each equation holds.
(a)
\(\sin z=\cosh 4\text{.}\)
(b)
\(\cos z=2\text{.}\)
(c)
\(\sin z=i\sinh 1\text{.}\)
(d)
\(\sinh z=\frac{i}{2}\text{.}\)
(e)
\(\cosh z=1\text{.}\)
11.
Show that the zeros of \(\sin z\) are at \(z=n\pi\) where \(n\) is an integer.
Solution.
By Identity (5.4.4), \(\sin z=0\text{,}\) if and only if \(\sin x \cosh y + i\cos x \sinh y = 0\text{.}\) Equate real and imaginary parts to show this occurs iff \(x=k\pi\text{,}\) where \(k\) is an integer.
12.
Use Equation (5.4.7) to show that, for \(z=x+iy\text{,}\)
\begin{equation}
|\sinh y| \le |\sin z| \le \cosh y\text{.}\tag{5.4.16}
\end{equation}
13.
\begin{equation*}
|\cos z|^2 + |\sin z|^2 \ge 1\text{,}
\end{equation*}
and show that equality holds iff \(z\) is a real number.
Solution.
Combining (5.4.7) and (5.4.8), and letting \(z=x+iy\text{,}\) we get \(|\sin z|^2 + |\cos z|^2 = \sin ^2x + \sinh^2y + \cos^2x + \sinh^2y = 1 + 2\sinh^2y\text{.}\) This quantity equals 1 iff \(y=0\) (when \(z\) is a real number), and is greater than 1 otherwise.
14.
Show that the mapping \(w=\sin z\)
(a)
maps the \(y\) axis one-to-one and onto the \(v\) axis;
(b)
maps the ray given \(\{(x,y) : x=\frac{\pi}{2},\,y>0\}\) one-to-one and onto the ray defined by \(\{(u,v) :u>1,\,v=0\}\text{.}\)
15.
Given an elegant argument that explains why the following functions are harmonic.
(a)
\(h(x,y) = \sin x\cosh y\text{.}\)
Solution.
Consider the real part of Identity (5.4.4), and appeal to Theorem 3.3.1.
(b)
\(h(x,y) = \cos x\sinh y\text{.}\)
(c)
\(h(x,y) = \sinh x\cos y\text{.}\)
Solution.
Consider the imaginary part of \(\sin (iz)\text{,}\) and appeal to Theorem 3.3.1.
(d)
\(h(x,y) = \cosh x\sin y\text{.}\)
16.
Establish the following identities.
(a)
\(e^{iz}=\cos z+i\sin z\text{.}\)
(b)
\(\cos z=\cos x\cosh y-i\sin x\sinh y\text{.}\)
(c)
\(\sin(z_1+z_2) =\sin z_1\cos z_2+\cos z_1\sin z_2\text{.}\)
(d)
\(|\cos z|^2 =\cos^2x+\sinh^2y\text{.}\)
(e)
\(\cosh z=\cosh x\cos y+i\sinh x\sin y\text{.}\)
(f)
\(\cosh^2 z-\sinh^2z=1\text{.}\)
(g)
\(\cosh(z_1+z_2) =\cosh z_1\cosh z_2+\sinh z_1\sinh z_2\text{.}\)
17.
Find the complex impedance \(Z\) if
(a)
\(R=10, \, L=10, \, C=0.05\text{,}\) and \(\omega =2\text{.}\)
Solution.
\(Z=10+10i\text{.}\)
(b)
\(R=15, \, L=10, \, C=0.05\text{,}\) and \(\omega =4\text{.}\)
18.
Explain how \(\sin z\) and the function \(\sin x\) that you studied in calculus are different. How are they similar?
