The inverse Z-transform

IZT Method 2: Long tail division

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The second method to calculate the IZT of a sequence is by using long tail division. In many practical cases $X(z)$ is rational, with numerator polynomial $N(z)$ and denominator polynomial $D(z)$, both polynomials are nonzero, and usually the first coefficient $a_0$ is chosen as 1. $$\begin{eqnarray*} X(z) &=& \frac{N(z)}{D(z)} = \frac{\sum_{p=0}^{N} b_p z^{-p}}{ \sum_{p=0}^{M} a_p z^{-p}}= \frac{\sum_{p=0}^{N} b_p z^{-p}}{ 1+ \sum_{p=1}^{M} a_p z^{-p}} \end{eqnarray*}$$ $$\begin{eqnarray*} N \geq 0 &:& b_N \neq 0 \newline M \geq 0 &:& M=0 \quad \Rightarrow \quad D(z)=a_0 \text{ usually } a_0=1 \newline & & M>0 \quad\Rightarrow \quad a_M \neq 0 \end{eqnarray*}$$ In the long tail division method we write the rational expression of $X(z)$ as a power series expansion. In general, this can be achieved by applying long tail division which results in, a possible infinite length, polynomial expression in powers of $z$ with coefficients $c_k$. $$ X(z) = \frac{N(z)}{D(z)} \overset{\textit{Long tail division}}{=\!=\!=\!=\!=\!=\!=} \sum_{k }^{} c_k z^{-k} $$ Since $z^{-k}$ $\circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ$ $\delta[n-k]$, the coefficient $c_k$ multiplying the term $z^{-k}$ is the $k^{th}$ sample of the sequence $x[n]$: $$ X(z) = \sum_{k }^{} c_k z^{-k} \hspace{3mm} \circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ \hspace{3mm} x[n] = \sum_{k }^{} c_k \delta[n-k] $$ The IZT $x[n]$ is a, possible infinite length, sequence of samples with weights $c_k$.

Long tail division: Both $N(z),D(z)$ descending order

The long tail division can be performed in different ways, resulting in different polynomial expressions. In this first approach both numerator and denominator polynomials are represented in descending order: $$\begin{eqnarray*} X(z) &=& (b_0 + b_1 z^{-1} + \cdots + b_N z^{-N}) \cdot \left ( \frac{1}{a_0 + a_1 z^{-1} + \cdots + a_M z^{-M}}\right ) \newline &=&\left ( \sum_{p=0}^{N} b_p z^{-p} \right ) \cdot \left (\frac{1}{\sum_{p=0}^{M} a_p z^{-p}}\right ) \end{eqnarray*}$$ Applying long tail division to the second term results in an infinite long polynomial from which all powers of the variable $z$ are negative or zero. The weights of these terms are denoted by $\tilde{a}_k$. $$\begin{eqnarray*} X(z) &=&\left ( \sum_{p=0}^{N} b_p z^{-p} \right ) \cdot \left ( \sum_{k=0}^{\infty} \tilde{a}_k z^{-k} \right ) = \sum_{k=0}^{\infty} c_k z^{-k} \end{eqnarray*}$$ Multiplying this infinite long polynomial with the numerator polynomial results in an infinite long polynomial with powers of the variable $z$ which are all smaller or equal to zero, from which the weights are denoted by $c_k$. From the fact that the resulting polynomial contains only negative powers of $z$, the ROC is outside a circle with radius $r$. It can be shown that this radius $r$ depends on the values of $\tilde{a}_k$.

The IZT of the rational function $X(z)$ results in the sequence $x[n]$, which is a right sided causal sequence from which the FTD exist when the value of radius $r$ is smaller than one. $$\begin{eqnarray*} X(z)=\frac{N(z)}{D(z)}=\sum_{k=0}^{\infty} c_k z^{-k} & \circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ & x[n] =\sum_{k=0}^{\infty} c_k \delta[n-k] \end{eqnarray*}$$

