Proofs for Z-transform properties, presented in the Z-Transforms article.\(\)
Proofs for pairs
Impulse
The discrete impulse function \(\delta[t]\) is different from the continuous impulse function. The impulse function is commonly used as an theoretical input signal to study the filter’s behavior.
The definition is
$$
\delta[n] =
\begin{cases}
1, & n=0 \\
0, & n\neq0
\end{cases}\label{eq:impuls_def1}
$$
Apply the unilateral Z-transforms definition to equation \(\eqref{eq:impuls_def1}\)
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\delta[n] \ztransform \Delta(z)=\sum_{n=0}^{\infty}z^{-n}\ \delta[n]
$$
Since the impulse is \(0\) everywhere but at \(n=0\), the summation simplifies to
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\delta[n] \ztransform \Delta(z)= \cancelto{1}{z^{-0}}\ \cancelto{1}{\delta[0]}
$$
The unilateral Z-Transform of the Unit Impulse Function follows as
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
\shaded{\delta[n]
\ztransform
1\triangleq\Delta(z)
},
&& \text{all }z,\text{ including }\infty
\end{align}
\label{eq:impulse}
$$
This is very similar to the Laplace transform of the continuous impulse function.
Delayed Impulse
Consider the discrete delayed impulse function \(\delta[n-a]\)
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\lfzraised#1{\raise{10mu}{#1}}
\def\laplace{\lfz{\mathscr{L}}}
\def\fourier{\lfz{\mathcal{F}}}
\def\ztransform{\lfz{\mathcal{Z}}}
\newcommand\ccancel[2][black]{\color{#1}{\cancel{\color{black}{#2}}}}
\newcommand\ccancelto[3][black]{\color{#1}{\cancelto{#2}{\color{black}{#3}}}}
\delta[n-a]=\begin{cases}
1, & n=a \\
0, & n\neq0
\end{cases}\label{eq:delayedimpuls_def1}
$$
Sorry proof missing. Similar to delay proof? Have the proof, please share it in the comments.
The unilateral Z-Transform of the Delayed Unit Impulse Function follows as
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
\shaded{\delta[n-a]\,\ztransform\, \shaded{\begin{cases}z^{-a},&a\geq0\\0,&a\lt0\end{cases}}},&&z\neq0
\end{align}
\label{eq:delayedimpulse}
$$
Unit Step
The discrete unit or Heaviside step function, denoted with \(\gamma[n]\) is defined as
$$
\gamma[n] =
\begin{cases}
0, & n\lt 0 \\
1, & n\geq 0
\end{cases}
\label{eq:unitstep_def}
$$
The unilateral Z-transforms of \(\eqref{eq:unitstep_def}\) follows
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
\gamma[n] \ztransform \Gamma(z) &= \sum_{n=0}^\infty z^{-n}\,\cancelto{1}{\gamma[t] } =\sum_{n=0}^\infty\,z^{-n} \nonumber \\
&= \sum_{n=0}^\infty\,\underbrace{\left(z^{-1}\right)^n}_{\color{blue}{r^n}} \label{eq:unitstep0}
\end{align}
$$
Apply the power series
$$
\begin{align}
\sum_{n=0}^{\infty}r^n = \frac{1}{1-r},&&|r|\lt1 \nonumber
\end{align} \nonumber
$$
to \(\eqref{eq:unitstep0}\)
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
\gamma[n] \ztransform \Gamma(z) = \frac{1}{1-z^{-1}},&&|z^{-n}|\lt1
\end{align}
$$
The unilateral Z-transform for the step function follows as
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
\shaded{
\gamma[n]
\ztransform
\frac{z}{z-1}\triangleq\Gamma(z)
}
, && |z|\gt1
\end{align}
\label{eq:step}
$$
Scaled
Consider the discrete power function starting at \(n=0\)
$$
f[n] =
\begin{cases}
0, & n\lt0 \\
a^n, & n\geq 0
\end{cases}
\label{eq:scaled_def}
$$
Apply the unilateral Z-transforms and the power series
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
a^n\,\color{grey}{\gamma[n]}
\ztransform
&\sum_{n=0}^\infty z^{-n}\,a^n\nonumber\\
\ztransform
&\sum_{n=0}^\infty (az^{-1})^n\nonumber\\
\ztransform
&\frac{1}{1-az^{-1}},&|z|\gt|a|
\end{align}
$$
The unilateral Z-Transform of the scaled function follows
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
\shaded{
a^n\,\color{grey}{\gamma[n]} \ztransform {\frac{z}{z-a}}
}
,&&|z|\gt|a|\label{eq:scaled}
\end{align}
$$
The Binomial scaled proof comes to this same transform.
