Exercises — Applications — solving polynomial equations with complex roots

(a) False. No real solution, yes — but $x^2=-9$ gives $x=\pm 3i$. Complex numbers always supply the answers. (b) True. Real coefficients force complex roots into conjugate pairs (Conjugate Root Theorem). (c) False. Complex roots come in pairs (an even count). Five roots cannot all pair up, so at least one is real.

(a) Quadratic formula — it is degree 2 with (as we will see) a negative discriminant. (b) De Moivre / Roots of Unity — the form $z^n=c$ means "$n$ equally spaced points on a circle." (c) Conjugate pairs + Polynomial Division. Here is what that recipe means in one concrete pass, so the name is not just a label:

1. Pair: real coefficients ⇒ $\overline{1+i}=1-i$ is also a root.
2. Form the real quadratic: $(z-(1+i))(z-(1-i))=z^2-2z+2$ (sum $2$, product $1+1=2$).
3. Divide: carry out $z^4-3z^3+\dots$ ÷ $z^2-2z+2$ to peel off this factor, leaving another quadratic to solve. (The actual long division is done in full at L3.1 and L4.1 — this is just the shape of the plan.)

Coefficients $a=1,b=-6,c=25$. Discriminant (the number $b^2-4ac$ under the root): $b^2-4ac=36-100=-64<0.$ Negative, so roots are a conjugate pair. Using the complex quadratic formula: $z=\frac{6\pm \sqrt{-64}}{2}=\frac{6\pm 8i}{2}=3\pm 4i.$ Check (Vieta): sum $=(3+4i)+(3-4i)=6=-b/a$ ✓; product $=9+16=25=c/a$ ✓.

$a=2,b=2,c=5$. Discriminant $=b^2-4ac=4-40=-36<0$. $z=\frac{-2\pm \sqrt{-36}}{2\cdot 2}=\frac{-2\pm 6i}{4}=-\frac{1}{2}\pm \frac{3}{2}i.$ Check: sum $=-1=-b/a=-2/2$ ✓; product $={\left(-\frac{1}{2}\right)}^2+{\left(\frac{3}{2}\right)}^2=\frac{1}{4}+\frac{9}{4}=\frac{10}{4}=\frac{5}{2}=c/a$ ✓.

A conjugate pair multiplies to a real quadratic (see Vieta's Formulas): $z^2-(\text{sum})z+(\text{product}).$ Sum $=(-1+2i)+(-1-2i)=-2$. Product $=(-1{)}^2+(2{)}^2=1+4=5.$ $\boxed{z^2+2z+5.}$

Step 1 — pair it. Real coefficients ⇒ $2-i$ is also a root. Step 2 — real quadratic factor. Sum $=(2+i)+(2-i)=4$; product $=4+1=5$. So $(z-(2+i))(z-(2-i))=z^2-4z+5.$ Step 3 — divide, by matching coefficients. We know the cubic equals $(z^2-4z+5)\times (\text{linear})$. Write the unknown linear factor as $z^2-4z+5$ times $(qz+s)$ and force the product to reproduce the cubic term by term — this is the matching-coefficients method: two polynomials are equal only if the coefficient of each power of $z$ agrees.

• Leading term ($z^3$): the product's top term is $z^2\cdot qz=q{z}^3$; the cubic's top term is $1\cdot z^3$. Matching forces $q=1$.
• Constant term ($z^0$): the product's constant is $5\cdot s$; the cubic's constant is $+5$. Matching forces $5s=5\Rightarrow s=1$. So the linear factor is $z+1$. Confirm the middle coefficients by expanding (every power must still agree): $(z^2-4z+5)(z+1)=z^3+z^2-4z^2-4z+5z+5=z^3-3z^2+z+5.✓$ The $z^2$ coefficient came out $1-4=-3$ and the $z$ coefficient $-4+5=+1$, both matching — so $z+1$ is correct. Roots: $\boxed{2+i,2-i,-1}$. Odd degree → one lonely real root, exactly as predicted.

Step 1 — third root. Real coefficients ⇒ the conjugate $1+2i$ is also a root. That is all three: $1-2i,1+2i,4$. Step 2 — rebuild the polynomial (leading coefficient $1$): $(z-4)(z-(1-2i))(z-(1+2i)).$ The conjugate pair gives $z^2-2z+5$ (sum $2$, product $1+4=5$). Then $(z-4)(z^2-2z+5)=z^3-2z^2+5z-4z^2+8z-20=z^3-6z^2+13z-20.$ So $\boxed{a=-6,b=13,c=-20}$. Vieta check: sum of roots $=(1-2i)+(1+2i)+4=6=-a$ ✓; product $=4\cdot (1+4)=20=(-1{)}^{3}c=-c$ ✓.

Because $q\ne 0$, the root $\alpha =p+qi$ sits off the real axis and its mirror image $\bar{\alpha }=p-qi$ sits on the other side, making them two distinct roots that merge into a single repeated real root at $p$ only in the limit $q\to 0$ when both dots slide onto the line.

Step 1 — pair. $-i$ is also a root ⇒ factor $(z-i)(z+i)=z^2+1$. Step 2 — divide. Split off $z^2+1$: $z^4-2z^3+6z^2-2z+5=(z^2+1)(z^2-2z+5).$ Check: $(z^2+1)(z^2-2z+5)=z^4-2z^3+5z^2+z^2-2z+5=z^4-2z^3+6z^2-2z+5$ ✓. Step 3 — solve the leftover $z^2-2z+5=0$: discriminant $=b^2-4ac=4-20=-16<0$, so $z=\frac{2\pm 4i}{2}=1\pm 2i.$ All roots: $\boxed{i,-i,1+2i,1-2i}$ — two conjugate pairs.

