How many roots does the polynomial z¹⁰⁰ − 50z³⁰ + 40z¹⁰ + 6z + 1 have in the open disc {z ∈ ℂ : |z| < 1}?

C. On |z| = 1, |−50z³⁰| = 50 > 1 + 40 + 6 + 1 = 48 ≥ |z¹⁰⁰ + 40z¹⁰ + 6z + 1|, so by Rouché the polynomial has as many zeros inside as −50z³⁰: thirty.

Part BCSIR NET December 2023rouche-dominant-term

Rouche dominant term

How many roots does the polynomial $z^{100} - 50z^{30} + 40z^{10} + 6z + 1$ have in the open disc { $z \in \mathbb{C}$ : |z| < 1}?

A.100
B.50
C.30✓ Correct
D.0

The idea, in plain words

On the unit circle one term is bigger than everything else put together. When one term dominates like that, the whole polynomial has exactly as many roots inside as that single term does — and a bare power has all its roots stacked at the origin.

Solution

On |z| = 1, | $-50z^{30}$ | = 50 > 1 + 40 + 6 + 1 = 48 ≥ | $z^{100} + 40z^{10} + 6z + 1$ |

so by Rouché the polynomial has as many zeros inside as $-50z^{30}$ : thirty.

Why each option is right or wrong

Checked against 2 symbolic computations

A.WrongExecution slip
False.

100 is the total number of roots in $\mathbb{C}$ , by the fundamental theorem of algebra — not the number inside the unit disc.

Rouché's theorem localises them, and it selects the dominant term on the circle, not the leading term of the polynomial.
B.WrongExecution slip
False.

50 is the coefficient of the dominant term, not its degree.

Rouché counts zeros of the dominant term $-50z^{30}$ , and a constant multiple contributes nothing to the zero count — only the exponent 30 does.
C.Correct
True.

On |z| = 1 every power has modulus 1

so | $-50z^{30}$ | = 50 while the remaining terms satisfy | $z^{100} + 40z^{10} + 6z + 1$ | ≤ 1 + 40 + 6 + 1 = 48 (verified)

leaving a margin of 2 (verified).

Since 48 < 50, Rouché's theorem says the polynomial has the same number of zeros inside as $-50z^{30}$

which has a zero of order 30 at the origin.

The trigger: on the unit circle, compare the single largest coefficient against the sum of all the others.
D.Wrong
False.

The polynomial takes the value 1 at z = 0 and is dominated by $-50z^{30}$ on the boundary

so it certainly changes character inside; Rouché gives exactly 30 zeros there, not none.

2 of these 3 wrong options correspond to a named reasoning error — the same errors recur across subjects.

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

Related counterexample: f′(z) ≠ 0 everywhere implies f is injective

Rouche dominant term

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