Long tail division: Both $N(z),D(z)$ ascending order

When representing both numerator and denominator polynomials in ascending order, which writes in short hand notation as the given product, we first divide the numerator by $z^{-N}$ and the denominator by $z^{-M}$ . $$\begin{eqnarray*} X(z) &=& (b_N z^{-N} + \cdots + b_1 z^{-1} + b_0) \cdot \left ( \frac{1}{a_M z^{-M} + \cdots + a_1 z^{-1} + a_0}\right )\newline &=& \left ( \sum_{q=0}^{N} b_{N-q} z^{-N+q} \right ) \cdot \left (\frac{1}{\sum_{q=0}^{M} a_{M-q} z^{-M+q}}\right )\newline &=& \left ( z^{-N} \cdot \sum_{q=0}^{N} b_{N-q} z^{q} \right ) \cdot \left ( z^{M} \cdot \frac{1}{\sum_{q=0}^{M} a_{M-q} z^{q}}\right ) \end{eqnarray*}$$ These two components can be combined into a term $z^{M-N}$ and the first summation can be copied. Applying long tail division to the second term, results in an infinite long polynomial containing powers of $z$ which are all greater or equal to zero, from which the weights are denoted by $\hat{a}_k$ as follows: $$\begin{eqnarray*} X(z) &=& z^{M-N} \cdot \left ( \sum_{q=0}^{N} b_{N-q} z^{q} \right ) \cdot \left (\sum_{k=0}^{\infty} \hat{a}_k z^{k} \right ) = z^{M-N} \cdot \sum_{k=0}^{\infty} d_k z^k \end{eqnarray*}$$ The resulting polynomial writes as an infinite length polynomial in $z$ with exponents that are greater or equal than zero. The weights of this polynomial are denoted by $d_k$ and it is pre-multiplied with $z^{M-N}$. From the fact that the resulting polynomial contains only positive powers of $z$, it follows that the ROC is inside a circle with radius $r$. It can be shown that this radius $r$ depends on the values of the weights $\hat{a}_k$. $$\begin{eqnarray*} X(z)=\frac{N(z)}{D(z)}=z^{M-N} \cdot \sum_{k=0}^{\infty} d_k z^{k} & \circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ & x[n] =\sum_{k=0}^{\infty} d_k \delta[n+(M-N)+k] \end{eqnarray*}$$ The IZT now results in the sequence $x[n]$, which is a left sided sequence which is partially causal. The sequence $x[n]$ has a causal part when the order $N$ of the numerator is larger than the order $M$ of the denominator. Finally, depending on the value of the radius $r$, the FTD exist.

IZT Method 3: Partial fraction

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The third method to calculate the IZT of a sequence is by using the partial fraction method. This makes use of the fundamental algebraic property that the order $M$ polynomial of the denominator can be written as a product of $M$ first order polynomials. The roots of the denominator are called the poles of $X(z)$ and in first instance we assume simple first order poles. $$\begin{eqnarray*} X(z) &=& \frac{N(z)}{D(z)} = \frac{\sum_{p=0}^{N} b_p z^{-p}}{1+ \sum_{p=1}^{M} a_p z^{-p}} = \frac{\sum_{p=0}^{N} b_p z^{-p}}{\prod_{k=1}^{M} (1 - \alpha_k z^{-1})} \end{eqnarray*}$$ In case the order $M$ of the denominator polynomial is larger than the order $N$ of the nominator polynomial, $X(z)$ can be written as the following sum of $M$ first order terms, each weighted by scalars $A_k$.

Case $M>N$

$$\begin{eqnarray*} X(z) =\sum_{k=1}^{M} \frac{A_k}{1 - \alpha_k z^{-1}} & \circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ & x[n] =\sum_{k=1}^{M} A_k (\alpha_k)^n u[n] \end{eqnarray*}$$ From the table with known ZT pairs it follows that the IZT of each individual first order term is an exponential decaying sequence. The sequence $x[n]$ consists of a summation of $M$ of such terms. The weights $A_k$ can be obtained by either coefficient matching or via the so called residual approach: $$ A_k = \left [ (1 - \alpha_k z^{-1}) X(z) \right ] \vert_{z=\alpha_k} $$

We will show both approaches by computing the IZT of the following rational function: $$ X_6(z) = \frac{1 + 2 z^{-1}}{1 + 0.4 z^{-1} - 0.12 z^{-2}} $$