Write $8=8(\cos 0+i\sin 0)$, so modulus $r=8$, angle $\theta =0$. By De Moivre's Theorem the three cube roots have modulus $r^{1/3}=2$ and angles $\frac{0+2\pi k}{3}$ for $k=0,1,2$: $k=0:2(\cos 0+i\sin 0)=2,$ $k=1:2\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)=2\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)=-1+\sqrt{3}i,$ $k=2:2\left(\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3}\right)=-1-\sqrt{3}i.$ The figure below plots these as three amber position-vectors on the circle of radius $2$; read it as a force-balance diagram — it is the visual proof that the three roots cancel. Look at how the single root $2$ pointing right is exactly opposed by the two upper-left roots, whose sideways ($\sqrt{3}i$ and $-\sqrt{3}i$) components cancel each other while their leftward parts ($-1$ and $-1$) add to $-2$, killing the $+2$.

Write $-16=16(\cos \pi +i\sin \pi )$, so $r=16$, $\theta =\pi$. Modulus of each root $=16^{1/4}=2$. Angles $\frac{\pi +2\pi k}{4}$ for $k=0,1,2,3$: $\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$ (i.e. $45^{\circ },135^{\circ },225^{\circ },315^{\circ }$). Since $\cos$ and $\sin$ of these are $\pm \frac{\sqrt{2}}{2}$, each root is $2\cdot \left(\pm \frac{\sqrt{2}}{2}\pm \frac{\sqrt{2}}{2}i\right)=\pm \sqrt{2}\pm \sqrt{2}i$: $\boxed{\sqrt{2}+\sqrt{2}i,-\sqrt{2}+\sqrt{2}i,-\sqrt{2}-\sqrt{2}i,\sqrt{2}-\sqrt{2}i.}$ The figure below plots the four amber dots and lets you see the conjugate-pair symmetry directly — the two upper dots are exact mirror images (across the real axis) of the two lower dots, which is why $z^4+16$ has real coefficients. Notice each dot sits at a diagonal $45^{\circ }$ offset, so none lands on an axis (the starting angle was $45^{\circ }$, not $0^{\circ }$).

Step 1 — all four roots. Real coefficients force both conjugates in: roots are $3+i,3-i,2-i,2+i$. Step 2 — two real quadratics. Pair $3\pm i$: sum $6$, product $9+1=10\Rightarrow z^2-6z+10$. Pair $2\pm i$: sum $4$, product $4+1=5\Rightarrow z^2-4z+5$. Step 3 — multiply. $(z^2-6z+10)(z^2-4z+5).$ Expand term by term: $z^4-4z^3+5z^2-6z^3+24z^2-30z+10z^2-40z+50$ $=z^4-10z^3+(5+24+10)z^2+(-30-40)z+50=z^4-10z^3+39z^2-70z+50.$ So $\boxed{b=-10,c=39,d=-70,e=50}$. Vieta checks: sum of roots $=6+4=10=-b$ ✓; product $=10\cdot 5=50=e$ (since $(-1{)}^4=1$) ✓.

Algebraic view. The solutions are the roots of $z^n-c=0$, a polynomial whose $z^{n-1}$ coefficient is $0$ (there is no $z^{n-1}$ term). By Vieta's Formulas, sum of roots $=-\frac{a_{n-1}}{a_n}=-\frac{0}{1}=0$ for all $n\ge 2$. Geometric view. Every solution equals $r^{1/n}$ times an $n$-th root of unity. Those $n$ unit vectors are equally spaced by $\frac{2\pi }{n}$ around the circle — a symmetric star. Summing them is like adding the position vectors of the corners of a regular $n$-gon centred at the origin: by rotational symmetry (rotating the whole set by $\frac{2\pi }{n}$ leaves it unchanged, but a nonzero sum would have to rotate too — contradiction), the total is $0$. Scaling by $r^{1/n}$ keeps the sum zero.

Step 1 — conjugate. $1-2i$ is also a root. Their real quadratic: sum $2$, product $1+4=5\Rightarrow z^2-2z+5$. Step 2 — the third root $\rho$. Sum of all three roots must be $0$ (coefficient of $z^2$ is $0$). So $(1+2i)+(1-2i)+\rho =0\Rightarrow 2+\rho =0\Rightarrow \rho =-2.$ Step 3 — build $p,q$. The cubic is $(z+2)(z^2-2z+5)=z^3-2z^2+5z+2z^2-4z+10=z^3+z+10.$ The $z^2$ terms cancel — good, that is the built-in check. So $\boxed{p=1,q=10}$, real root $\rho =-2$. Vieta confirm: sum of roots $=(1+2i)+(1-2i)+(-2)=0$ ✓; product $=(-2)(1+4)=-10=(-1{)}^{3}q=-q$ ✓.

• Negative discriminant $b^2-4ac<0$ → conjugate pair via $\frac{-b\pm i\sqrt{4ac-b^2}}{2a}$ (denominator $2a$!).
• One complex root given → pair first, form real quadratic, then divide.
• $z^n=c$ → modulus $r^{1/n}$, angles $\frac{\theta +2\pi k}{n}$, $k=0,\dots ,n-1$ (all $n$ of them).
• Always Vieta-check: sum $=-a_{n-1}/a_n$, product $=(-1{)}^n{a}_0/a_n$.
• Missing $z^{n-1}$ term ⇔ roots sum to $0$.