Introduction

In the following chapters, we will see that, in order to define complex functions and their features, we can extend some properties of real functions; but, in some cases the mere extension will not be enough.

We start with the following:

Consider ${\mathcal {D}}\subset \mathbb {C}$ $\mathcal{D}\subset \mathbb {C}$ . A complex-valued function is a function:

f:\mathcal{D}\rightarrow \mathbb {C}  .

We see that $\forall \,z=x+iy\in {\mathcal {D}}$ $\forall \, z=x+iy\in \mathcal{D}$ the function $f$ $f$ can be decomposed as:

f(z)=u(z)+iv(z) \ ,

where the functions

u,v

are real-valued functions. In particular, we define:

u=Re(f)\; \& \; v=Im(f) \ .

Alternatively, we can use the isomorphism between

\mathbb{C}

and

\mathbb{R}^2

and look at

u

and

v

as two variables real functions:

f(z)=f(x+iy)=u(x,y)+iv(x,y) \ .

In fact, by calling ${\mathcal {\tilde {D}}}\subset \mathbb {R} ^{2}$ $\mathcal{\tilde{D}}\subset \mathbb {R}^{2}$ , we have that: $u,v:{\mathcal {{\tilde {D}}\rightarrow \mathbb {R} }}\ .$ $u,v:\mathcal{\tilde{D}\rightarrow \mathbb {R}} \ .$

Now we define some complex functions as extensions of known real functions:

Trigonometric Functions: We recall that, thanks to the Euler's formula, we can write:

\cos x=\frac{e^{ix}+e^{-ix}}{2} \ ,

So, the natural extension to the complex field is:

\cos z=\frac{e^{iz}+e^{-iz}}{2} \ ,

Analogously, we have that:

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

by setting

z=x+iy

we see that:

\sin (z)=sin(x+iy)=

=\frac{e^{i(x+iy)}+e^{-i(x+iy)}}{2i}=

=\frac{e^{-y}e^{ix}-e^{y}e^{-ix}}{2i} \ .

We underline that $\mid e^{\pm ix}\mid =e^{\pm ix}(e^{\pm ix})^{*}=e^{\pm ix}e^{\mp ix}=1.$ $\mid e^{\pm ix}\mid =e^{\pm ix}(e^{\pm ix})^{*}=e^{\pm ix}e^{\mp ix}=1.$ All the complex exponentials of the form $e^{if(x)}{\mbox{, where f(x) can be also a constant function,}}$ $e^{if(x)}\mbox{, where f(x) can be also a constant function,}$ are called phases and have unitary modulus.

The following properties (simple to prove) are valid:

$\cos ^{2}z+\sin ^{2}z=1\ ,$ $\cos ^{2}z+\sin ^{2}z=1 \ ,$
$\cos(iy)=\cosh y\ ,$ $\cos (iy)=\cosh y \ ,$
$\sin(iy)=i\sinh y\ ,$ $\sin (iy)=i\sinh y \ ,$
$\cos(z)=\cos(x+iy)=\cos x\cosh y-i\sin x\sinh y\ ,$ $\cos (z)=\cos (x+iy)=\cos x\cosh y-i\sin x\sinh y \ ,$
$\sin(z)=\sin(x+iy)=\sin x\cosh y+i\cos x\sinh y\ .$ $\sin (z)=\sin (x+iy)=\sin x\cosh y+i\cos x\sinh y \ .$

Logarithm function in $\mathbb {C}$ $\mathbb {C}$ : In order to define the complex logarithm, we use the polar form of complex numbers. So, we have:

\log (z)=\log (\mid z\mid e^{i\theta })=\log (\mid z\mid e^{iArg(z)})=

=\log (\mid z\mid )+iArg(z) \ .

Actually, we have to keep in mind that complex numbers are defined with a certain ambiguity in their polar form, so we have that:

\log (z)=\log (\mid z\mid )+i(Arg(z)+2k\pi ),\; k\in \mathbb {Z} \ .

The following properties are true:

$\log(-1)=i(\pi +2k\pi )\ ,$ $\log (-1)=i(\pi +2k\pi ) \ ,$
$\log(i)=i({\frac {\pi }{2}}+2k\pi )\ ,$ $\log (i)=i(\frac{\pi }{2}+2k\pi ) \ ,$
$\forall \,z_{1},z_{2}\in \mathbb {C} {\mbox{ ,we have that }}\log(z_{1}z_{2})=\log(z_{1})+\log(z_{2})+2k\pi i\ ,$ $\forall \, z_{1},z_{2}\in \mathbb {C}\mbox{ ,we have that }\log (z_{1}z_{2})=\log (z_{1})+\log (z_{2})+2k\pi i \ ,$
$\forall \,z_{1},z_{2}\in \mathbb {C} {\mbox{ ,we have that }}\log({\frac {z_{1}}{z_{2}}})=\log(z_{1})-\log(z_{2})+2k\pi i\ .$ $\forall \, z_{1},z_{2}\in \mathbb {C}\mbox{ ,we have that }\log (\frac{z_{1}}{z_{2}})=\log (z_{1})-\log (z_{2})+2k\pi i \ .$