Example using coefficient matching

The first step is to spilt the second order polynomial of $X_6(z)$ into two first order polynomials. $$\begin{eqnarray*} X_6(z) &=& \frac{{\color{red}{1}} + {\color{blue}{2}} z^{-1}}{(1 - 0.2 z^{-1})(1 + 0.6 z^{-1})} \end{eqnarray*}$$ For the coefficient matching approach, we assume that $X_6(z)$ can be written as the sum of two terms, each weighted by an unknown weight $A_1$ and $A_2$ respectively. $$\begin{eqnarray*} X_6(z)&=& \frac{A_1}{1 - 0.2 z^{-1}} + \frac{A_2}{1 + 0.6 z^{-1}} = \frac{{\color{red}{(A_1 + A_2)}} + {\color{blue}{( 0.6 A_1 - 0.2 A_2)}} z^{-1}}{(1 - 0.2 z^{-1})(1 + 0.6 z^{-1})} \end{eqnarray*}$$ These two terms can be combined again into one expression, as denoted at the right hand side of the equation, which contains the unknown parameters in the numerator polynomial. When comparing (matching) the coefficients of all terms with the original numerator polynomial results in this case in two equations with two unknowns $A_1$ and $A_2$, which can be solved as follows: $$ \begin{array}{lcl} {\color{red}{A_1 + A_2}} &=& {\color{red}{1}} \newline {\color{blue}{0.6 A_1 -0.2 A_2}} &=& {\color{blue}{2}} \end{array} \hspace{3mm} \Rightarrow \hspace{3mm} A_1=2.7 \text{ ; } A_2=-1.75 $$ These results can be used in the original expression of $X_6(z)$ $$\begin{eqnarray*} X_6(z) &=&\frac{1 + 2 z^{-1}}{1 + 0.4 z^{-1} - 0.12 z^{-2}} = \frac{2.75}{1 - 0.2 z^{-1}} + \frac{-1.75}{1 + 0.6 z^{-1}} \end{eqnarray*}$$ from which the IZT results into the following expression for the sequence $x_6[n]$: $$\begin{eqnarray*} X_6(z) & \circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ & x_6[n] = 2.75 (0.2)^n u[n] - 1.75 (-0.6)^n u[n] \end{eqnarray*}$$

Example using residuals

The left hand side of the equation represents the original expression from which the denominator is split in first order terms. $$\begin{eqnarray*} \frac{1 + 2 z^{-1}}{(1 - 0.2 z^{-1})(1 + 0.6 z^{-1})} &=&\frac{A_1}{1 - 0.2 z^{-1}} + \frac{A_2}{1 + 0.6 z^{-1}} \end{eqnarray*}$$ The right hand side of the equal sign represents the two first order terms with unknown weights which can be solved via the so called residual approach: $$ A_k = \left [ (1 - \alpha_k z^{-1}) X(z) \right ] \vert_{z=\alpha_k} $$ This approach can be explained as follows:

From the right hand side of the representation of $X_6(z)$ it follows that the weight $A_1$ can obtained by multiplying the right hand side with $1-0.2 z^{-1}$ and then evaluating this result at $z=0.2$. Thus $$\begin{eqnarray*} \left ( 1 - 0.2 z^{-1} \right ) \cdot \left ( \frac{A_1}{1 - 0.2 z^{-1}} + \frac{A_2}{1 + 0.6 z^{-1}} \right ) = A_1 + \frac{A_2 \cdot (1 - 0.2 z^{-1})}{1 + 0.6 z^{-1}} \overset{z=0.2}{=} A_1 \end{eqnarray*}$$ Applying the same procedure to the left hand side of the representation of $X_6(z)$, that is first multiply with $1-0.2 z^{-1}$ and evaluating the result at $z=0.2$ leads to the value of the weight $A_1$: $$\begin{eqnarray*} A_1 &=& (1 - 0.2 z^{-1}) X_6(z)|_{z=0.2} = \frac{1 + 2 z^{-1}}{1 + 0.6 z^{-1}}|_{z=0.2} =2.75 \end{eqnarray*}$$ Applying the same procedure results in the value of the weight $A_2$: $$\begin{eqnarray*} A_2 &=& (1 + 0.6 z^{-1}) X_6(z)|_{z=-0.6} = \frac{1 + 2 z^{-1}}{1 - 0.2 z^{-1}}|_{z=-0.6} =-1.75 \end{eqnarray*}$$ Thus $X_6(z)$ can be split in two first order terms from which the IZT lead to the same resulting sequence $x_6[n]$ as before.

