Hyperbolic functions and complex exponential expressions

Exponentials encode growth and decay, trigonometric functions encode oscillation, and hyperbolic functions combine opposite exponential rates. Complex exponentials place these families in a common algebra without eliminating the need to choose a complex logarithm value or branch. This chapter develops the identities used in differential equations and coordinate models rather than assuming a symbolic substitution is always safe.

1 The complex exponential from an absolutely convergent series

For $z\in\mathbb C$, define $e^z=\sum_{n=0}^\infty z^n/n!$. Absolute convergence follows from the ratio $|z|/(n+1)\to0$, and convergence is uniform on every bounded disk by the same scalar exponential majorant. The absolutely convergent Cauchy product and binomial theorem give

$$e^{z+w}=e^ze^w.$$

Also $e^h=1+h+O(|h|^2)$ as $h\to0$, with the remainder bounded by $|h|^2e^{|h|}/2$. Thus the difference quotient of $e^{z+h}=e^ze^h$ proves derivative $e^z$ even for complex increments. The product identity gives $e^ze^{-z}=1$, so the exponential never vanishes.

Separate even and odd terms for real $t$ to obtain Euler's identity $e^{it}=\cos t+i\sin t$, using the power-series definitions of the real sine and cosine. Hence

$$e^{x+iy}=e^x(\cos y+i\sin y),\qquad |e^{x+iy}|=e^x.$$

Real and imaginary parts of $e^{(-\gamma+i\omega)t}$ are the familiar damped oscillations $e^{-\gamma t}\cos\omega t$ and $e^{-\gamma t}\sin\omega t$. A complex notation can therefore simplify a real linear calculation without changing the physical variable into a complex measured quantity.

2 Hyperbolic functions and their identities

For real $x$, define

$$\cosh x=\frac{e^x+e^{-x}}2,\qquad\sinh x=\frac{e^x-e^{-x}}2.$$

The first is even, the second odd. Multiplying the expressions gives

$$\cosh^2x-\sinh^2x=1.$$

Their derivatives are $(\sinh x)'=\cosh x$ and $(\cosh x)'=\sinh x$. The exponential addition law gives

$$\sinh(x+y)=\sinh x\cosh y+\cosh x\sinh y,$$

with the corresponding plus-sign formula for $\cosh(x+y)$. The plus sign differs from the circular cosine addition identity because the defining quadratic relation has a minus sign.

Since $e^x>0$ on the reals, $\cosh x\ge1$, with equality only at zero by the arithmetic-geometric mean inequality or $(e^{x/2}-e^{-x/2})^2\ge0$. Therefore $\tanh x=\sinh x/\cosh x$ is defined for all real $x$, and

$$\frac d{dx}\tanh x=\frac{\cosh^2x-\sinh^2x}{\cosh^2x}=\operatorname{sech}^2x>0.$$

The curve $(\cosh t,\sinh t)$ lies on $X^2-Y^2=1$, explaining the term hyperbolic. No claim is made that its parameter is ordinary circular angle.

3 One complex variable links the two families

For complex $z$, define

$$\cos z=\frac{e^{iz}+e^{-iz}}2,\qquad \sin z=\frac{e^{iz}-e^{-iz}}{2i},$$

and use the same exponential formulas for $\cosh z,\sinh z$. These agree with their real power series on the real axis. Substitution gives

$$\cos(ix)=\cosh x,\quad\sin(ix)=i\sinh x,\quad \cosh(ix)=\cos x,\quad\sinh(ix)=i\sin x.$$

For $z=x+iy$, expanding by the addition identities gives

$$\sin(x+iy)=\sin x\cosh y+i\cos x\sinh y,$$ $$\cos(x+iy)=\cos x\cosh y-i\sin x\sinh y.$$

Thus complex sine is not bounded in magnitude by 1: at $z=iy$ its magnitude is $|\sinh y|$, unbounded as $|y|\to\infty$. A real-variable range statement cannot be exported to complex arguments.

4 Inverse hyperbolic functions and logarithm domains

The real function $\sinh$ is strictly increasing because its derivative $\cosh$ is positive, and its limits are opposite infinities. Solving $x=(u-u^{-1})/2$ with $u=e^y>0$ gives

$$\operatorname{arsinh}x=\log(x+\sqrt{x^2+1}),\qquad x\in\mathbb R.$$

The logarithm argument is positive for every real $x$. Differentiation gives derivative $1/\sqrt{1+x^2}$. Similarly, restricting $\cosh$ to $y\ge0$ gives

$$\operatorname{arcosh}x=\log(x+\sqrt{x^2-1}),\qquad x\ge1,$$

with derivative $1/\sqrt{x^2-1}$ only for $x>1$. Solving for $\tanh$ gives $\operatorname{artanh}x=\tfrac12\log[(1+x)/(1-x)]$ on $|x|<1$.

For complex numbers, the exponential is periodic: $e^{z+2\pi i}=e^z$. If $z\ne0$ has magnitude $r$ and angle $\theta$, every value $\log r+i(\theta+2\pi k)$, $k\in\mathbb Z$, exponentiates to $z$. Since the exponential is not injective, it has no two-sided inverse on its full domain.

