2.3 Complex Trigonometric Functions

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We define and discuss the complex trigonometric functions.

1 The Complex Cosine

To define $f(z) = \cos z$ we will use Maclaurin series and the sum identity for the cosine.

The series of interest are: \begin{equation*} \begin{aligned}[c] &\sin (x) =\sum _{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!}\\[15pt] &\sinh (x)=\sum _{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!}\\ \end{aligned} \quad \quad \quad \begin{aligned}[c] &\cos (x) =\sum _{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}\\[15pt] &\cosh (x)=\sum _{n=0}^\infty \frac{x^{2n}}{(2n)!}\\ \end{aligned} \end{equation*}

The sum identity for the cosine is: $\cos (\alpha + \beta ) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$ We begin by allowing an imaginary term in the sum identity: $\cos (x+iy) = \cos (x) \cos (iy) - \sin (x) \sin (iy)$ Next, we define the sine and cosine of a purely imaginary input by using their respective Maclauren series: $\cos (iy) = \sum _{n=0}^\infty (-1)^n \frac {(iy)^{2n}}{(2n)!}$ and $\sin (iy) = \sum _{n=0}^\infty (-1)^n \frac {(iy)^{2n+1}}{(2n+1)!}$ These power series can be simplified into the Maclauren series for the hyperbolic functions(!) by noting that $i^{2n} = \left (i^2\right )^n = (-1)^n$ so that $(-1)^n i^{2n} = 1$ for all $n$ : $\cos (iy) = \sum _{n=0}^\infty \frac {y^{2n}}{(2n)!} = \cosh (y)$ and $\sin (iy) = i\sum _{n=0}^\infty \frac {y^{2n+1}}{(2n+1)!}= i\sinh (y)$

Substituting these into the sum identity establishes: $\cos (x+iy) = \cos (x) \cosh (y) - i\sin (x) \sinh (y)$ We use this as our definition of the complex cosine.

The Complex Cosine For all $z = x+iy \in \C$ we define the complex cosine by $\cos (z) = \cos (x+iy) = \cos (x) \cosh (y) - i\sin (x) \sinh (y)$

2 The Complex Sine

The sum identity for the sine function states that: $\sin (\alpha + \beta ) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$ for all (real) inputs $\alpha$ and $\beta$ .

$\sin {z} = \sin (x+iy) = \sin (x) \cos (iy) + \cos (x) \sin (iy)$

$\cos (iy) = \cosh (y)$

$\sin (iy) = i \sinh (y)$

Based on the results obtained in the method for defining the complex cosine, which of the following is the definition of the complex sine function $f(z) = \sin z$ ?

$\sin (x) \sinh (y) - i\cos (x) \cosh (y)$
$\sin (x) \sinh (y) + i\cos (x) \cosh (y)$
$\sin (x) \cosh (y) - i\cos (x) \sinh (y)$
$\sin (x) \cosh (y) + i\cos (x) \sinh (y)$

example 1 Solve for $z: \; \sin (z) = 2$ .
Since $\sin (z) = \sin (x) \cosh (y) + i\cos (x) \sinh (y)$ we need $\sin (x) \cosh (y) =2 \;\; \mbox {and} \;\; \cos (x) \sinh (y) = 0$ simultaneously. The second equation is eeasier to work with because we can use the zero product principle. It gives: $x = \pi /2 + k\pi \;\; \mbox {or} \;\; y = 0$ We substitute these results into the first equation one at at time. If $y = 0$ , the first equation becomes: $\sin (x)\cosh 0 = 2$ Since $\cosh (0) = 1$ , this leads to the equation $\sin (x) = 2$ which has no solutions.
Next, if $x = \pi /2 + k\pi$ then $\sin (x) = \pm 1$ . Thus we consider cases. If where $k$ is even, then $\sin (x) = \sin \left (\frac {\pi }{2} + k\pi \right ) = 1$ and we arrive at $\cosh (y) = 2$ which has two solutions (verify) $y = \ln \left (2\pm \sqrt 3 \right )$ Finally, if $x = \pi /2 + k\pi$ where $k$ is odd, then $\sin (x) = \sin \left (\frac {\pi }{2} + k\pi \right ) = -1$ and we arrive at $\cosh (y) = -2$ which has no solutions (verify). Thus the equation $\sin (z) = 2$ has as its solutions $z = \left [\frac {\pi }{2} + k\pi \right ] + i\left [\ln \left (2\pm \sqrt 3\right )\right ]$ where $k$ is an even integer.