Case $M \leq N$

In case of simple poles the IZT procedure of the partial fraction method needs one extra step when the order of the denominator $M$ is smaller or equal than the order $N$ of the nominator. First split $X(z)$ by using long tail division and stop this long division procedure when the resulting rational polynomial function $F(z)$ is such that the order of the denominator is larger than the order of the nominator. $$\begin{eqnarray*} X(z) = \sum_{k=0}^{N-M}B_k z^{-k} + F(z) & \circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ & x[n] =\sum_{k=0}^{N-M}B_k \delta[n-k] +f[n] \end{eqnarray*}$$ The coefficients $B_k$ are calculated via the first steps of the ascended ordered long tail division method, while the partial fraction method can be applied to the rational polynomial function $F(z)$. The IZT results in a sequence $x[n]$ which consists of two parts: The first part is a sequence of delayed delta pulses with weights $B_k$ and the second part $f[n]$ is the IZT of $F(z)$.

Modification for multiple-order poles

Finally the partial fraction method needs to be adjusted in case of multiple order poles. $$ X(z) = \frac{\sum_{p=0}^{N} b_p z^{-p}}{\left ( \prod_{k=1}^{M-K} (1 - \alpha_k z^{-1})\right ) \cdot \left ( 1 - \tilde{\alpha} z^{-1} \right )^K } $$ The equation shows an example of a rational polynomial $X(z)$ in which there are $M-K$ simple poles $\alpha_k$ and one $K$ fold pole denoted by $\tilde{\alpha}$. Depending on the order $M$ in relation to the order $N$ we can split $X(z)$ into three different parts: $$ \boxed{X(z) = \sum_{k=0}^{N-M}B_k z^{-k} +\sum_{k=1}^{M-K} \frac{A_k}{1 - \alpha_k z^{-1}}+ \sum_{p=1}^{K} \frac{C_p}{(1 - \tilde{\alpha} z^{-1})^p}} $$ When the order $M$ is smaller or equal to the order $N$ the first part is obtained by (ascending ordered) long tail division. The second part, which belongs to the $M-K$ simple poles, can be written as a sum of $M-K$ first order polynomials by using coefficient matching or residuals as discussed before. The third term consists of $K$ higher order polynomials from which the weights can be found by coefficient matching or via the given generalized equation for residuals: $$ \boxed{ C_p = \frac{1}{(K-p)!\cdot (-\tilde{\alpha})^{(K-p)}} \cdot \frac{\text{d}^{K-p}}{\text{d} (z^{-1})^{K-p}} \left [ (1-\tilde{\alpha} z^{-1})^K \cdot F(z) \right ] \bigg\vert_{z=\tilde{\alpha}} } $$ This procedure will be explained by finding the IZT of the following function with one simple pole $z=\alpha_1=\frac{1}{2}$ and one second-order pole $z=\tilde{\alpha}= - \frac{1}{3}$: $$ X_8(z) = \frac{3 + \frac{8}{3} z^{-1}}{(1-\frac{1}{2} z^{-1})(1 + \frac{1}{3} z^{-1})^2} $$

Example using coefficient matching

The first step is to split the function $X_8(z)$ into the following three terms with unknown coefficients $A$, $C_1$ and $C_2$ respectively: $$ X_8(z) = \frac{A}{1-\frac{1}{2} z^{-1}} + \frac{C_1}{1 + \frac{1}{3} z^{-1}} + \frac{C_2}{(1 + \frac{1}{3} z^{-1})^2} $$ Coefficient matching of the numerator results in the following three equations: $$ 3 + \frac{8}{3} z^{-1} = \left ( A + C_1 + C_2\right ) + \left ( \frac{2}{3} A - \frac{1}{6} C_1 - \frac{1}{2} C_2 \right ) z^{-1}+ \left ( \frac{1}{9} A - \frac{1}{6} C_1\right ) z^{-2} $$ which lead to the values for $A$, $C_1$ and $C_2$. $$ \begin{array}{lcl} A + C_1 + C_2 &=& 3 \newline \frac{2}{3} A - \frac{1}{6} C_1 - \frac{1}{2} C_2 &=& \frac{8}{3} \newline \frac{1}{9} A - \frac{1}{6} C_1 &=& 0 \end{array} \hspace{3mm} \Rightarrow \hspace{3mm} A=3 \text{ ; } C_1=2 \text{ ; } C_2=-2 $$ The IZT of the three terms of $X_8(z)$ result in the following sequence $x_8[n]$: $$\begin{eqnarray*} &&X_8(z) =\frac{3}{1-\frac{1}{2} z^{-1}} + \frac{2}{1 + \frac{1}{3} z^{-1}} - \frac{2}{(1 + \frac{1}{3} z^{-1})^2}\newline &\circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ & \hspace{2mm} x_8[n] = 3 \left ( \frac{1}{2} \right )^n u[n] + 2 \left ( -\frac{1}{3} \right )^n u[n] -2 (n+1) \left ( - \frac{1}{3} \right ) ^n u[n] \end{eqnarray*}$$ from which the last two parts are found by using the table of ZT pairs.