Scaled delayed
Recall the delay property, and the scaled pair
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
f[n-b]\,\color{grey}{\gamma[n}-b\color{grey}{]}\, & \ztransform z^{-b}F(z) \nonumber \\
a^n\ \color{grey}{\gamma[n]} & \ztransform \frac{z}{z-a},&|z|\gt|a| \nonumber
\end{align}
\nonumber
$$
The unilateral Z-Transform of the scaled delayed function follows
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
\shaded{
a^{n-1}\gamma[n-1] \ztransform \dfrac{1}{z-1}
}
,&&|z|\gt|a|
\end{align}
$$
\(n\) scaled
Consider the discrete ramp function starting at \(n=0\)
$$
\begin{cases}
0, & n\lt0 \\
n\,a^n, & n\geq 0
\end{cases}
\label{eq:nscaled_def}
$$
Apply the unilateral Z-transforms definition and expand
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
n\,a^n
\ztransform
F(z)=&\sum_{n=0}^\infty z^{-n}\ n\,a^n\nonumber\\
\ztransform
&\left[\cancel{0}+\cancel{1}az^{-1}+2a^2z^{-2}+3a^3z^{-3}+\cdots\right]\nonumber\\
\ztransform
&\left[az^{-1}+2a^2z^{-2}+3a^3z^{-3}+\cdots\right]\label{eq:ramp1}
\end{align}
$$
To counter the infinite series, introduce \(az^{-1}F(z)\)
$$
az^{-1}\,F(z)=\left[a^2z^{-2}+2a^3z^{-3}+3a^4z^{-4}+\cdots\right]
\label{eq:ramp2}
$$
Subtract \(\eqref{eq:ramp2}\) from \(\eqref{eq:ramp1}\)
$$
\newcommand\ccancel[2][black]{\color{#1}{\cancel{\color{black}{#2}}}}
\newcommand\ccancelto[3][black]{\color{#1}{\cancelto{#2}{\color{black}{#3}}}}
\begin{align}
F(z)-az^{-1}\,F(z)
&=\left(az^{-1}+\ccancel[red]{2}a^2z^{-2}+\ccancel[green]{3}a^3z^{-3}+\cdots\right)-\\
&\quad\quad\left(\ccancel[red]{a^2z^{-2}}+\ccancel[green]{2a^3z^{-3}}+\ccancel[orange]{3a^4z^{-4}}+\cdots\right)\Rightarrow \nonumber \\[12mu]
(1-az^{-1})\,F(z)
&=\left(az^{-1}+a^2z^{-2}+a^3z^{-3}+\cdots\right) \nonumber \\[6mu]
&=az^{-1}\left(1+az^{-1}+a^2z^{-2}+\cdots\right) \nonumber \\[6mu]
&=az^{-1}\sum_{n=0}^{\infty}\underbrace{\left(az^{-1}\right)^n}_{\color{blue}{r^n}} \label{eq:ramp3}
\end{align}
$$
Apply the power series
$$
\begin{align}
\sum_{n=0}^{\infty}r^n=\frac{1}{1-r},&&|r|\lt1\nonumber
\end{align}
\nonumber
$$
to \(\eqref{eq:ramp3}\) where \(r=az^{-1}\)
$$
\begin{align}
(1-az^{-1})\,F(z) &= az^{-1}\,\frac{1}{1-az^{-1}},&|z|\gt1 \nonumber \\
F(z) &= \frac{az^{-1}}{(1-az^{-1})^2},&|z|\gt1
\end{align}
$$
The unilateral Z-Transform of the \(n\) scaled function follows
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
\shaded{
n\,a^n\ \color{grey}{\gamma[n]} \ztransform \frac{az}{\left(z-a\right)^2}
}
,&&|z|\gt1\label{eq:timescaled}
\end{align}
$$
The Binomial scaled proof comes to this same transform.
Ramp
This is a special case for the \(n\) scaled \(z\)-transform, where \(a=1\)
Substitute \(a=1\) in
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
n\,a^n\ \color{grey}{\gamma[n]} \ztransform \frac{az}{\left(z-a\right)^2}, && |z|\gt1 \nonumber
\end{align}
\nonumber
$$
The unilateral Z-Transform of the discrete ramp function follows
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\lfzraised#1{\raise{10mu}{#1}}
\def\laplace{\lfz{\mathscr{L}}}
\def\fourier{\lfz{\mathcal{F}}}
\def\ztransform{\lfz{\mathcal{Z}}}
\newcommand\ccancel[2][black]{\color{#1}{\cancel{\color{black}{#2}}}}
\newcommand\ccancelto[3][black]{\color{#1}{\cancelto{#2}{\color{black}{#3}}}}
\begin{align}
\shaded{
n\ \color{grey}{\gamma[n]} \ztransform \frac{z}{\left(z-1\right)^2}
}
,&&|z|\gt1
\end{align}
\label{eq:ramp}
$$
Binomial scaled, \(|z|\gt |a|\)
We will do this proof starting from the \(z\)-domain
$$
F(z) = \frac{z^m}{(z-a)^m}=\frac{1}{(1-az^{-1})^m}
$$
where \(n\) is an integer, and \(a\) is a constant possibly complex.
Recall the Negative Binomial Series
$$
\begin{align}
(1-x)^{-m}&=\sum_{n=0}^{\infty}{n+m-1 \choose n}\,x^n,&|x|\lt1\nonumber\\
\end{align}
\nonumber
$$
Work towards the form \((1-x)^{-m}\)
$$
\begin{align}
F(z) &= (1-\underbrace{az^{-1}}_{\color{blue}{=x}})^{-m}
\end{align}
$$
Use the Binomial Series where \(x=az^{-1}\)
$$
\begin{align}
F(z)=\left(1-(az^{-1})\right)^{-m}&=\sum_{n=0}^{\infty}{n+m-1 \choose n}\,(az^{-1})^k, \quad |az^{-1}|\lt1\nonumber\\
&=\sum_{n=0}^{\infty}{n+m-1 \choose n}\,a^n\,z^{-n}, \quad |z|\gt|a|
\end{align}
$$
Apply the Symmetry Rule for Binomial Coefficients \({n \choose k}={n \choose n-k}\)
$$
\begin{align}
F(z) &= \sum_{n=0}^{\infty}\underbrace{{n+m-1 \choose m-1}\,a^n}_{\color{blue}{=f[n]}}\,z^{-n},&|z|\gt|a|
\end{align}
$$
The unilateral Z-Transform follows
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
\shaded{{n+m-1 \choose m-1}\,a^n\,\color{grey}{\gamma[n]}
\ztransform
\frac{z^m}{(z-a)^m}=\frac{1}{(1-az^{-1})^m}}, \quad |z|\gt |a|\label{eq:binomialscaled}
\end{align}
$$
Recall the delay property
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
f[n-b]\,\color{grey}{\gamma[n}-b\color{grey}{]}\,\ztransform\,z^{-b}F(z)\nonumber
$$
Combining equation \(\eqref{eq:binomialscaled}\) with the delay property
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
{n+m-1 \choose m-1}\,a^{n-m+1}\,\color{grey}{\gamma[n-m+1]}
\,\ztransform\,
&z^{-(m-1)}\frac{z^m}{(z-a)^m}&|z|\gt |a| \nonumber\\
\end{align}
$$
So that in general
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
\shaded{
n+m-1 \choose m-1
}\,a^{n}\,\color{grey}{\gamma[n-m+1]
\,\ztransform\,\frac{a^{m-1}z}{(z-a)^m}}&&|z|\gt |a| \nonumber \\
\end{align}
$$
For the case where \(m=1\)
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
{n \choose 0}\,a^n\,\gamma[n]
\,\ztransform\,
&\frac{z}{z-a}&|z|\gt |a|\nonumber\\
\frac{\ccancel[red]{n!}}{0!\ccancel[red]{(n-0)!}}\,a^n\,\gamma[n]
\,\ztransform
\nonumber\\
a^n\,\gamma[n]
\,\ztransform
\end{align}
$$
so that
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
\shaded{
a^n\,\gamma[n]
\,\ztransform
\frac{z}{z-a}
}
&&|z|\gt |a|
\end{align}
$$
For the case where \(m=2\)
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
{n \choose 1}\,a^{n-1}\,\gamma[n-1]
\,\ztransform\,
&\frac{z}{(z-a)^2}&|z|\gt |a|\nonumber\\
\frac{\ccancelto[red]{n}{n!}}{1!\ccancel[red]{(n-1)!}}\,a^{n-1}\,\gamma[n]
\,\ztransform\nonumber\\
n\,a^{n-1}\,\gamma[n]
\,\ztransform
\end{align}
$$
so that
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
\shaded{n\,a^n\,\gamma[n]
\,\ztransform
\frac{az}{(z-a)^2}}&&|z|\gt |a|
\end{align}
$$
For the case where \(m=3\)
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
{n \choose 3}\,a^{n-2}\,\gamma[n-2]
\,\ztransform\,
&\frac{z}{(z-a)^3}&|z|\gt |a|\nonumber\\
\frac{\ccancelto[red]{n(n-1)}{n!}}{2!\ccancel[red]{(n-2)!}}\,a^{n-2}\,\gamma[n]
\,\ztransform\nonumber\\[6mu]
\frac{n(n-1)}{2}\,a^{n-2}\,\gamma[n]
\,\ztransform
\end{align}
$$
so that
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
\shaded{\tfrac{1}{2}{n(n-1)}\,a^n\,\gamma[n]
\,\ztransform
\frac{a^2z}{(z-a)^3}}&&|z|\gt |a|
\end{align}
$$
Binomial scaled, \(|z|\lt |a|\)
Similar to the previous proof, we will do this starting from the \(z\)-domain
$$
F(z) = \frac{z^m}{(z-a)^m}=\frac{1}{(1-az^{-1})^m}
$$
where \(n\) is an integer, and \(a\) is a constant possibly complex.
Recall the Binomial Series
$$
\begin{align}
(1+x)^r&=\sum_{k=0}^{\infty}{r \choose k}\,x^k,&|x|\lt1\nonumber\\
\end{align}\nonumber
$$
Work towards the form \((1-x)^{-m}\)
$$
\begin{align}
F(z) &= (1\underbrace{-az^{-1}}_{\color{blue}{=x}})^{-m}
\end{align}
$$
Use the Binomial Series where \(x=-az^{-1}\)
$$
\begin{align}
F(z) = (1-az^{-1})^{-m}&=\sum_{n=0}^{\infty}{-m \choose n}\,(-az^{-1})^n \nonumber \\
&=\sum_{n=0}^{\infty}{-m \choose n}\,(-1)^n\,a^nz^{-n} \nonumber
\end{align}
$$
Apply the Moving Top Index to Bottom in Binomial Coefficient \({n \choose m}=(-1)^{n-m}\,{-(m+1) \choose n-m}\)
$$
\begin{align}
F(z) = (1-az^{-1})^{-m} &= \sum_{n=0}^{\infty}{-(n+1) \choose -m-n}\,(-1)^{-m-n}\,(-1)^n\,a^nz^{-n} \nonumber \\
&= \sum_{n=0}^{\infty}{-n-1 \choose -m-n}\,(-1)^{-m}\,a^nz^{-n} \nonumber
\end{align}
$$
Apply the Symmetry Rule for Binomial Coefficients \({n \choose k}={n \choose n-k}\)
$$
\begin{align}
F(z)=(1-az^{-1})^{-m} &= \sum_{n=0}^{\infty}{-n-1 \choose \cancel{-n}-1+m\cancel{+n})}\,(-1)^{-m}\,a^nz^{-n} \nonumber \\
&=\sum_{n=0}^{\infty}\,\underbrace{(-1)^{-m}\,{-n-1 \choose m-1}\,a^n}_{\color{blue}{=f[n]}}\,z^{-n} \nonumber
\end{align}
$$
The unilateral Z-Transform follows
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
\shaded{
(-1)^{-m}\,{-n-1 \choose m-1}\,a^n\,\color{grey}{\gamma[n]} \ztransform \frac{z^m}{(z-a)^m}=\frac{1}{(1-az^{-1})^m}
}
&& |z|\lt |a|
\end{align}
$$
Exponential
Consider the discrete exponential function starting at \(n=0\)
$$
f[n]=\begin{cases}
0, & n\lt 0 \\
\mathrm{e}^{-anT}, & n\geq 0
\end{cases}
\label{eq:exponential_def}
$$
Apply the unilateral Z-transforms definition
$$
\begin{align}
\mathrm{e}^{-anT}\ \gamma[n] \ztransform & \sum_{n=0}^\infty z^{-n}\ \mathrm{e}^{-anT} = \nonumber \\
&\sum_{n=0}^\infty {\underbrace{\left(z^{-1}\ \mathrm{e}^{-aT}\right)}_a}^n
\label{eq:exponential0}
\end{align}
$$
Apply the power series
$$
\begin{align}
\sum_{n=0}^{\infty}r^n=\frac{1}{1-r},&&|r|\lt1 \nonumber
\end{align} \nonumber
$$
to \(\eqref{eq:exponential0}\), where \(a=z^{-1}\ \mathrm{e}^{-aT}\)
$$
\begin{align}
\mathrm{e}^{-anT}\ \gamma[n]
\ztransform
&\frac{1}{1-z^{-1}\ \mathrm{e}^{-aT}}\ ,&|\mathrm{e}^{-aT}|\lt z
\end{align}
$$
The unilateral Z-Transform of the exponential function follows
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
\shaded{
\mathrm{e}^{-anT}\ \gamma[n]
\ztransform
\frac{z}{z-\mathrm{e}^{-aT}}
},\ \ &&{|\mathrm{e}^{-aT}|\lt |z|}
\end{align}
\label{eq:exponential}
$$
Sine
The Z-transforms of the sine is similar to that of the cosine function, except that it uses the Euler identity for sine
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
\shaded{
\sin(\omega n)\ztransform\frac{z\sin(\omega T)}{z^2-2z\cos(\omega T)+1}
}
,&& |z|\gt1\\
\end{align}
\label{eq:sine}
$$
Cosine
A common notation is to use \(\Omega\) to represent frequency in the \(z\)-domain, and \(\omega\) for frequency in the \(s\)-domain. Here we use \(\omega\) to represent both types of frequency. Another notation that you may encounter is \(\omega_0\) to represent \(\omega T\).
