Complex Number Formulas - Cheatsheet | Last Minute Notes

Conjugate of a Complex Number: If $z = x + iy$, then the conjugate $\overline{z}$ is $x - iy$.

Example: Let $z = 3 + 4i$. The conjugate is $\overline{z} = 3 - 4i$.

Modulus of a Complex Number: If $z = x + iy$, then the modulus $|z|$ is $\sqrt{x^2 + y^2}$.

Example: Let $z = 3 + 4i$. The modulus is $|z| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5$.

Argument of a Complex Number: If $z = x + iy$, then the argument $\theta$ is given by $\theta = \tan^{-1}\left(\frac{y}{x}\right)$.

Example: Let $z = 3 + 4i$. The argument is $\theta = \tan^{-1}\left(\frac{4}{3}\right)$.

Polar Form of a Complex Number: $z = r(\cos \theta + i \sin \theta)$, where $r = |z|$ and $\theta$ is the argument.

Example: Let $z = 3 + 4i$. The polar form is $z = 5(\cos(\tan^{-1}\frac{4}{3}) + i \sin(\tan^{-1}\frac{4}{3}))$.

Multiplication of Complex Numbers in Polar Form:
If $z_1 = r_1 (\cos \theta_1 + i \sin \theta_1)$ and $z_2 = r_2 (\cos \theta_2 + i \sin \theta_2)$,
$z_1 z_2 = r_1 r_2 \left[\cos(\theta_1 + \theta_2) + i \sin(\theta_1 + \theta_2)\right]$.

Example: Let $z_1 = 2(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})$ and $z_2 = 3(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6})$. Then $z_1 z_2 = 6(\cos \left(\frac{\pi}{4} + \frac{\pi}{6}\right) + i \sin \left(\frac{\pi}{4} + \frac{\pi}{6}\right))$.

De Moivre's Theorem

For any integer $n$,
$(r(\cos \theta + i \sin \theta))^n = r^n (\cos(n \theta) + i \sin(n \theta))$.

Example Question

Question: Use De Moivre's Theorem to find $(1 + i)^4$.

Solution:
Step 1: Express $1 + i$ in polar form.
$z = 1 + i$ can be written as $r(\cos \theta + i \sin \theta)$, where $r = |z|$ and $\theta = \arg(z)$.
Calculate $r$:
$r = \sqrt{1^2 + 1^2} = \sqrt{2}$.
Calculate $\theta$:
$\theta = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4}$.
Therefore, $1 + i = \sqrt{2} \left( \cos\left(\frac{\pi}{4}\right) + i \sin\left(\frac{\pi}{4}\right) \right)$.

Step 2: Apply De Moivre's Theorem with $r = \sqrt{2}$, $\theta = \frac{\pi}{4}$, and $n = 4$.
$(1 + i)^4 = \left(\sqrt{2} \left( \cos\left(\frac{\pi}{4}\right) + i \sin\left(\frac{\pi}{4}\right) \right)\right)^4$.

Step 3: Calculate the magnitude and argument.
$(\sqrt{2})^4 = 4$.
$4 \left( \cos\left(4 \cdot \frac{\pi}{4}\right) + i \sin\left(4 \cdot \frac{\pi}{4}\right) \right)$.

Step 4: Simplify the argument.
$4 \left( \cos(\pi) + i \sin(\pi) \right)$.
$4 \left( -1 + 0i \right) = -4$.

Therefore, $(1 + i)^4 = -4$.

Cube Roots of Unity

The cube roots of unity are $1, \omega, \omega^2$, where $\omega = -\frac{1}{2} + i \frac{\sqrt{3}}{2}$ and $\omega^2 = -\frac{1}{2} - i \frac{\sqrt{3}}{2}$.

Example: Consider the cube roots of unity. Verify the value of $\omega$:

$\omega = -\frac{1}{2} + i \frac{\sqrt{3}}{2}$
Compute $\omega^3$:
$\omega^3 = \left(-\frac{1}{2} + i \frac{\sqrt{3}}{2}\right)^3$
This simplifies to $1$.

Properties of Cube Roots of Unity:

• $1 + \omega + \omega^2 = 0$
• $\omega^3 = 1$
• $\omega^2 \cdot \omega = \omega^3 = 1$

Example: Verify the property $1 + \omega + \omega^2 = 0$:

Substitute the values:
$1 + \left(-\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) + \left(-\frac{1}{2} - i \frac{\sqrt{3}}{2}\right)$
Simplify:
$1 - \frac{1}{2} - \frac{1}{2} + i \frac{\sqrt{3}}{2} - i \frac{\sqrt{3}}{2}$
This results in:
$0$.

Trigonometric Identities involving Complex Numbers

Sum of Angles:
$\cos(A + B) = \cos A \cos B - \sin A \sin B$
$\sin(A + B) = \sin A \cos B + \cos A \sin B$.

Double Angle Formulas:
$\cos(2A) = \cos^2 A - \sin^2 A$
$\sin(2A) = 2 \sin A \cos A$.

Triple Angle Formulas:
$\cos(3A) = 4\cos^3 A - 3\cos A$
$\sin(3A) = 3\sin A - 4\sin^3 A$.

Example Question

Question: Verify the identity $\cos(3A) = 4\cos^3 A - 3\cos A$ using De Moivre's Theorem.

Solution:
Step 1: Express $z = \cos A + i \sin A$ in exponential form using Euler's formula:
$z = e^{iA}$.

Step 2: Use De Moivre's Theorem to find $z^3$:
$z^3 = (e^{iA})^3 = e^{i(3A)}$.

Step 3: Convert back to trigonometric form:
$e^{i(3A)} = \cos(3A) + i \sin(3A)$.

Step 4: Expand $z^3$ using binomial expansion:
$z^3 = (\cos A + i \sin A)^3$
$= \cos^3 A + 3i \cos^2 A \sin A - 3 \cos A \sin^2 A - i \sin^3 A$
$= (\cos^3 A - 3 \cos A \sin^2 A) + i (3 \cos^2 A \sin A - \sin^3 A)$.

Step 5: Equate the real parts:
$\cos(3A) = \cos^3 A - 3 \cos A \sin^2 A$.
Using the Pythagorean identity $\sin^2 A = 1 - \cos^2 A$:
$\cos(3A) = \cos^3 A - 3 \cos A (1 - \cos^2 A)$
$= \cos^3 A - 3 \cos A + 3 \cos^3 A$
$= 4 \cos^3 A - 3 \cos A$.

Therefore, the identity $\cos(3A) = 4 \cos^3 A - 3 \cos A$ is verified.