Example using residuals

The values for $A$, $C_1$ and $C_2$ can also be found by using the general expression for the residuals. For the first simple pole the general residual expression reduces to the expression as discussed before: $$\begin{eqnarray*} A &=& (1-\frac{1}{2} z^{-1}) \cdot X_8(z) \vert_{z = \frac{1}{2}}=3 \end{eqnarray*}$$ For the second order pole the general residual expression leads to the following two equations: $$\begin{eqnarray*} C_1 &=& \frac{1}{1/3} \cdot \frac{\text{d}}{\text{d} (z^{-1})} \left [ (1 + \frac{1}{3} z^{-1})^2 \cdot X_8(z)\right ] \vert_{z=-\frac{1}{3}} = 2 \newline C_2 &=& (1 + \frac{1}{3} z^{-1})^2 \cdot X_8(z) \vert_{z=-\frac{1}{3}} = -2 \end{eqnarray*}$$ which leads to the same result as with the coefficient matching approach as discussed above.

IZT Formal method

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In this section we will derive the formal definition of the Inverse $Z$-transform (IZT). For this let's see if we intuitively can derive the IZT from the IFTD, which is defined by the following integral: $$\begin{eqnarray*} \text{IFTD: } x[n] &=& \frac{1}{2\pi} \int_{-\pi}^{\pi} X(e^{j\theta}) e^{jn \theta} \text{d} \theta \end{eqnarray*}$$ For this intuitive derivation we replace in the IFTD definition the exponent $e^{j\theta}$ by the complex variable $z$. Differentiating the variable $z$ to the variable $\theta$ now leads to the following expression for d$\theta$: $$ \frac{\text{d} }{\text{d} \theta } { z } = j e^{j\theta} \hspace{3mm} \Rightarrow \hspace{3mm} \text{d} \theta = \frac{1}{j} z^{-1} \text{d} z $$ Furthermore, the integral over variable $\theta$ has to be replaced by an integral in the z plane over a closed contour on the unit circle, denoted by $|z|=1$, thus: $$ \int_{-\pi}^{\pi}\hspace{3mm} \Rightarrow \hspace{3mm} \text{encircle unit circle counterclockwise: } \oint_{|z|=1} $$ Using these two steps the IFTD integral can be generalized, which can be rewritten as the following equation: $$\begin{eqnarray*} x[n] &=& \frac{1}{2\pi} \oint_{|z|=1} X(z) z^{n} \cdot \frac{1}{j} z^{-1} \text{d} z = \frac{1}{2 \pi j} \oint_{{\color{red}{|z|=1}}} X(z) z^{n-1} \text{d} z \end{eqnarray*}$$ The formal definition of the IZT is almost the same. $$ \boxed{ x[n] = \frac{1}{2 \pi j} \oint_{{\color{red}{C}}} X(z) z^{n-1} \text{d} z } $$ This equation represents an integral in the z plane over a counterclockwise closed contour $C$ inside the ROC of $X(z)$ encircling the origin $z=0$. The IZT is equivalent to the IFTD, when the IZT is evaluated on the unit circle.

In the following subsections we will derive the IZT in a more formal way, via the so called Cauchy residue theorem.

