# 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.

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

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

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

${\displaystyle f(z)=u(z)+iv(z)\ ,}$
where the functions ${\displaystyle u,v}$ are real-valued functions. In particular, we define:
${\displaystyle u=Re(f)\;\&\;v=Im(f)\ .}$
Alternatively, we can use the isomorphism between ${\displaystyle \mathbb {C} }$ and ${\displaystyle \mathbb {R} ^{2}}$ and look at ${\displaystyle u}$ and ${\displaystyle v}$ as two variables real functions:
${\displaystyle f(z)=f(x+iy)=u(x,y)+iv(x,y)\ .}$

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

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:

${\displaystyle \cos x={\frac {e^{ix}+e^{-ix}}{2}}\ ,}$
So, the natural extension to the complex field is:
${\displaystyle \cos z={\frac {e^{iz}+e^{-iz}}{2}}\ ,}$
Analogously, we have that:
${\displaystyle \sin z={\frac {e^{iz}-e^{-iz}}{2i}}\ ,}$
by setting ${\displaystyle z=x+iy}$ we see that:

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

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

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

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

The following properties (simple to prove) are valid:

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

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

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

${\displaystyle =\log(\mid z\mid )+iArg(z)\ .}$

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

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

The following properties are true:

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