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Number and Algebra - Complex Numbers (HL)

Grade 11IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Imaginary Unit and Cartesian Form: A complex number zz is expressed as z=a+biz = a + bi, where aa is the real part Re(z)\text{Re}(z) and bb is the imaginary part Im(z)\text{Im}(z), with i=−1i = \sqrt{-1}. Visually, complex numbers are represented on an Argand Diagram, a 2D plane where the horizontal axis is the Real axis and the vertical axis is the Imaginary axis. The number z=a+biz = a + bi corresponds to the point (a,b)(a, b) or a vector from the origin to that point.

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Modulus and Argument: The modulus ∣z∣|z| represents the magnitude or distance of the complex number from the origin (0,0)(0,0). The argument arg(z)\text{arg}(z) is the angle θ\theta measured counter-clockwise from the positive real axis to the vector representing zz. In a geometric view, if you draw a line from the origin to zz, the length of that line is the modulus and the slope/direction is determined by the argument.

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The Complex Conjugate: For any z=a+biz = a + bi, the conjugate is z∗=a−biz^* = a - bi. Geometrically, z∗z^* is a reflection of zz across the real (horizontal) axis. This operation is crucial for division, as multiplying a complex number by its conjugate z⋅z∗z \cdot z^* always results in a purely real number equal to the square of the modulus ∣z∣2|z|^2.

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Polar and Exponential Forms: Complex numbers can be written in polar form z=r(cos⁡θ+isin⁡θ)z = r(\cos \theta + i \sin \theta) or exponential form z=reiθz = re^{i\theta}, where r=∣z∣r = |z| and θ=arg(z)\theta = \text{arg}(z). This visualizes the number as a point on a circle of radius rr. These forms are significantly more efficient for multiplication and division, where you multiply/divide the radii and add/subtract the angles.

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De Moivre’s Theorem: This theorem states that for any integer nn, (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)). Visually, raising a complex number to a power nn results in scaling its distance from the origin by rnr^n and rotating its position around the origin by nn times its original angle.

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Roots of Complex Numbers: To find the nn-th roots of a complex number ww, we solve zn=wz^n = w. There are exactly nn distinct roots. Geometrically, these roots are perfectly symmetrical; they all lie on a circle of radius ∣w∣1/n|w|^{1/n} and form the vertices of a regular nn-sided polygon centered at the origin on the Argand diagram.

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Euler’s Identity and Relation: The relation eiθ=cos⁡θ+isin⁡θe^{i\theta} = \cos \theta + i \sin \theta links trigonometry and complex exponentials. The famous case where θ=π\theta = \pi leads to eiπ+1=0e^{i\pi} + 1 = 0, which visually represents a 180∘180^{\circ} rotation on the unit circle from the point (1,0)(1,0) to (−1,0)(-1,0).

📐Formulae

i2=−1i^2 = -1

z=a+biz = a + bi

∣z∣=a2+b2|z| = \sqrt{a^2 + b^2}

arg(z)=θ=arctan⁡(ba)\text{arg}(z) = \theta = \arctan(\frac{b}{a}) (adjusted for quadrant)

z∗=a−biz^* = a - bi

zz∗=∣z∣2z z^* = |z|^2

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

z1z2=r1r2ei(θ1+θ2)z_1 z_2 = r_1 r_2 e^{i(\theta_1 + \theta_2)}

z1z2=r1r2ei(θ1−θ2)\frac{z_1}{z_2} = \frac{r_1}{r_2} e^{i(\theta_1 - \theta_2)}

(reiθ)n=rneinθ(re^{i\theta})^n = r^n e^{in\theta}

zk=r1/nei(θ+2kπn)z_k = r^{1/n} e^{i(\frac{\theta + 2k\pi}{n})} for k=0,1,…,n−1k = 0, 1, \dots, n-1

💡Examples

Problem 1:

Given z=1+i3z = 1 + i\sqrt{3}, express zz in polar form and find z6z^6.

Solution:

  1. Find the modulus: r=∣z∣=12+(3)2=1+3=2r = |z| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2.
  2. Find the argument: θ=arctan⁡(31)=π3\theta = \arctan(\frac{\sqrt{3}}{1}) = \frac{\pi}{3}.
  3. Polar form: z=2(cos⁡π3+isin⁡π3)z = 2(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}).
  4. Use De Moivre's Theorem for z6z^6: z6=26(cos⁡(6⋅π3)+isin⁡(6⋅π3))z^6 = 2^6(\cos(6 \cdot \frac{\pi}{3}) + i \sin(6 \cdot \frac{\pi}{3})).
  5. Simplify: z6=64(cos⁡2π+isin⁡2π)=64(1+0i)=64z^6 = 64(\cos 2\pi + i \sin 2\pi) = 64(1 + 0i) = 64.

Explanation:

We first convert the Cartesian coordinates to polar coordinates (r,θ)(r, \theta) to make exponentiation easier. Applying De Moivre's Theorem allows us to raise the magnitude to the power of 6 and multiply the angle by 6, resulting in a real number.

Problem 2:

Solve the equation z3=8iz^3 = 8i and describe the positions of the roots on the Argand diagram.

Solution:

  1. Convert 8i8i to exponential form: 8i=8eiπ28i = 8e^{i\frac{\pi}{2}} (since it lies on the positive imaginary axis).
  2. Apply the root formula: zk=81/3ei(π/2+2kπ3)z_k = 8^{1/3} e^{i(\frac{\pi/2 + 2k\pi}{3})} for k=0,1,2k = 0, 1, 2.
  3. For k=0k=0: z0=2eiπ6=2(cos⁡π6+isin⁡π6)=3+iz_0 = 2e^{i\frac{\pi}{6}} = 2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}) = \sqrt{3} + i.
  4. For k=1k=1: z1=2ei(π/2+2π3)=2ei5π6=2(cos⁡5π6+isin⁡5π6)=−3+iz_1 = 2e^{i(\frac{\pi/2 + 2\pi}{3})} = 2e^{i\frac{5\pi}{6}} = 2(\cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6}) = -\sqrt{3} + i.
  5. For k=2k=2: z2=2ei(π/2+4π3)=2ei9π6=2ei3π2=−2iz_2 = 2e^{i(\frac{\pi/2 + 4\pi}{3})} = 2e^{i\frac{9\pi}{6}} = 2e^{i\frac{3\pi}{2}} = -2i.

Explanation:

To solve for nn-th roots, we represent the complex number in exponential form and use the general root formula. Geometrically, the roots z0,z1,z2z_0, z_1, z_2 all have a modulus of 2 and are separated by 2π3\frac{2\pi}{3} radians (120∘120^{\circ}), forming an equilateral triangle inscribed in a circle of radius 2.