3.1 Complex numbers and the roots of polynomials
A complex number is \(z = x+iy\) with \(x,y\) real and \(i^2 = -1\); \(x = \operatorname {Re}z\) is its real part and \(y = \operatorname {Im}z\) its imaginary part. Complex numbers are added and multiplied by the usual rules of algebra, replacing \(i^2\) by \(-1\): \[(a+ib)+(c+id) = (a+c)+i(b+d),\qquad (a+ib)(c+id) = (ac-bd)+i(ad+bc).\] The conjugate of \(z = x+iy\) is \(\bar z = x-iy\) and its modulus is \(|z| = \sqrt {x^2+y^2}\), so that \(z\bar z = |z|^2\). Division uses the conjugate of the denominator: \[\frac {a+ib}{c+id} = \frac {(a+ib)(c-id)}{c^2+d^2}.\] Every \(z\neq 0\) has a polar form \(z = r(\cos \theta +i\sin \theta ) = re^{i\theta }\), with \(r = |z|\) and \(\theta \) the angle from the positive real axis. In polar form, multiplication multiplies moduli and adds angles: \(r_1e^{i\theta _1}\,r_2e^{i\theta _2} = r_1r_2e^{i(\theta _1+\theta _2)}\).
Solution. \((2+3i)(1-i) = 2-2i+3i-3i^2 = 5+i\). For the quotient, \[\frac {2+3i}{1-i} = \frac {(2+3i)(1+i)}{(1-i)(1+i)} = \frac {2+2i+3i+3i^2}{2} = \frac {-1+5i}{2} .\] For the cube roots, write \(z = re^{i\theta }\); then \(z^3 = r^3e^{3i\theta } = 1 = e^{2\pi ik}\) gives \(r = 1\) and \(\theta = \frac {2\pi k}{3}\), \(k = 0,1,2\). The roots are \[1,\qquad e^{2\pi i/3} = -\frac 12+\frac {\sqrt 3}{2}i,\qquad e^{4\pi i/3} = -\frac 12-\frac {\sqrt 3}{2}i,\] equally spaced on the unit circle.
Theorem 3.1.2 (Fundamental Theorem of Algebra). Every polynomial of degree \(n\geq 1\), with real or complex coefficients, has at least one complex root.
The proof belongs to complex analysis (it follows, for example, from Liouville’s Theorem) and is given in the complex variables course. Applied repeatedly, it gives the factorization of every polynomial.
Corollary 3.1.3. If \(P\) has degree \(n\geq 1\), there are complex numbers \(x_1,\dots ,x_k\) (the distinct roots) and positive integers \(m_1,\dots ,m_k\) with \(m_1+\dots +m_k = n\) such that \[P(x) = a_n(x-x_1)^{m_1}(x-x_2)^{m_2}\cdots (x-x_k)^{m_k} .\] So \(P\) has exactly \(n\) roots counted with multiplicity. Consequently, if two polynomials of degree at most \(n\) agree at \(n+1\) distinct points, they are identical.
Proof. By Theorem 3.1.2, \(P\) has a root \(x_1\), and by the factor theorem (Theorem 3.2.1 below, with \(P(x_1) = 0\)), \(P(x) = (x-x_1)Q(x)\) with \(\deg Q = n-1\). Repeating the argument on \(Q\) gives the factorization. If \(P_1\) and \(P_2\) have degree at most \(n\) and agree at \(n+1\) points, then \(P_1-P_2\) has degree at most \(n\) and \(n+1\) roots; by the factorization it must be the zero polynomial. □
Theorem 3.1.4. If \(P\) has real coefficients and \(z\) is a root of \(P\), then so is \(\bar z\). Hence the non-real roots of a real polynomial occur in conjugate pairs, each pair contributing a real quadratic factor \[(x-z)(x-\bar z) = x^2-2\operatorname {Re}(z)\,x+|z|^2 .\]
Proof. Conjugation respects sums and products: \(\overline {z+w} = \bar z+\bar w\) and \(\overline {zw} = \bar z\,\bar w\). Since the \(a_k\) are real, \[P(\bar z) = \sum _ka_k\bar z^k = \sum _k\overline {a_kz^k} = \overline {P(z)} = \bar 0 = 0 .\] □
Solution. Check: \((1+i)^2 = 2i\) and \((1+i)^3 = 2i(1+i) = -2+2i\), so \(P(1+i) = -2+2i-2i+2 = 0\). By Theorem 3.1.4, \(1-i\) is also a root, and \((x-(1+i))(x-(1-i)) = x^2-2x+2\) divides \(P\). Long division gives \(x^3-x^2+2 = (x^2-2x+2)(x+1)\), so the third root is \(-1\).
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