IZT via Cauchy's residue theorem

Cauchy's residue theorem states that for a rational function $F(z)$ the integral in the $z$-plane of the function $F(z)$ over a counterclockwise closed contour $C$ can be calculated as a sum of the residues of $F(z)$ at all poles of $F(z)$ inside the closed contour $C$: $$ \boxed{ \frac{1}{2 \pi j} \oint_{C} F(z) \text{d} z = \sum \left [ \text{Residues $F(z)$ @} {\color{red}{\text{poles inside } C}} \right ]} $$

The procedure to calculate the residue of $F(z)$ is as follows: In case of a single pole at $z=z_0$ inside $C$, the residue can be found by first cancelling the pole of $F(z)$ at $z=z_0$ with a zero at $z=z_0$, that is multiply $F(z)$ with $z-z_0$, and then evaluating the result at the $z=z_0$. $$\begin{eqnarray*} \text{Res} \left [ F(z) \text{ @ } z=z_0 \right ] &=& \left\{ (z-z_0) \cdot F(z) \right\} \bigg\vert_{z=z_0} \end{eqnarray*}$$ In case of a two-fold pole at $z=z_0$ inside $C$: First cancel the two-fold pole by a two-fold zero at $z=z_0$, that is multiply $F(z)$ with $(z-z_0)^2$. Then differentiate to the variable $z$ and evaluate the result at $z=z_0$. $$\begin{eqnarray*} \text{Res} \left [ F(z) \text{ @ } z=z_0 \right ] &=& \frac{\text{d}}{\text{d} z} \left\{ (z-z_0)^2 \cdot F(z) \right\} \bigg\vert_{z=z_0} \end{eqnarray*}$$ In general in case of a $K$-fold pole inside $C$, the procedure to calculate the residue of $F(z)$ is as follows: First cancel the $K$-fold pole by a $K$-fold zero at $z=z_0$, that is multiply $F(z)$ with $(z-z_0)^K$. Then take the $K-1$$^{th}$ differentiate and divide by $K-1$-faculty. Evaluate this result at $z=z_0$. $$ \boxed{ \text{Res} \left [ F(z) \text{ @ } z=z_0 \right ] = \frac{1}{(K-1)!} \cdot \frac{\text{d}^{K-1}}{\text{d} z^{K-1}} \left [ (z-z_0)^K \cdot F(z) \right ] \bigg \vert_{z=z_0} } $$

Example

In this example we will calculate the IZT of the function $X(z)=1$ by using Cauchy's residue theorem. Applying the IZT definition to the function $X(z)=1$ leads to an integral in the $z$-plane over a counterclockwise closed contour $C$ of the function $z^{n-1}$. $$\begin{eqnarray*} X(z)=1 \quad \overset{\text{IZT}}{\circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ} \quad x[n] &=& \frac{1}{2 \pi j} \oint_{C} X(z) z^{n-1} \text{d} z =\frac{1}{2 \pi j} \oint_{C} z^{n-1} \text{d} z \end{eqnarray*}$$ This integral can be computed for the following different cases:

When the index $n$ is larger or equal than 1 the function $z^{n-1}$ has no poles inside any closed contour $C$ and thus from Cauchy's residue theorem it follows that the result $x[n]=0$ for all indices $n \geq 1$.

$\color{blue}{n = 0}$: For index $n=0$ the function $z^{n-1}$ has one poles at $z=0$. From the procedure of Cauchy's residue theorem it follows that in this case the integral results into the value 1 and thus $x[n]=1$ for index $n=0$. $$\begin{eqnarray*} x[n] &=& (z-0) \cdot z^{-1} \vert_{z=0} =1 \text{ for } n=0 \end{eqnarray*}$$

$\color{blue}{n = -1}$: For index $n=-1$ the function $z^{n-1}$ has two poles at $z=0$. Applying Cauchy's residue theorem results in the value 0 and thus $x[n]=0$ for index $n=-1$. $$\begin{eqnarray*} x[n] &=& \frac{\text{d}}{\text{d} z} \left [ (z-0)^2 \cdot z^{-2} \right ] \vert_{z=0} =\frac{\text{d}}{\text{d} z} \left [ 1 \right ] \vert_{z=0} =0 \text{ for } n=-1 \end{eqnarray*}$$.

