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 \(z =\)

• A. 1 and \((2 - i)\)
• B. \(2i\) and \(-2i\)
• C. 1 and \((2 + i)\)
• D. \((2 + i)\) only

Hint:

Factor the numerator and denominator first. Check whether any factor cancels between them before you decide which points are truly singular.

Answer 1

The Correct Option is B

Step 1: Test $z=1$ as a root of the denominator.
Write $g(z) = z^3 - z^2 + 4z - 4$ (the denominator, rearranged in standard order). Put $z=1$ into $g(z)$: $1 - 1 + 4 - 4 = 0$. So $z=1$ is a root of $g(z)$, and $(z-1)$ must be one of its factors.

Step 2: Divide out the known root.
Divide $g(z)$ by $(z-1)$ using synthetic division with coefficients $1, -1, 4, -4$. Bring down $1$. Multiply by $1$, add to $-1$, get $0$. Multiply by $1$, add to $4$, get $4$. Multiply by $1$, add to $-4$, get $0$ (the remainder, confirming the root). The quotient is $z^2 + 0z + 4 = z^2 + 4$, so $g(z) = (z-1)(z^2+4)$.

Step 3: Check the numerator at $z=1$.
The numerator is $h(z) = z^2 - 5z + 4$. At $z=1$: $1 - 5 + 4 = 0$. Since both the numerator and the denominator vanish at $z=1$, this point is not a genuine pole; the singularity is removable once the common factor is cancelled, because the limit of $f(z)$ as $z \to 1$ stays finite.

Step 4: Solve for the remaining zeros of the denominator.
Set $z^2 + 4 = 0$, so $z^2 = -4$, giving $z = \pm\sqrt{-4} = \pm 2i$. Since the numerator $h(z) = z^2-5z+4$ does not vanish at $z = 2i$ or $z = -2i$, both points are genuine poles of $f(z)$.

Step 5: Eliminate the remaining options.
Any option keeping $z=1$ as a singular point (A, C) is wrong because that factor cancels. Option (D) misses the second root $-2i$ of $z^2+4=0$, so it is incomplete.

Final Answer:
The singular points are $z = 2i$ and $z = -2i$, option (B).\[ \boxed{z = 2i,\ -2i} \]