Question:
The function \(f(z)\) of the complex variable \(z\) given below,\[ f(z) = \dfrac{z^{2} - 5z + 4}{z^{3} + 4z - z^{2} - 4} \]has singular points at \(z =\)
The function \(f(z)\) of the complex variable \(z\) given below,\[ f(z) = \dfrac{z^{2} - 5z + 4}{z^{3} + 4z - z^{2} - 4} \]has singular points at \(z =\)
Show Hint
Factor the numerator and denominator first. Check whether any factor cancels between them before you decide which points are truly singular.
Updated On: Jul 28, 2026
- 1 and \((2 - i)\)
- \(2i\) and \(-2i\)
- 1 and \((2 + i)\)
- \((2 + i)\) only
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The Correct Option is B
Solution and Explanation
Step 1: Factor the numerator.
The numerator is \(z^2 - 5z + 4\). We need two numbers that multiply to 4 and add to -5, which are -1 and -4. So the numerator splits as \(z^2 - 5z + 4 = (z-1)(z-4)\).
Step 2: Factor the denominator by grouping.
Write the denominator in standard order first: \(z^3 + 4z - z^2 - 4 = z^3 - z^2 + 4z - 4\). Group the first two terms and the last two terms separately.
\[ z^3 - z^2 + 4z - 4 = z^2(z-1) + 4(z-1) \]
Both groups share the factor \((z-1)\), so pull it out.
\[ z^3 - z^2 + 4z - 4 = (z-1)(z^2+4) \]
Step 3: Simplify the function.
Put the factored numerator and denominator together.
\[ f(z) = \dfrac{(z-1)(z-4)}{(z-1)(z^2+4)} \]
The factor \((z-1)\) appears in both the top and the bottom, so it cancels for every \(z\) apart from \(z=1\) itself. This means \(z=1\) is a removable singularity, not a genuine pole. Once the common factor is removed we are left with \(f(z) = \dfrac{z-4}{z^2+4}\) away from \(z=1\).
Step 4: Find the true singular points.
The remaining denominator \(z^2+4\) vanishes when \(z^2 = -4\), which gives \(z = 2i\) and \(z = -2i\). At these two points the numerator \(z-4\) is not zero, so these are genuine poles of \(f(z)\).
Step 5: Why the other options are wrong.
Options (A) and (C) both keep \(z=1\) as a singular point, but that factor cancels out, so \(z=1\) is not a true singularity of \(f(z)\); this rules out both. Option (D) drops \(-2i\) even though \(z^2+4=0\) has two roots, \(2i\) and \(-2i\), so it is incomplete.
Final Answer:
The genuine singular points of \(f(z)\) are \(z=2i\) and \(z=-2i\), which is option (B).\[ \boxed{z = 2i,\ -2i} \]
The numerator is \(z^2 - 5z + 4\). We need two numbers that multiply to 4 and add to -5, which are -1 and -4. So the numerator splits as \(z^2 - 5z + 4 = (z-1)(z-4)\).
Step 2: Factor the denominator by grouping.
Write the denominator in standard order first: \(z^3 + 4z - z^2 - 4 = z^3 - z^2 + 4z - 4\). Group the first two terms and the last two terms separately.
\[ z^3 - z^2 + 4z - 4 = z^2(z-1) + 4(z-1) \]
Both groups share the factor \((z-1)\), so pull it out.
\[ z^3 - z^2 + 4z - 4 = (z-1)(z^2+4) \]
Step 3: Simplify the function.
Put the factored numerator and denominator together.
\[ f(z) = \dfrac{(z-1)(z-4)}{(z-1)(z^2+4)} \]
The factor \((z-1)\) appears in both the top and the bottom, so it cancels for every \(z\) apart from \(z=1\) itself. This means \(z=1\) is a removable singularity, not a genuine pole. Once the common factor is removed we are left with \(f(z) = \dfrac{z-4}{z^2+4}\) away from \(z=1\).
Step 4: Find the true singular points.
The remaining denominator \(z^2+4\) vanishes when \(z^2 = -4\), which gives \(z = 2i\) and \(z = -2i\). At these two points the numerator \(z-4\) is not zero, so these are genuine poles of \(f(z)\).
Step 5: Why the other options are wrong.
Options (A) and (C) both keep \(z=1\) as a singular point, but that factor cancels out, so \(z=1\) is not a true singularity of \(f(z)\); this rules out both. Option (D) drops \(-2i\) even though \(z^2+4=0\) has two roots, \(2i\) and \(-2i\), so it is incomplete.
Final Answer:
The genuine singular points of \(f(z)\) are \(z=2i\) and \(z=-2i\), which is option (B).\[ \boxed{z = 2i,\ -2i} \]
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