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Exponential in basis

Exponential in basis .

Definition 4.1.1 If

, with real

, we define:

$e^z=e^x ( \cos y +i \sin y )$ .

For example, $e^{2+\pi i}= e^2 ( \cos \pi + i \sin \pi) = -e^2$ and $e^{3+i}=e^3 ( \cos 1 + i \sin 1 )$ .

Note that the main requirement is fulfilled:

$\displaystyle \forall x \in \mathbb{R}, \; e^{x+0 \; i}=e^x\; (\cos 0 + i \; \sin 0)=\underbrace{e^x}_{\text{real exponential}}.$

Proposition 4.1.2

(i) The exponential is defined on $\mathbb{C}$ and $\forall z \in \mathbb{C}, \; e^z \neq 0$ .
(ii) The exponential is an entire function.
(iii) $\forall z_1,z_2 \in \mathbb{C}, \; e^{z_1+z_2}=e^{z_1} \cdot e^{z_2}$ .
(iv) $\forall z_1,z_2 \in \mathbb{C}, \; e^{z_1-z_2}=e^{z_1} / e^{z_2}$ .
(v) $\forall z \in \mathbb{C}, \; e^{z+2 \pi i} = e^z$ .

Proof.

(i) Let $z=x+iy, \; x,y \in \mathbb{R}$ . Then $e^z=e^x \; (\cos y + i \; \sin y)$ . As for any real number

, $e^x \neq 0$ , we have $e^z \neq 0$ .
(ii) As above, let $z=x+iy, \; x,y \in \mathbb{R}$ . Then we have:

$\displaystyle e^z=\underbrace{e^x\; \cos y}_{=u(x,y)} + i \; \underbrace{e^x \; \sin y}_{=v(x,y)}$

It can be easily shown that the functions

and

verify the Cauchy-Riemann equations over the whole plane. As the plane is open, the exponential function is differentiable at every point of an open set, whence analytic at every point. This means that the function is an entire function.
(iii) Denote

and

, with $x_1,x_2,y_1,y_2 \in \mathbb{R}$ . Then we have:

$\displaystyle z_1+z_2$	$\displaystyle = x_1+x_2 + i \; (y_1 +y_2)$
$\displaystyle e^{z_1+z_2}$	$\displaystyle = e^{x_1+x_2} \; \left( \cos (y_1+y_2)+i \; \sin (y_1+y_2) \right)$
$\displaystyle \quad$	$\displaystyle = e^{x_1} \cdot e^{x_2} \left[ \cos y_1 \cos y_2 - \sin y_1 \sin y_2 + i \left( \sin y_1 \cos y_2 + \cos y_1 \sin y_2 \right) \right]$
$\displaystyle \quad$	$\displaystyle = \left[ e^{x_1} \; \left( \cos y_1 + i \; \sin y_1 \right) \right] \cdot \left[ e^{x_2} \; \left( \cos y_2 + i \; \sin y_2 \right) \right]$
$\displaystyle \quad$	$\displaystyle = e^{z_1} \cdot e^{z_2}.$

(iv) Proceed as for (iii).
(v) For any $z \in \mathbb{C}$ , we have:

$\displaystyle e^{z+2k \pi \; i}=e^{x+i\; (y+2k \pi)}=e^x\; (\cos (y+ 2k \pi) +i \; \sin (y+2k \pi ) =e^x\; (\cos y +i \; \sin y)=e^z.$

$\qedsymbol$

Example 4.1.3 Let

and

. Then:

$\displaystyle \frac {e^{1+2i}}{e^{2-i}} = e^{(1+2i)-(2-i)} = e^{-1+3i} = \frac 1e (\cos 3 + i \sin 3 ).$

Example 4.1.4 Solve the equation

in $\mathbb{C}$ .

Let , where are real numbers. We have:

$\displaystyle e^z=1 \Longleftrightarrow e^x(\cos y + i \sin y )=1 \Longleftrightarrow \begin{cases}e^x \cos y = 1 \ e^x \sin y=0 \end{cases}$

As $e^x \neq 0$ , for any $x \in \mathbb{R}$ , we have $\sin y =0$ , i.e. $y=k \pi, \; k \in \mathbb{Z}$ . We consider now two cases:

(i) If $y=2k \pi$ , with $k \in \mathbb{Z}$ , we have $\cos y =1$ . The first equation has one solution, given by

.
(ii) If $y=(2k+1) \pi$ , with $k \in \mathbb{Z}$ , we have $\cos y =-1$ . The first equation implies now that

, and has no solution.
We conclude: the solution set of the given equation in $\mathbb{C}$ is $\{ 2k \pi i, \; k \in \mathbb{Z} \}$ .

Example 4.1.5 Solve the equation

in $\mathbb{C}$ . Let

, where

are real numbers. We have:

$\displaystyle e^z=i \Longleftrightarrow e^x(\cos y + i \sin y )=i \Longleftrightarrow \begin{cases}e^x \cos y = 0 \ e^x \sin y=1 \end{cases}$

As $e^x \neq 0$ , for any $x \in \mathbb{R}$ , we have $\cos y =0$ , i.e. $y=k \pi, \; k \in \mathbb{Z}$ . We consider now two cases:

(i) If $y= \pi /2 + 2k \pi$ , for $k \in \mathbb{Z}$ , we have $\sin y = 1$ . The second equation implies

, i.e.

.
(ii) If $y= \pi /2 + (2k+1) \pi$ , for $k \in \mathbb{Z}$ , we have $\sin y = -1$ .The second equation implies

,which has no real solution.
We conclude: the solution set of the given equation in $\mathbb{C}$ is $\{ (\pi /2 + 2k \pi)i , \; k \in \mathbb{Z} \}$ .

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