Chapter 5.7, Problem 6WE

To determine

To simplify the given expression.

Answer to Problem 6WE

  $z+1$

Explanation of Solution

Given:

Expression: $\frac{z-\frac{1}{z}}{1-\frac{1}{z}}$

Concept used:

To simplify the given expression:

1. Find the Least Common Multiple of the fraction and write in reduced form.
2. Take out common terms in the given expression.
3. Cancel out the terms that are in common in numerator and the denominator.
4. Find if any algebraic identities can simplify the given expression.

Calculation:

Using the above concept,

Consider the numerator and simplify,

  $\begin{array}{l}\Rightarrow z-\frac{1}{z}\\ \Rightarrow \frac{z}{1}-\frac{1}{z}\end{array}$

Least Common Multiple of (1, z ) is z .

  $\begin{array}{l}\Rightarrow \frac{z\times z}{1\times z}-\frac{1\times 1}{z\times 1}\\ \Rightarrow \frac{z^2}{z}-\frac{1}{z}\\ \Rightarrow \frac{z^2-1}{z}\end{array}$

Consider the denominator and simplify,

  $\begin{array}{l}\Rightarrow 1-\frac{1}{z}\\ \Rightarrow \frac{1}{1}-\frac{1}{z}\end{array}$

Least Common Multiple of (1, z ) is z .

  $\begin{array}{l}\Rightarrow \frac{1\times z}{1\times z}-\frac{1\times 1}{z\times 1}\\ \Rightarrow \frac{z}{z}-\frac{1}{z}\\ \Rightarrow \frac{z-1}{z}\end{array}$

Substitute in the numerator and denominator of the given expression.

  $\begin{array}{l}\Rightarrow \frac{\frac{z^2-1}{z}}{\frac{z-1}{z}}\\ \Rightarrow \frac{(z^2-1)\times z}{z\times (z-1)}\\ \Rightarrow \frac{(z^2-1)\times \cancel{z}}{\cancel{z}\times (z-1)}\\ \Rightarrow \frac{(z^2-1)}{(z-1)}\\ \Rightarrow \frac{(z-1)(z+1)}{(z-1)}[\because (a^2-b^2)=(a-b)(a+b)]\\ \Rightarrow \frac{\cancel{(z-1)}(z+1)}{\cancel{(z-1)}}\\ \Rightarrow (z+1)\end{array}$

Conclusion:

Therefore, the simplified form of the expression is $z+1.$