Consider the cosine function starting at \(n=0\)
$$
f[n] = \cos(\omega nT)\,\gamma[n]
\label{eq:cos_def}
$$
The unilateral Z-transforms of the cosine is
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\cos(\omega nT)\ \gamma[n] \ztransform \sum_{n=0}^\infty z^{-n}\ \cos(\omega nT)
$$
Recall the Euler identity for cosine
$$
\cos\varphi = \frac{\mathrm{e}^{j\varphi}+\mathrm{e}^{-j\varphi}}{2} \nonumber
$$
Apply the identify for cosine
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
\cos(\omega nT)\ \gamma[n] \ztransform & \frac{1}{2}\left(\frac{1}{1-\left(z^{-1}\ e^{j\omega T}\right)}+\frac{1}{1-\left(z^{-1}\ e^{-j\omega T}\right)}\right),& |z|\leq |e^{j\omega T}|=1 \nonumber \\
=\, & \frac{1}{2}\left(\frac{z}{z-e^{j\omega T}}+\frac{z}{z-e^{-j\omega T} }\right),& |z|\lt1
\end{align}
$$
Recall the power series
$$
\begin{align}
\sum_{n=0}^{\infty}r^n=\frac{1}{1-r},&&|r|\lt1\nonumber
\end{align}\nonumber
$$
Apply the geometric power series
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\lfzraised#1{\raise{10mu}{#1}}
\def\laplace{\lfz{\mathscr{L}}}
\def\fourier{\lfz{\mathcal{F}}}
\def\ztransform{\lfz{\mathcal{Z}}}
\newcommand\ccancel[2][black]{\color{#1}{\cancel{\color{black}{#2}}}}
\newcommand\ccancelto[3][black]{\color{#1}{\cancelto{#2}{\color{black}{#3}}}}
\begin{align}
\cos(\omega nT)\ \gamma[n] \ztransform &\frac{1}{2}\left(\frac{1}{1-\left(z^{-1}\ e^{j\omega T}\right)}+\frac{1}{1-\left(z^{-1}\ e^{-j\omega T}\right)}\right),& |z|\leq |e^{j\omega T}|=1 \nonumber \\
=\, &\frac{1}{2}\left(\frac{z}{z-e^{j\omega T}}+\frac{z}{z-e^{-j\omega T}}\right),& |z|\lt1
\end{align}
$$
Bring over a common denominator and regroup
$$
\begin{align}
F(z) &= \frac{1}{2}\left(\frac{z(z-e^{-j\omega})}{(z-e^{j\omega n})(z-e^{-j\omega})}+\frac{z(z-e^{j\omega})}{(z-e^{-j\omega})(z-e^{-j\omega})}\right),& |z|\lt 1 \nonumber \\
&= \frac{1}{2}\left(\frac{z(z-e^{-j\omega})+z(z-e^{j\omega})}{z^2-ze^{-j\omega}+ze^{j\omega}+e^{j\omega}e^{-j\omega}}\right),& |z|\lt 1 \nonumber \\
&= \frac{1}{2}\left(\frac{z^2-ze^{-j\omega}+z^2-ze^{j\omega}}{z^2-z\left(e^{-j\omega}+e^{j\omega}\right)+e^0}\right),& |z|\lt 1 \nonumber \\
&= \frac{z^2-z\frac{1}{2}\left(e^{-j\omega}+e^{j\omega}\right)}{z^2-z\left(e^{-j\omega}+e^{j\omega}\right)+1},& |z|\lt 1
\end{align}
$$
Once more, apply the Euler identity for cosine
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
\shaded{
\cos(\omega n)\ztransform\frac{z^2-z\cos(\omega)}{z^2-2z\cos(\omega T)+1}
}
,&& |z|\gt1
\end{align}
\label{eq:cosine}
$$
Decaying Sine
Consider a decaying sine function for \(n\geq0\)
$$
a^n \sin(\omega n)\,\color{grey}{\gamma(n)}
$$
Sorry proof is missing. If you have the proof, please help make this list complete and share it in the comments.
The unilateral Z-Transform of the decaying sine follows as
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
\shaded{
a^n \sin(\omega n)\,\color{grey}{\gamma(n)}
\ztransform
\dfrac{az\sin(\omega)}{z^2-2az\cos(\omega)+a^2}
}
,&&|z|\gt|a|
\end{align}
$$
Decaying Cosine
Consider a decaying cosine function for \(n\geq0\)
$$
a^n \cos(\omega n)\,\color{grey}{\gamma(n)}
$$
Sorry the proof is missing. If you have the proof, please share it in the comments.
The unilateral Z-Transform of the decaying cosine follows as
$$
\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\ztransform{\lfz{\mathcal{Z}}}
\begin{align}
\shaded{
a^n \cos(\omega n)\,\color{grey}{\gamma(n)} \ztransform \dfrac{1-az\cos(\omega)}{z^2-2az\cos(\omega)+a^2}
}
,&&|z|\gt|a|
\end{align}
$$
