Complex Number Formulas - Cheatsheet | Last Minute Notes

Introduction to Complex Numbers

Complex numbers are an extension of the real number system and play a crucial role in various branches of mathematics and engineering. They are numbers of the form $z = a + bi$, where $a$ and $b$ are real numbers, and $i$ is the imaginary unit defined as $i^2 = -1$. The real part ($a$) represents the number's horizontal component, while the imaginary part ($b$) represents its vertical component. Complex numbers provide a comprehensive framework for solving equations that have no solutions within the set of real numbers, such as $x^2 + 1 = 0$.

Alzebra of Complex Number

Algebra of complex numbers deals with the basic operations that can be performed on complex numbers. A complex number is a number of the form $a + bi$, where $a$ is the real part, and $bi$ is the imaginary part, with $b$ being the imaginary coefficient and $i$ being the imaginary unit such that $i ^ 2 = -1$.

The algebra of complex numbers involves several operations:

Addition

To add two complex numbers $z_1 = a + bi$ and $z_2 = c + di$, add the real parts and the imaginary parts separately:

$z_1 + z_2 = (a + c) + (b + d)i$

Subtraction

To subtract two complex numbers $z_1 = a + bi$ and $z_2 = c + di$, subtract the real parts and the imaginary parts separately:

$z_1 - z_2 = (a - c) + (b - d)i$

Multiplication

To multiply two complex numbers $z_1 = a + bi$ and $z_2 = c + di$, use the distributive property and simplify:

$z_1 \cdot z_2 = (a + bi)(c + di) = ac + adi + bci + bdi^2$

$z_1 \cdot z_2 = (ac - bd) + (ad + bc)i$

Division

To divide two complex numbers $z_1 = a + bi$ and $z_2 = c + di$, multiply the numerator and denominator by the conjugate of $z_2$:

$z_1 / z_2 = \frac{a + bi}{c + di} \cdot \frac{c - di}{c - di}$

$z_1 / z_2 = \frac{(a + bi)(c - di)}{(c + di)(c - di)} = \frac{(ac + bd) + (bc - ad)i}{c^2 + d^2}$

$z_1 / z_2 = \frac{ac + bd}{c^2 + d^2} + \frac{bc - ad}{c^2 + d^2}i$

Conjugate of a Complex Number

The conjugate of a complex number is obtained by changing the sign of the imaginary part while keeping the real part unchanged. If $z = x + iy$, then the conjugate is denoted as $\overline{z}$, and is given by:

The complex conjugate plays a significant role in simplifying expressions involving complex numbers and in calculating the modulus. It is also useful when performing division involving complex numbers.

Properties of Complex Conjugates

• $\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}$: The conjugate of the sum of two complex numbers is equal to the sum of their conjugates.
• $\overline{z_1 z_2} = \overline{z_1} \cdot \overline{z_2}$: The conjugate of the product of two complex numbers is equal to the product of their conjugates.
• $\overline{\overline{z}} = z$: The conjugate of the conjugate of a complex number is the number itself.
• $\overline{\frac{z_1}{z_2}} = \frac{\overline{z_1}}{\overline{z_2}}$: The conjugate of the quotient of two complex numbers is equal to the quotient of their conjugates.
• $|\overline{z}| = |z|$: The modulus (or absolute value) of a complex number and its conjugate are equal.

Modulus of a Complex Number

The modulus of a complex number is the distance of the number from the origin in the complex plane. If $z = x + iy$, then the modulus $|z|$ is calculated as:

This is essentially the length of the vector that represents the complex number in the complex plane. The modulus is important in many operations, such as dividing complex numbers and converting to polar form.

