2.3 Elementary functions

1.

(a) At this level, use $e^{z} = e^{x}(\cos{y} + i\sin{y})$.

$e^{z} = 1$ implies $e^{x}(\cos{y} + i\sin{y}) = 1$. Note that two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. Thus we obtain the following system of equations.

$\left\{\begin{array}{l} e^{x}\cos{y} = 1\\ e^{x}\sin{y} = 0 \end{array}\right.$

Note that $\cos^{2}{y}+\sin^{2}{y} = 1$. Then

$$e^{2x}\cos^{2}{y} + e^{2x}\sin^{2}{y} = e^{2x} = 1$$

Thus, $2x = 0$. That is $x = 0$. Putting this into the above equation, we have $\cos{y} = 1, \sin{y} = 0$. Thus, $y = 2n\pi (n = 0, \pm 1, \pm 2, \ldots )$. Therefore, $z$ saitisfies

$$z = x + iy = 2n\pi i$$

(b) At this level, use $e^{z} = e^{x}(\cos{y} + i\sin{y})$.

$e^{z} = i$ implies $e^{x}(\cos{y} + i\sin{y}) = i$. Note that two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. Then we have the following system of equations.

$\left\{\begin{array}{l} e^{x}\cos{y} = 0\\ e^{x}\sin{y} = 1 \end{array}\right.$

Note that $\cos^{2}{y}+\sin^{2}{y} = 1$. Then

$$e^{2x}\cos^{2}{y} + e^{2x}\sin^{2}{y} = e^{2x} = 1$$

Thus, $2x = 0$. That is $x = 0$. Putting this into the above equation, we have $\cos{y} = 1, \sin{y} = 0$. Thus, $y = \frac{1}{2} + 2n\pi (n = 0, \pm 1, \pm 2, \ldots )$. Therefore $z$ is

$$z = x + iy = (\frac{1}{2} + 2n\pi)i$$

(c) At this level, use $e^{z} = e^{x}(\cos{y} + i\sin{y})$.

$e^{z} = -2$ implies $e^{x}(\cos{y} + i\sin{y}) = -2$. Note that two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. Then we have the following system of equations.

$\left\{\begin{array}{l} e^{x}\cos{y} = -2\\ e^{x}\sin{y} = 0 \end{array}\right.$

Note that $\cos^{2}{y}+\sin^{2}{y} = 1$.

$$e^{2x}\cos^{2}{y} + e^{2x}\sin^{2}{y} = e^{2x} = 4$$

Then $2x = \log{4}$. That is $x = \log{2}$. Putting this into the above equation, we have $\cos{y} = -1, \sin{y} = 0$. Thus, $y = \pi + 2n\pi (n = 0, \pm 1, \pm 2, \ldots )$. Therefore, $z$ is

$$z = x + iy = \log{2} + (\pi + 2n\pi) i$$

(a) Trigonometric functions are once rewritten using exponential functions. After that, the polar form can be changed to the orthogonal form.

When $\sin{2i}$ is expressed using an exponential function, we have

$$\sin{z} = \frac{e^{iz} - e^{-iz}}{2i}$$

Then

$$\sin{2i} = \frac{e^{-2} - e^{2}}{2i} = i\frac{e^2 - e^{-2}}{2}$$

(b) Simplify using the addition theorem.

$$\sin{\left(\frac{\pi}{2} + i\right)} = \sin{\frac{\pi}{2}}\cos{i} + \cos{\frac{\pi}{2}}\sin{i}$$

$$= \cos{i} = \frac{e^{-1} + e}{2}$$

(c) Simplify using the addition theorem.

$$\cos{\left(\frac{\pi}{3} - i\right)} = \cos{\frac{\pi}{3}}\cos{i} + \sin{\frac{\pi}{3}}\sin{i}$$

$$= \frac{1}{2}\cos{i} + \frac{\sqrt{3}}{2}\sin{i}$$

$$= \frac{1}{2}\cdot \frac{e^{-1} + e}{2} + \frac{\sqrt{3}}{2}\cdot \frac{e^{-1} - e}{2i}$$

$$= \frac{e^{-1} + e}{4} - i\frac{\sqrt{3}(e - e^{-1})}{4}$$

(d) $tan{z} = \frac{\sin{z}}{\cos{z}}$

$$\tan{\left(\frac{\pi}{6} + 2i\right)} = \frac{\sin{\frac{\pi}{6}} + 2i}{\cos{\frac{\pi}{6}} + 2i}$$

$$= \frac{\sin{\frac{\pi}{6}}\cos{2i} + \cos{\frac{\pi}{6}}\sin{2i}} {\cos{\frac{\pi}{6}}\cos{2i} - \sin{\frac{\pi}{6}}\sin{2i}}$$

$$= \frac{\frac{1}{2}\cdot\frac{e^{-2} + e^2}{2} + \frac{\sqrt{3}}{2}... ...{3}}{2}\cdot \frac{e^{-2} + e^2}{2} - \frac{1}{2}\cdot i\frac{e^2 - e^{-2}}{2}}$$