(problem 1) Solve the following equations for $z$ . \begin{align*} i) & \; \sin (z) = -5\\ ii) & \; \sin (z) = i\\ iii) & \; \cos (z) = 5\\ iv) & \; \cos (z) = -2i\\ \end{align*}

Here is a video solution of one part of problem 1, part iv:

3 Identities

3.1 Periodicity

Since the real sine and cosine functions are $2\pi$ periodic, so are their complex extensions.

$\sin (z + 2\pi ) = \sin (z) \;\; \mbox {and} \;\; \cos (z+2\pi ) = \cos (z)$

The periodicity follows immediately from the definition: $\sin (z + 2\pi ) = \sin (x+2\pi )\cosh (y) + i\cos (x+2\pi ) \sinh (y) = \sin (x)\cosh (y) + i\cos (x) \sinh (y) =\sin (z)$ The periodicity of the cosine is proved similarly (verify).

3.2 Evenness and Oddness

Recall that $\sin (x)$ and $\sinh (x)$ are odd functions and that $\cos (x)$ and $\cosh (x)$ are even functions. As a result, $\sin (z)$ is an odd function and $\cos (z)$ is an even function.

$\sin (-z) = -\sin (z) \;\; \mbox {and} \;\; \cos (-z) = \cos (z)$

To obtain the first equation, we have \begin{align*} \sin (-z) &= \sin (-x-iy) = \sin (-x)\cosh (-y) + i\cos (-x) \sinh (-y)\\ &= -\sin (x)\cosh (y) -i\cos (x) \sinh (y) = -\sin (z) \end{align*}

The evenness of the complex cosine is demonstrated similarly (verify).

Shift by $\pi$

$\sin (z + \pi ) = -\sin (z) \;\; \mbox {and} \;\; \cos (z+\pi ) = -\cos (z)$

The verification of these identities follows directly from their real counterparts (verify).

3.4 Relation to the Complex Exponential

The complex Cosine and sine functions are related to the complex exponential function!

For all $z \in \C$ , we have the following identities which relate the complex cosine and sine to the complex exponential function: $\cos (z) = \frac {e^{iz} + e^{-iz}}{2} \;\; \mbox {and} \;\; \sin (z) = \frac {e^{iz} - e^{-iz}}{2i}$

We will verify the second identity and leave the first to the reader. Using $z = x+iy$ , the right hand side yields: $e^{iz} = e^{-y+ix} = e^{-y}\cos (x) + ie^{-y}\sin (x)$ and $e^{-iz} = e^{y-ix} = e^{y}\cos (x) - ie^{y}\sin (x)$ Subtracting the second of these from the first, we get \begin{align*} e^{iz} - e^{-iz} &= \left [e^{-y}-e^y\right ]\cos (x) + i\left [e^{-y}+e^y\right ]\sin (x)\\ &=-2\cos (x) \sinh (y) +2i \sin (x)\cosh (y)\\ & = 2i\left [\sin (x)\cosh (y) + i \cos (x)\sinh (y)\right ]\\ &=2i\sin (z) \end{align*}

The result follows by dividing by $2i$ . The proof of the second equation is similar (verify).