Example: Let $z = 3 + 4i$. The modulus is calculated as:

Properties of Modulus

1. $|z|^2 = z\overline{z}$ - The square of the modulus equals the product of a complex number and its conjugate
2. $|z| = 0 \iff z = 0$ - The modulus is zero if and only if the complex number is zero
3. $\frac{1}{z} = \frac{\overline{z}}{|z|^2}$ if $z \neq 0$ - Reciprocal property
4. $|z_1z_2| = |z_1||z_2|$ - The modulus of a product equals the product of the moduli
5. $|\frac{z_1}{z_2}| = \frac{|z_1|}{|z_2|}$ if $z_2 \neq 0$ - The modulus of a quotient equals the quotient of the moduli
6. $-|z| \leq z \leq |z|$ - Bounds property
7. $|z_1 + z_2|^2 = |z_1|^2 + |z_2|^2 + z_1\overline{z_2} + \overline{z_1}z_2$
   $= |z_1|^2 + |z_2|^2 + 2\Re(z_1\overline{z_2})$ - Squared sum property

Note: These properties are fundamental in complex analysis and are frequently used in solving problems involving complex numbers.

Example Questions - Conjugate and Modulus of Complex Number

Question 1:
Let $Z = x + iy$ and $\omega = \frac{1-iZ}{Z-1}$. If $|\omega| = 1$, show that Z is purely real.

Solution:
Let $Z = x + iy$
$|\omega| = 1$
$\Rightarrow |\omega|^2 = 1$
$\Rightarrow \omega \cdot \overline{\omega} = 1$
$\omega = \frac{1 - iZ}{Z - 1}$
$\overline{\omega} = \frac{\overline{1 - iZ}}{\overline{Z - 1}} = \frac{1 + i\overline{Z}}{\overline{Z} - 1}$
$\Rightarrow \frac{1 - iZ}{Z - 1} \cdot \frac{1 + i\overline{Z}}{\overline{Z} - 1} = 1$
$\Rightarrow \frac{(1 - iZ)(1 + i\overline{Z})}{(Z - 1)(\overline{Z} - 1)} = 1$
Numerator: $(1 - iZ)(1 + i\overline{Z}) = 1 - Z\overline{Z}$
Denominator: $(Z - 1)(\overline{Z} - 1) = |Z - 1|^2$
$\Rightarrow \frac{1 - Z\overline{Z}}{|Z - 1|^2} = 1$
$\Rightarrow 1 - Z\overline{Z} = |Z - 1|^2$
$\Rightarrow 1 - (x^2 + y^2) = (x - 1)^2 + y^2$
Expand: $1 - x^2 - y^2 = x^2 - 2x + 1 + y^2$
$\Rightarrow 0 = 2x - 2$
$\Rightarrow x = 1$
Substitute $x = 1$: $y^2 = 0$
$\therefore Z = x$, Z is purely real.

Question 2:
If $|Z| = 1$, $Z \neq -1$, show that $\frac{Z-1}{Z+1}$ is purely imaginary.

Solution:
Let $Z = x + iy$
$|Z| = 1$
$\Rightarrow x^2 + y^2 = 1$
$\frac{Z-1}{Z+1} = a + ib$
$\Rightarrow (Z-1) = (a + ib)(Z + 1)$
Substitute $Z = x + iy$:
$(x - 1 + iy) = (a + ib)(x + 1 + iy)$
Expand:
$(x - 1) + iy = ax + a + iby + ib$
Separate real and imaginary parts:
Real: $x - 1 = ax + a - by$
Imaginary: $y = ay + bx + b$
Solve for $a$:
$a = 0$
$\therefore \frac{Z-1}{Z+1}$ is purely imaginary.

Question 3:
If $|z - i| = |z + i|$, show that $\Im(z) = 0$.

Solution:
Let $z = x + iy$
$|z - i| = |z + i|$
$\Rightarrow |x + i(y - 1)| = |x + i(y + 1)|$
Squaring both sides:
$x^2 + (y - 1)^2 = x^2 + (y + 1)^2$
Expand:
$x^2 + y^2 - 2y + 1 = x^2 + y^2 + 2y + 1$
Cancel terms:
$-2y = 2y$
$\Rightarrow y = 0$
$\therefore \Im(z) = 0$.

