Complex Numbers

Work with complex numbers for IB Maths HL. Perform operations, find modulus-argument form, and apply De Moivre's theorem.

Complex numbers extend the real number system with i=−1i = \sqrt{-1}. They're central to IB Math AA HL.

Core Concepts

z=a+biz = a + bi where aa = real part, bb = imaginary part.

Operations

Addition, subtraction, multiplication (FOIL with i2=−1i^2 = -1), division (multiply by conjugate).

Conjugate

zˉ=a−bi\bar{z} = a - bi. zzˉ=a2+b2=∣z∣2z\bar{z} = a^2 + b^2 = |z|^2.

Modulus-Argument (Polar) Form

z=r(cos⁡θ+isin⁡θ)=reiθz = r(\cos\theta + i\sin\theta) = re^{i\theta}

r=∣z∣=a2+b2r = |z| = \sqrt{a^2 + b^2}, θ=arg⁡(z)=arctan⁡(ba)\theta = \arg(z) = \arctan(\frac{b}{a}).

De Moivre's Theorem

[r(cos⁡θ+isin⁡θ)]n=rn(cos⁡nθ+isin⁡nθ)[r(\cos\theta + i\sin\theta)]^n = r^n(\cos n\theta + i\sin n\theta)

Worked Examples

(3+2i)(1−i)=3−3i+2i−2i2=5−i(3+2i)(1-i) = 3 - 3i + 2i - 2i^2 = 5 - i.

∣3+4i∣=5|3+4i| = 5. arg⁡(3+4i)=arctan⁡(43)≈53.1°\arg(3+4i) = \arctan(\frac{4}{3}) \approx 53.1°.

Practice Problems

    1. Express 2+3i1−i\frac{2+3i}{1-i} in the form a+bia + bi.
    1. Convert 2+2i2 + 2i to polar form.
    1. Use De Moivre's to find (1+i)8(1+i)^8.

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Key Takeaways

  • ✓

    i2=−1i^2 = -1. Conjugate of a+bia+bi is a−bia-bi.

  • ✓

    Polar form: reiθr e^{i\theta}. Modulus: a2+b2\sqrt{a^2+b^2}.

  • ✓

    De Moivre's for powers and roots.

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