Here is a video demonstration of the equation $\cos (z) = \frac {e^{iz}+e^{-iz}}{2}$ :

(Pythagorean Identity) For all $z \in \C$ $\cos ^2(z) + \sin ^2(z) = 1$

Proof: Using the exponential forms of sine and cosine gives: \begin{align*} \cos ^2(z) + \sin ^2(z) &= \left (\frac{e^{iz} + e^{-iz}}{2}\right )^2 + \left (\frac{e^{iz} - e^{-iz}}{2i} \right )^2\\ &= \frac{e^{2iz} + 2 + e^{-2iz}}{4} + \frac{e^{2iz} - 2 + e^{-2iz}}{-4}\\ &= \frac{e^{2iz} + 2 + e^{-2iz}}{4} - \frac{e^{2iz} - 2 + e^{-2iz}}{4}\\ &= \frac 44 = 1. \end{align*} $\blacksquare$

3.5 Co-functions

$\sin \left (\frac {\pi }{2}-z\right ) = \cos (z) \;\; \mbox {and} \;\; \cos \left (\frac {\pi }{2}-z\right ) = \sin (z)$

These identities follow directly from their counterparts for the real sine and cosine functions.

3.6 Sum and Difference Identities

$\sin (z_1 + z_2) = \sin (z_1)\cos (z_2) + \cos (z_1)\sin (z_2)$ $\sin (z_1 - z_2) = \sin (z_1)\cos (z_2) - \cos (z_1)\sin (z_2)$ $\cos (z_1 + z_2) = \cos (z_1)\cos (z_2) - \sin (z_1)\sin (z_2)$ $\cos (z_1 - z_2) = \cos (z_1)\cos (z_2) + \sin (z_1)\sin (z_2)$

To prove the first equation, we rewrite the right hand side using the complex exponential. The first term is \begin{align*} \sin (z_1)\cos (z_2) &= \frac{e^{iz_1} - e^{- iz_1}}{2i} \cdot \frac{e^{iz_2} + e^{-iz_2}}{2}\\ &= \frac{1}{4i}\left [e^{i(z_1+ z_2)} + e^{i(z_1 - z_2)} - e^{-i(z_1 - z_2)} - e^{-i(z_1+ z_2)}\right ] \end{align*}

The second term is \begin{align*} \cos (z_1)\sin (z_2) &= \frac{e^{iz_1} + e^{- iz_1}}{2} \cdot \frac{e^{iz_2} - e^{-iz_2}}{2i}\\ &= \frac{1}{4i}\left [e^{i(z_1+ z_2)} - e^{i(z_1 - z_2)} + e^{-i(z_1 - z_2)} - e^{-i(z_1+ z_2)}\right ] \end{align*}

Adding these, we get $\sin (z_1)\cos (z_2) + \cos (z_1)\sin (z_2) = \frac {1}{4i}\left [2e^{i(z_1+ z_2)} - 2e^{-i(z_1+ z_2)}\right ]$ $= \sin (z_1 + z_2)$

The second equation follows from the first by replacing $z_2$ with $-z_2$ and using evenness and oddness. The third and fourth equations are proved in the same manner as the first and second (verify).

4 The Other Trig functions

The other four trigonometric functions are defined in terms of the sine and cosine.

$\tan z = \frac {\sin (z)}{\cos (z)} \;\; \mbox {and} \;\; \cot (z) = \frac {\cos (z)}{\sin (z)}$ $\sec z = \frac {1}{\cos (z)} \;\; \mbox {and} \;\; \csc (z) = \frac {1}{\cos (z)}$

The functions $\tan (z)$ and $\cot (z)$ are $\pi$ -periodic and the functions $\sec (z)$ and $\csc (z)$ are $2\pi$ -periodic (verify). The Pythagorean Identity for the sine and cosine gives rise to two other Pythagorean identities: $1+ \tan ^2(z) = \sec ^2(z)$ and $1 + \cot ^2(z) = \csc ^2(z)$ 2024-09-27 14:06:01

1 The Complex Cosine

2 The Complex Sine

3 Identities

3.1 Periodicity

3.2 Evenness and Oddness

3.3 Shift by \pi

3.4 Relation to the Complex Exponential

3.5 Co-functions

3.6 Sum and Difference Identities

4 The Other Trig functions

Shift by $\pi$