Calcul 1 Z 2 Z 1 Substition

Use this premium calculator to evaluate the expression 1 / (z² – 2z + 1), rewrite it with the substitution u = z – 1, and visualize how the denominator and reciprocal behave around the singular point z = 1. This is especially useful for algebra review, precalculus, and calculus preparation.

Calculator Inputs

Identity used:
z² – 2z + 1 = (z – 1)²
Therefore:
1 / (z² – 2z + 1) = 1 / (z – 1)²
With substitution u = z – 1, the expression becomes 1 / u².

Results

Understanding calcul 1 / (z² – 2z + 1) by substitution

The expression 1 / (z² – 2z + 1) appears simple, but it is an excellent example for learning algebraic substitution, factor recognition, and function behavior near a critical point. Many students first try to evaluate it directly by plugging in a value of z. That works for most values, but it does not immediately reveal the deeper structure of the denominator. Once you recognize that z² – 2z + 1 is a perfect square trinomial, the entire problem becomes more transparent.

The key algebraic identity is:

z² – 2z + 1 = (z – 1)²

That means the original expression can be rewritten as:

1 / (z² – 2z + 1) = 1 / (z – 1)²

Now apply the substitution u = z – 1. Under that change of variable, the expression becomes:

1 / u²

This is the heart of the substitution approach. Instead of treating the denominator as a quadratic every time, you transform it into a squared linear quantity. That makes evaluation, graphing, and analysis much easier. It is also a valuable bridge into calculus, where substitution is often used to simplify derivatives, integrals, and rational expressions.

Why substitution matters here

Substitution is not just a formal trick. It reveals structure. When you let u = z – 1, you are centering the problem around the critical point z = 1. This is the point where the denominator becomes zero. In the transformed variable, that same issue becomes u = 0, which is even easier to recognize and discuss.

• The denominator is always nonnegative because it is a square.
• The denominator equals zero only when z = 1.
• The reciprocal grows very large as z approaches 1 from either side.
• The expression is undefined at z = 1.

These points are not merely symbolic observations. They tell you about the graph, the domain, and the numerical sensitivity of the function. Near z = 1, even tiny changes in z can create very large changes in the output. This is one reason a substitution-based view is so powerful in both teaching and computation.

Step by step method

1. Start with the denominator: z² – 2z + 1.
2. Recognize the perfect square pattern: a² – 2ab + b² = (a – b)².
3. Rewrite it as (z – 1)².
4. Substitute u = z – 1.
5. The expression becomes 1 / u².
6. Evaluate with your chosen value of z, as long as z ≠ 1.

Worked examples

Suppose z = 2. Then:

• Direct denominator: 2² – 2(2) + 1 = 4 – 4 + 1 = 1
• Function value: 1 / 1 = 1
• Substitution: u = 2 – 1 = 1
• Transformed value: 1 / u² = 1 / 1² = 1

Now let z = 1.5. Then:

• Denominator: 1.5² – 2(1.5) + 1 = 2.25 – 3 + 1 = 0.25
• Function value: 1 / 0.25 = 4
• Substitution: u = 1.5 – 1 = 0.5
• Transformed value: 1 / 0.5² = 1 / 0.25 = 4

As expected, both approaches produce the same answer. The substitution method simply exposes the inner logic more clearly.

z	z² – 2z + 1	(z – 1)²	1 / (z² – 2z + 1)	Interpretation
0	1	1	1.0000	Regular point, easy to evaluate.
0.5	0.25	0.25	4.0000	Closer to z = 1, output rises sharply.
0.9	0.01	0.01	100.0000	Near the singular point, very sensitive.
1.1	0.01	0.01	100.0000	Symmetric behavior on the right side.
2	1	1	1.0000	Same value as z = 0 due to symmetry around z = 1.

What the graph tells you

Graphically, the denominator (z – 1)² is a parabola shifted one unit to the right from z². Its vertex is at (1, 0). Since the denominator touches zero there, the reciprocal 1 / (z – 1)² has a vertical asymptote at z = 1. This means the function is not defined at that point and increases without bound as z approaches 1 from either side.

There is another important graph insight: the function is symmetric about the vertical line z = 1. In practical terms, values such as z = 0.8 and z = 1.2 give the same output because both are the same distance from 1. The substitution u = z – 1 makes that symmetry obvious.

Numerical sensitivity near z = 1

Because the denominator is squared, the function changes very quickly near the singular point. The next table shows how rapidly the output grows as z moves closer to 1.

z	Distance from 1, |z – 1|	Squared distance, (z – 1)²	Reciprocal value	Growth factor vs z = 0.9
0.9	0.1	0.01	100	1x
0.99	0.01	0.0001	10,000	100x
0.999	0.001	0.000001	1,000,000	10,000x
1.01	0.01	0.0001	10,000	100x
1.001	0.001	0.000001	1,000,000	10,000x

These are real computed values, and they show a critical fact: every time the distance from 1 shrinks by a factor of 10, the reciprocal increases by a factor of 100. That is because the denominator is proportional to the square of the distance from 1.

Domain, restrictions, and common mistakes

The domain includes all real values of z except z = 1. Students often make one of the following errors:

• Forgetting to factor the denominator as a perfect square.
• Ignoring the undefined point at z = 1.
• Using the wrong substitution, such as u = z + 1 instead of u = z – 1.
• Assuming the reciprocal is negative somewhere. It is not, because the denominator is a square and therefore nonnegative.

Important: Since (z – 1)² ≥ 0 for every real z, the expression 1 / (z – 1)² is always positive wherever it is defined.

Why this pattern is useful in calculus

Expressions of this type often appear in derivative work, limit problems, and integration exercises. For example, if you need to integrate 1 / (z – 1)², substitution can simplify setup. If you are studying limits, the point z = 1 becomes the natural center of analysis. If you are graphing rational functions, this is a clean model for understanding vertical asymptotes and symmetry.

To deepen your conceptual understanding, review materials from authoritative academic sources such as MIT OpenCourseWare and Lamar University calculus and algebra notes. For broader mathematical standards and computational reference resources, you can also consult NIST.

Direct form vs substitution form

Direct evaluation is useful when you already know the numerical value of z. Substitution is better when you want insight. Both are correct, but they serve different purposes.

• Direct form: Efficient for straightforward numerical plugging.
• Substitution form: Better for pattern recognition, graph interpretation, and extension to calculus.
• Factored form: Best for discussing the domain and asymptotic behavior.

When to prefer substitution

1. When you suspect a quadratic can be simplified into a square.
2. When you want to study the expression near a critical value.
3. When the same structure may later appear in derivatives or integrals.
4. When you need a more stable conceptual form for teaching or checking work.

Final takeaway

The problem of calcul 1 / (z² – 2z + 1) by substitution is a compact but very rich exercise. It reinforces algebraic factoring, variable substitution, graph reading, and domain analysis all at once. The transformation from z² – 2z + 1 to (z – 1)² is the key move. Once that is recognized, the substitution u = z – 1 turns the expression into the much simpler form 1 / u².

In other words, the real lesson is not only how to compute one rational expression. It is how to detect hidden structure and use substitution to make mathematics easier, clearer, and more reliable. That is exactly the habit that supports stronger performance in algebra, precalculus, and calculus.

This calculator is for educational use. Values very close to z = 1 produce extremely large outputs because the denominator approaches zero.