$\color{blue}{n \leq -1}$: This result can be generalized to all indices $n \leq -1$. With $r=-n$ the function $z^{-r-1}$ has $r+1$ poles at $z=0$. Cauchy's residual theorem results in the value 0 for this case. $$\begin{eqnarray*} x[n] &=& \frac{1}{r!} \cdot \frac{\text{d}^{r}}{\text{d} z^{r}} \left [ (z-0)^{r+1} \cdot z^{-r-1} \right ] \bigg \vert_{z=0} = \frac{\text{d}^{r}}{\text{d} z^{r}} \left [ 1 \right ] \bigg\vert_{z=0} =0 \text{ for } n \leq -1 \end{eqnarray*}$$ Combining the results for all indices $n$ implies that $x[n]$ is always 0, except for index $n=0$. Concluding it follows that the IZT of the function $X(z)=1$ is equal to the delta pulse $x[n]=\delta[n]$, which is known as the following so called ${\color{red}{\text{Cauchy's integral theorem}}}$: $$ \boxed{\frac{1}{2 \pi j} \oint_{C} z^{n-1} \text{d} z = \delta[n]} $$ This theorem is the counterpart of the Fourier integral which states that the IFTD of the function 1 is equal to $\delta[n]$. Intuitively this can be shown by substituting for the complex variable $z$ the complex exponent $e^{j\theta}$ and evaluating the integral on the unit circle $|z|=1$ which indeed leads to the following result: $$ \frac{1}{2 \pi j} \oint_{|z|=1} z^{n-1} \text{d} z \quad \Rightarrow \quad \frac{1}{2 \pi j} \int_{-\pi}^{\pi} \left ( e^{j\theta} \right )^{n-1} j e^{j\theta} \text{d} \theta = \frac{1}{2\pi} \int_{-\pi}^{\pi} e^{jn \theta} \text{d} \theta \overset{\color{red}{\text{IFTD}}}{=} \delta[n] $$

Proof IZT via Cauchy's integral theorem

In this subsection we will use Cauchy's integral theorem to derive a more formal proof of the IZT equation. First we replace the function $X(z)$ by its ZT definition and than interchange the summation with the integral: $$\begin{eqnarray*} \frac{1}{2 \pi } \oint_{C} X(z) z^{n-1} \text{d} z &=& \frac{1}{2 \pi j} \oint_{C} \left ( \sum_{p=-\infty}^{\infty} x[p] z^{-p} \right ) z^{n-1} \text{d} z \
&=& \sum_{p=-\infty}^{\infty} x[p] \left\{ {\color{red}{\frac{1}{2 \pi j} \oint_{C} z^{n-1-p} \text{d} z }} \right\} \end{eqnarray*}$$ The integral (in red) can be evaluated by using Cauchy's integral theorem, which leads to the function $\delta[n-p]$, which is only 1 when index $p$ is equal to $n$: $$\begin{eqnarray*} \frac{1}{2 \pi j} \oint_{C} X(z) z^{n-1} \text{d} z &=& \sum_{p=-\infty}^{\infty} x[p] \left\{ {\color{red}{\delta[n-p]}} \right\} = x[n] \end{eqnarray*}$$ In other words the result is $x[n]$ which finalized the proof of the derivation of the IZT equation.

Example

In this example we will compute the following ZT-pair: $$ X(z)= \frac{1}{1 - a z^{-1}} \hspace{3mm} \circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ \hspace{3mm} x[n]=a^n u[n] $$ with Cauchy's Residual theorem for the case $|a|<1$. Applying the formal definition the IZT results in the following equation: $$\begin{eqnarray*} X(z) = \frac{1}{1-a z^{-1}} \quad \overset{\text{IZT}}{\circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ} \quad x[n] &=& \frac{1}{2 \pi j} \oint_{C} \left ( \frac{1}{1-a z^{-1}} \right ) z^{n-1} \text{d} z =\frac{1}{2 \pi j} \oint_{C} \frac{z^n}{z-a}\text{d} z \end{eqnarray*}$$ When using the unit circle for contour $C$, the result of the IZT is equivalent to the result of the IFTD: $$\begin{eqnarray*} X(z) \overset{\text{IZT=IFTD}}{\circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ} \quad x[n] &=& =\frac{1}{2 \pi j} \oint_{{\color{red}{|z|=1}}} \frac{z^n}{z-a}\text{d} z \end{eqnarray*}$$ The derivation for the case that the absolute value of the parameter $a$ is smaller than 1 goes in the following steps:

Case $\color{blue}{n \geq 0}$:

Case $\color{blue}{n < 0}$: