Question bank — Applications — solving polynomial equations with complex roots

False in general — this only holds when the coefficients are real. A polynomial like $(z-(2+3i))$ with a complex coefficient has $2+3i$ as its only root and nowhere for the conjugate to appear.

True. Complex roots arrive in conjugate pairs (an even count), so five roots cannot all be non-real — at least one must be left over and real.

False. Four is even, so all four roots can pair up as two conjugate couples, e.g. $z^4+1=0$ whose roots are all non-real.

True. The quadratic formula gives $\frac{-b}{2a}\pm \frac{i\sqrt{4ac-b^2}}{2a}$ — same real part, opposite imaginary parts, which is exactly the conjugate relationship.

False. Over the complex numbers a cube produces three roots ($2$, $-1\pm \sqrt{3}i$), equally spaced on a circle of radius $2$. "One cube root" is only the real-number habit.

True. The polynomial $z^n-1$ has no $z^{n-1}$ term, so by Vieta's Formulas the sum of roots $=-a_{n-1}/a_n=0$. Geometrically the Roots of Unity are symmetric arrows around a circle; equal-length arrows spread evenly point in balanced directions, so tip-to-tail they return to the origin.

True. $(p+qi)(p-qi)=p^2+q^2$, which has no imaginary part — this is exactly why conjugate pairs can "hide" inside a real polynomial as a real quadratic factor.

False. Distinct and counted-with-multiplicity differ: $(z-1{)}^3$ has degree $3$ but only one distinct root, $z=1$, repeated three times.

The error is the hidden word real. Over the reals there is no solution, but over $\mathbb{C}$ the roots are $\pm i$. The Fundamental Theorem of Algebra forbids "no solutions" entirely.

A cubic has three roots. Only two are accounted for; you must divide out $z^2-4z+13$ to find the remaining real root $z=1$.

This finds only one of four roots. The fix is the $+2\pi k$ principle: $16$ points along angle $0$, but angle $0$, $2\pi$, $4\pi$, $6\pi$ are the same arrow — dividing each by $4$ gives four different directions $0,\frac{\pi }{2},\pi ,\frac{3\pi }{2}$, i.e. the roots $2,2i,-2,-2i$ spaced $90°$ apart. Taking a single real root discards three of them.

The arrow points the wrong way. Real coefficients imply conjugate roots, but a polynomial can have both $3+i$ and $3-i$ as roots and still carry complex coefficients elsewhere; the pairing alone doesn't force realness.

Dropping the $+2\pi k$ loses $n-1$ roots. Because $\theta$ and $\theta +2\pi k$ point the same way before dividing, they become $n$ distinct angles $\frac{\theta +2\pi k}{n}$ for $k=0,\dots ,n-1$ after dividing.

Sign slip: $-4i^2=-4(-1)=+4$, so the product is $z^2+4$, a real quadratic. Forgetting $i^2=-1$ is the classic conjugate-pair trap.

No mismatch. With multiplicity it counts as two roots (both equal to $3$), keeping the total at degree $2$ exactly as the theorem requires.

Think of conjugation as a mirror flip across the real axis. Flipping the whole equation $P(\alpha )=0$ flips every coefficient too — but real coefficients sit on the mirror line and don't move, so the flipped equation is still $P$. Its solution is the flipped root $\bar{\alpha }$, so $\bar{\alpha }$ must also be a root: the picture is symmetric top-to-bottom.

Vieta's Formulas give sum of roots $=-a_{n-1}/a_n$, so a missing $z^{n-1}$ ($a_{n-1}=0$) forces the sum to $0$. Visually the roots are equal-length arrows spread evenly around a circle; place them tip-to-tail and they close back to the origin because no direction is favoured.

Factoring over the reals searches for two real numbers multiplying to a negative — geometrically it asks for two points on the real line, which don't exist here. Completing the square instead isolates one squared term equal to a negative number, and its root steps off the line into the imaginary direction, producing the $i$ cleanly every time.

Non-real roots come as mirror-twins (an even count), so an odd total always leaves one root unpartnered — and a lonely root has no mirror image only if it lies on the mirror, i.e. is real. A real root is an $x$-intercept, so the curve must cut the axis.

Dividing by the real quadratic keeps every intermediate coefficient real (no complex arithmetic to slip on) and clears both mirror-twin roots in a single Polynomial Division — you never even have to write down the conjugate separately.

Before dividing, $\theta$ and $\theta +2\pi k$ are the same arrow (a full turn lands you back where you started). After dividing by $n$, a full turn shrinks to a fraction $\frac{2\pi }{n}$ of a turn, so each $k$ nudges the arrow to a genuinely new direction — that fractional step is exactly the even spacing you see in the ring.

One distinct root, $z=0$, with multiplicity 2. Degree still equals total root count; the two roots simply coincide.

No — its coefficients $-(3+2i)$ and $1+i$ are not real, so the theorem gives no guarantee. Its roots ($i$ and $3+i$) are not a conjugate pair, which is perfectly allowed.

All $n$ roots collapse to the single value $z=0$ (with multiplicity $n$). Here the length of $c$ is $r=0$, which forces each root's length $\rho =r^{1/n}=0$ — every arrow has zero length, so the "ring of points" shrinks to the origin.

The base angle is now $\frac{\pi }{n}$, so the ring is rotated off the real axis. Example: $z^2=-1$ gives $\pm i$ — no real root at all — whereas $z^3=-8$ gives one real root $-2$ (from $k=1$) plus a conjugate pair. Never assume angle $0$; negative $c$ means start from $\pi$.

Zero. All three roots are real (one counted twice), so there is no conjugate pair to form — a real cubic can be entirely real.

No. Non-real roots must pair with their conjugates, so their count is always even; "exactly one" is impossible.

If $n$ is even, two roots are real ($\pm r^{1/n}$, where $r=|c|$); if $n$ is odd, exactly one root is real ($+r^{1/n}$). Either way at least one real root exists because the index $k=0$ gives angle $0$, an arrow lying flat along the positive real axis.