$$= \frac{e^{-2} + e^2 + i\sqrt{3}(e^2 - e^{-2})}{\sqrt{3}(e^{-2} + e^2) - i (e^2 - e^{-2})}$$

$$= \frac{(e^{-2} + e^2 + i\sqrt{3}(e^2 - e^{-2}))(\sqrt{3}(e^{-2} + ... ...}(e^{-2} + e^2) - i (e^2 - e^{-2}))(\sqrt{3}(e^{-2} + e^2) + i (e^2 - e^{-2}))}$$

$$= \frac{\sqrt{3}(e^{-4} + 2 + e^4) - \sqrt{3}(e^4 - 2 + e^{-4}) + 3i(e^4 - e^{-4}) + i(e^{4} - e^{-4})}{3(e^{-4} + 2 + e^{4}) + e^{4} - 2 + e^{-4}}$$

$$= \frac{4\sqrt{3} + 4i(e^{4} - e^{-4})}{4e^{-4} + 4 + e^{4}} = \frac{\sqrt{3} + i(e^{4} - e^{-4})}{e^{4} + 1 + e^{-4}}$$

(e)

$$\sin{iy} = \frac{e^{-y} - e^{y}}{2i} = i\frac{e^{y} - e^{-y}}{2}$$

(e)

$$\cos{iy} = \frac{e^{-y} + e^{y}}{2}$$

3.

(a) Note that $\cos^{2}{z} + \sin^{2}{z} = 1$.

Divide $\cos^{2}{z} + \sin^{2}{z} = 1$ by $\cos^{2}{z}$. Then

$$\frac{\cos^{2}{z}}{\cos^{2}{z}} + \frac{\sin^{2}{z}}{\cos^{2}{z}} = \frac{1}{\cos^{2}{z}}$$

Thus

$$1 + \tan^{2}{z} = \frac{1}{\cos^{2}{z}}$$

(b)

$$\sin{(-z)} = \frac{e^{-iz} - e^{iz}}{2i} = - \frac{e^{iz} - e^{-iz}}{2i} = - \sin{z}$$

(c)

$$\cos{(-z)} = \frac{e^{-iz} + e^{iz}}{2} = \frac{e^{iz} + e^{-iz}}{2} = \cos{z}$$

4.

(a) Of the $c$ that satisfies $f(z + c) = f(z)$, the one with the smallest $\vert c\vert$ is called the period of the function $f(z)$. Note also that $e^{(z + 2i\pi)} = e^{z}$.

Let $\sin{(z + c)} = \sin{z}$. Then find the value of $c$.

Since $\sin{(z+c)} = \frac{e^{i(z+c)} - e^{-i(z+c)}}{2i}$, let $\sin{(z + c)} = \sin{z}$. Then

$$e^{i(z+c)} - e^{-i(z+c)} = e^{iz} - e^{-iz}$$

$$e^{i(z+c)} - e^{iz} = -e^{-i(z+c)}e^{-iz}(e^{i(z+c)} - e^{iz})$$

Note that $e^{i(z+c)} - e^{iz}$ is the common term. Thus,

$$(e^{i(z+c)} - e^{iz})(1 + e^{-i(z+c)}e^{-iz}) = 0$$

and

$$e^{i(z+c)} = e^{iz}$$   or$$e^{i(z+c)} = - e^{-iz}$$

Now the period of the exponential function is $2\pi i$. Then $e^{ic} = 1$ implies $c = 2n\pi$. Also, $e^{i(z+c)} = - e^{-iz}$ implies $z+c = -z + n\pi$.

5.

(a) Set $\tan{z_{1}} = \tan{z_{2}}$. Then

$$\frac{\sin{z_{1}}}{\cos{z_{1}}} = \frac{\sin{z_{2}}}{\cos{z_{2}}}$$

$$\frac{\frac{e^{iz_{1}} - e^{-iz_{1}}}{2i}}{\frac{e^{iz_{1}} + e^{-iz_{1}}}{2}} = \frac{\frac{e^{iz_{2}} - e^{-iz_{2}}}{2i}}{\frac{e^{iz_{2}} + e^{-iz_{2}}}{2}}$$

$$\frac{i(e^{-iz_{1}} - e^{iz_{1}})}{e^{iz_{1}} + e^{-iz_{1}}} = \frac{i(e^{-iz_{2}} - e^{iz_{2}})}{e^{iz_{2}} + e^{-iz_{2}}}$$

$$\frac{i(1 - e^{2iz_{1}})}{e^{2iz_{1}} + 1} = \frac{i(1 - e^{iz_{2}}}{e^{2iz_{2}} + 1}$$

$$(1 - e^{2iz_{1}})(e^{2iz_{2}} + 1) = (e^{2iz_{1}} + 1)(1 - e^{2iz_{2}})$$

$$e^{2iz_{2}} - e^{2iz_{1}}e^{2iz_{2}} - e^{2iz_{1}}+ 1 = e^{2iz_{1}} - e^{2iz_{1}}e^{2iz_{2}} - e^{2iz_{2}}+ 1$$

$$e^{2iz_{2}} = e^{2iz_{1}}$$

Thus, $z_{2} = z_{1} + n \pi$. Therefore, $\tan{z}$ has the period $\pi$.