# Function proofs

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 $$n\geq0$$

$$a^n \sin(\omega n)\,\color{grey}{\gamma(n)}$$

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}

Proofs continue at Z-transform initial and final value proofs. Or, if you want to skip ahead, I suggest Discrete Transfer Functions.