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Number and Algebra - The complex plane (HL)

Grade 11IB_AA

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

🔑Concepts

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The complex plane (Argand diagram) represents complex numbers z=a+biz = a + bi as points (a,b)(a, b) or vectors from the origin.

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The Cartesian form is z=a+biz = a + bi, where a=Re(z)a = \text{Re}(z) and b=Im(z)b = \text{Im}(z).

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The Modulus of zz is its distance from the origin: ∣z∣=r=a2+b2|z| = r = \sqrt{a^2 + b^2}.

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The Argument of zz, denoted arg⁡(z)\arg(z), is the angle θ\theta measured from the positive real axis. The principal argument satisfies −π<θ≤π-\pi < \theta \le \pi.

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The Polar form of a complex number is z=r(cos⁡θ+isin⁡θ)z = r(\cos \theta + i \sin \theta), often abbreviated as z=r cis θz = r \text{ cis } \theta.

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The Euler form is z=reiθz = re^{i\theta}, which is derived from the identity eiθ=cos⁡θ+isin⁡θe^{i\theta} = \cos \theta + i \sin \theta.

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Multiplication and division are simplified in Euler form: z1z2=r1r2ei(θ1+θ2)z_1 z_2 = r_1 r_2 e^{i(\theta_1 + \theta_2)} and z1z2=r1r2ei(θ1−θ2)\frac{z_1}{z_2} = \frac{r_1}{r_2} e^{i(\theta_1 - \theta_2)}.

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De Moivre's 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).

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The nn-th roots of a complex number z=reiθz = re^{i\theta} are given by wk=r1/nei(θ+2kπn)w_k = r^{1/n} e^{i\left(\frac{\theta + 2k\pi}{n}\right)} for k=0,1,2,…,n−1k = 0, 1, 2, \dots, n-1. These roots form a regular nn-sided polygon centered at the origin on the complex plane.

📐Formulae

z=a+biz = a + bi

∣z∣=a2+b2|z| = \sqrt{a^2 + b^2}

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

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

z∗=a−bi=re−iθz^* = a - bi = re^{-i\theta}

zn=rneinθ=rn(cos⁡nθ+isin⁡nθ)z^n = r^n e^{in\theta} = r^n(\cos n\theta + i \sin n\theta)

z1/n=r1/nei(θ+2kπn), where k=0,1,…,n−1z^{1/n} = r^{1/n} e^{i\left(\frac{\theta + 2k\pi}{n}\right)}, \text{ where } k = 0, 1, \dots, n-1

💡Examples

Problem 1:

Given z=1−i3z = 1 - i\sqrt{3}, express zz in Euler form and find z4z^4 in Cartesian form.

Solution:

  1. Find modulus rr: r=12+(−3)2=1+3=2r = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = 2.
  2. Find argument θ\theta: Since zz is in the 4th quadrant, θ=arctan⁡(−31)=−π3\theta = \arctan\left(\frac{-\sqrt{3}}{1}\right) = -\frac{\pi}{3}.
  3. Euler form: z=2e−iπ3z = 2e^{-i\frac{\pi}{3}}.
  4. Use De Moivre's Theorem for z4z^4: z4=(2e−iπ3)4=24e−i4π3=16ei2π3z^4 = (2e^{-i\frac{\pi}{3}})^4 = 2^4 e^{-i\frac{4\pi}{3}} = 16 e^{i\frac{2\pi}{3}}
  5. Convert to Cartesian: 16(cos⁡2π3+isin⁡2π3)=16(−12+i32)=−8+8i316\left(\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}\right) = 16\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = -8 + 8i\sqrt{3}

Explanation:

We first identify the modulus and the quadrant to find the correct argument. Then we use exponent rules for the Euler form and convert back to Cartesian using trigonometric values.

Problem 2:

Find the three cube roots of w=8iw = 8i.

Solution:

  1. Convert w=8iw = 8i to polar form: r=8r = 8, θ=π2\theta = \frac{\pi}{2}, so w=8eiπ2w = 8e^{i\frac{\pi}{2}}.
  2. Apply the root formula zk=81/3ei(π/2+2kπ3)z_k = 8^{1/3} e^{i\left(\frac{\pi/2 + 2k\pi}{3}\right)} for k=0,1,2k=0, 1, 2.
  3. For k=0k=0: z0=2eiπ6=2(32+i12)=3+iz_0 = 2e^{i\frac{\pi}{6}} = 2\left(\frac{\sqrt{3}}{2} + i\frac{1}{2}\right) = \sqrt{3} + i.
  4. For k=1k=1: z1=2ei(π/2+2π3)=2ei5π6=2(−32+i12)=−3+iz_1 = 2e^{i\left(\frac{\pi/2 + 2\pi}{3}\right)} = 2e^{i\frac{5\pi}{6}} = 2\left(-\frac{\sqrt{3}}{2} + i\frac{1}{2}\right) = -\sqrt{3} + i.
  5. For k=2k=2: z2=2ei(π/2+4π3)=2ei9π6=2ei3π2=−2iz_2 = 2e^{i\left(\frac{\pi/2 + 4\pi}{3}\right)} = 2e^{i\frac{9\pi}{6}} = 2e^{i\frac{3\pi}{2}} = -2i.

Explanation:

To find nn-th roots, express the number in polar/Euler form and divide the argument by nn while adding multiples of 2π2\pi to find all distinct roots.

Problem 3:

Simplify (cos⁡θ+isin⁡θ)5(cos⁡ϕ+isin⁡ϕ)2\frac{(\cos \theta + i \sin \theta)^5}{(\cos \phi + i \sin \phi)^2} using De Moivre's Theorem.

Solution:

(cis θ)5(cis ϕ)2=cis(5θ)cis(2ϕ)\frac{(\text{cis } \theta)^5}{(\text{cis } \phi)^2} = \frac{\text{cis}(5\theta)}{\text{cis}(2\phi)} Using the property of division for complex numbers in polar form: cis(5θ−2ϕ)=cos⁡(5θ−2ϕ)+isin⁡(5θ−2ϕ)\text{cis}(5\theta - 2\phi) = \cos(5\theta - 2\phi) + i \sin(5\theta - 2\phi)

Explanation:

De Moivre's Theorem allows us to move powers into the argument of the cis function. Division then corresponds to the subtraction of arguments.