A single-valued right inverse can nevertheless be selected on all of $\mathbb C\setminus\{0\}$. Choose the principal argument $\operatorname{Arg}z\in(-\pi,\pi]$ and define $L(z)=\log|z|+i\operatorname{Arg}z$. Then $e^{L(z)}=z$, but $L$ is discontinuous along the negative real axis. In particular, a single-valued selection is not necessarily a continuous logarithm branch.

There is no global continuous logarithm branch on $\mathbb C\setminus\{0\}$, and hence no global holomorphic one. Here is the elementary circle obstruction. If a continuous $L$ satisfied $e^{L(z)}=z$ everywhere, then $h(t)=\operatorname{Im}L(e^{it})-t$ would be a continuous function from $[0,2\pi]$ into $2\pi\mathbb Z$, so the intermediate value theorem forces it to be constant. But $e^{i0}=e^{i2\pi}=1$ gives $h(2\pi)=h(0)-2\pi$, a contradiction.

Finally, writing $\log(zw)=\log z+\log w$ for a selected logarithm can fail by multiples of $2\pi i$. Real positive arguments have their familiar unambiguous real logarithm.

The complex completeness and product proof behind the exponential

A complex Cauchy sequence converges because its real and imaginary parts are real Cauchy sequences; recombining their limits gives a complex limit in absolute value. Thus if $\sum|a_n|<\infty$, its complex partial sums converge, since the absolute value of every tail is bounded by the corresponding real nonnegative tail. The same argument is uniform over a domain when one common scalar majorant controls those tails.

For $R\ge0$, the real series $\sum R^n/n!$ converges by an explicit geometric-tail test: after an index with $R/(n+1)\le1/2$, each following term is at most half its predecessor. This proves absolute convergence of the complex exponential and uniform convergence on $|z|\le R$ without applying a real-valued theorem directly to complex terms.

Now let $A=\sum|a_i|$, $B=\sum|b_j|$ be finite and set $c_n=\sum_{i=0}^n a_i b_{n-i}$. Finite triangular sums give $\sum_{n=0}^N|c_n|\le AB$, so the nonnegative partial sums converge and $\sum c_n$ converges absolutely. The square sum $S_N=\sum_{i,j=0}^N a_i b_j$ is the product of the two partial sums, hence tends to $(\sum a_i)(\sum b_j)$. Compare it with the triangular sum $C_N=\sum_{i+j\le N}a_i b_j$. If $i+j>N$, at least one of $i,j$ exceeds $\lfloor N/2\rfloor$. Consequently

$$|S_N-C_N|\le B\sum_{i>\lfloor N/2\rfloor}|a_i|+A\sum_{j>\lfloor N/2\rfloor}|b_j|\longrightarrow0.$$

Thus the Cauchy product equals the product of the sums. For $a_i=z^i/i!$, $b_j=w^j/j!$, its $n$th coefficient is $(z+w)^n/n!$ by the finite binomial theorem, proving the exponential addition law.

Finally $\sum_{n\ge2}|h|^n/n!\le(|h|^2/2)\sum_{m\ge0}|h|^m/m!$, because $(m+2)!\ge2m!$. This proves the local quadratic remainder used in the complex derivative. All rearrangements in Euler's even/odd split are justified by absolute tails; the real sine and cosine series supply the two resulting sums. The proof concerns finite complex scalars, not arbitrary operators or conditional series.

5 Graded practice

E1 Foundation. Evaluate $e^{\log2+i\pi/3}$ in rectangular complex form.

E2 Hyperbolic inversion. Solve $\sinh x=3$ and $\cosh x=3$ for real $x$.

E3 Differentiation. Differentiate $\tanh(2x)$ and verify $y=\cosh(3x)$ satisfies $y''=9y$.

E4 Complex identity. Derive the displayed formula for $\sin(x+iy)$ from the exponential definitions and use it to evaluate $\sin(i\log2)$.

6 Full solutions

S1. Euler's identity gives $2(1/2+i\sqrt3/2)=1+i\sqrt3$. Its modulus is 2, consistent with $|e^{x+iy}|=e^x$.

S2. The first has unique solution $x=\log(3+\sqrt{10})$. The second has two solutions $x=\pm\log(3+\sqrt8)$ because $\cosh$ is even and strictly increasing only after restriction to the nonnegative half-line.

S3. The chain rule gives $2\operatorname{sech}^2(2x)$. For the second, $y'=3\sinh3x$ and $y''=9\cosh3x=9y$.

S4. Insert $e^{i(x+iy)}=e^{-y}(\cos x+i\sin x)$ and the corresponding negative exponential into the sine definition; collecting real/imaginary parts yields the formula. At $x=0$, $y=\log2$, it gives $i\sinh(\log2)=i(2-1/2)/2=3i/4$.

Summary

Absolute power-series convergence justifies the complex exponential and its product law. Hyperbolic and trigonometric functions are related exponential combinations, but their real ranges and inverse domains differ. Complex logarithms need branch choices, and identities valid for positive real logs or bounded real trigonometric values cannot be transferred without checking those domains.