Complex Numbers (Pure 3)

The number $i$ and what it fixes, arithmetic with complex numbers, modulus–argument form, and reading a locus off an Argand diagram instead of solving for it.

หลักสูตรแกนกลาง / สสวท.: จำนวนและพีชคณิต (Number and Algebra)ตรงกับ ม.5 จำนวนเชิงซ้อน (เพิ่มเติม) — ส่วนแผนภาพอาร์กองด์และโลคัสไม่มีในหลักสูตรไทย
Common Core (US): N-CN
Cambridge: 9709 Pure Mathematics 3 — Complex numbers

12.1 Arithmetic with complex numbers

นิยาม

The imaginary unit

$i$ is defined by $i^{2} = - 1$ . It exists because $x^{2} + 1 = 0$ had no solution, and inventing one turns out to give every polynomial equation the number of roots its degree promises — which is the reason complex numbers are useful rather than merely consistent.

Multiplication and division

(a + bi) (c + d i) = (a c - b d) + (a d + b c) i \frac{1}{c + d i} = \frac{c - d i}{c ^{2} + d ^{2}}

Division works by multiplying top and bottom by the conjugate — the same move as rationalising a surd denominator, for the same reason: it makes the bottom real.

ตัวอย่าง

Express $\frac{3 + 2 i}{1 - 4 i}$ in the form $a + bi$ .

วิธีทำ

1.Multiply top and bottom by $1 + 4 i$
2.Numerator: $3 + 12 i + 2 i + 8 i^{2} = - 5 + 14 i$
3.Denominator: $1 + 16 = 17$

ตอบ $- \frac{5}{17} + \frac{14}{17} i$

ทำไมถึงเป็นแบบนั้น

Complex roots of a real-coefficient polynomial always come in conjugate pairs. So being told that $2 - 3 i$ is a root of a real cubic hands you $2 + 3 i$ for free, and with it a real quadratic factor — which is usually the whole first part of the question.

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