Question 4:
If $(a + bi)(3 + i) = (1 + i)(2 + i)$, find a and b.

Solution:
Expand right-hand side:
$(1 + i)(2 + i) = 2 + i + 2i + i^2 = 1 + 3i$
Expand left-hand side:
$(a + bi)(3 + i) = 3a + ai + 3bi + bi^2 = (3a - b) + (a + 3b)i$
Compare real and imaginary parts:
Real: $3a - b = 1$
Imaginary: $a + 3b = 3$
Solve the system of equations:
$a = 1, b = 2$.

Question 5:
If $(\cos \theta + i \sin \theta)^2 = x + iy$, show that $x^2 + y^2 = 1$.

Solution:
Apply De Moivre’s theorem:
$(\cos \theta + i \sin \theta)^2 = \cos 2\theta + i \sin 2\theta$
$x = \cos 2\theta$, $y = \sin 2\theta$
Use trigonometric identity:
$\cos^2 2\theta + \sin^2 2\theta = 1$
$\Rightarrow x^2 + y^2 = 1$.

Argument of a Complex Number

The argument of a complex number represents the angle between the positive real axis and the line representing the complex number in the complex plane. If $z = x + iy$, the argument $\theta$ is given by:

The argument is measured in radians, and it is important for converting complex numbers to polar form and performing various operations such as multiplication and division.

Example: Let $z = 3 + 4i$. The argument is:

Polar Form of a Complex Number

The polar form of a complex number expresses it in terms of its modulus and argument. If $z = x + iy$, then the polar form is given by:

The polar form is especially useful for multiplying and dividing complex numbers, as well as raising them to powers.

Example: Let $z = 3 + 4i$. First, calculate the modulus $r = 5$ and the argument $\theta = \tan^{-1}\left(\frac{4}{3}\right)$. The polar form is then:

Multiplication of Complex Numbers in Polar Form

Multiplying complex numbers in polar form involves multiplying their magnitudes and adding their arguments. If $z_1 = r_1 (\cos \theta_1 + i \sin \theta_1)$ and $z_2 = r_2 (\cos \theta_2 + i \sin \theta_2)$, then their product is:

This formula allows us to quickly multiply complex numbers when they are in polar form, avoiding the need to multiply and simplify their real and imaginary parts separately.

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 the product is:

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 the three solutions to the equation $x^3 = 1$. These roots are:

These are the cube roots of unity, where $1$ is the real cube root, and $\omega$ and $\omega^2$ are the complex cube roots.

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

$\omega = -\frac{1}{2} + i \frac{\sqrt{3}}{2}$
Compute $\omega^3$:
$\omega^3 = \left(-\frac{1}{2} + i \frac{\sqrt{3}}{2}\right)^3$
Expanding this expression, we simplify it to $1$.

This verifies that $\omega^3 = 1$, meaning $\omega$ is indeed a cube root of unity.

Properties of Cube Roots of Unity:

• $1 + \omega + \omega^2 = 0$: This property holds because the sum of the cube roots of unity is zero. These roots form an equilateral triangle in the complex plane, and their sum lies at the origin.
• $\omega^3 = 1$: This is the fundamental property of the cube roots of unity. Any power of $\omega$ that is a multiple of 3 will be 1, i.e., $\omega^3 = 1$, $\omega^6 = 1$, and so on.
• $\omega^2 \cdot \omega = \omega^3 = 1$: This shows that multiplying any cube root of unity by another gives 1, as $\omega^2 \cdot \omega = 1$.

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

Substitute the values of $1$, $\omega$, and $\omega^2$ into the equation:

This verifies that $1 + \omega + \omega^2 = 0$ is true.

Additional Properties: Cube roots of unity also satisfy the following properties:

• $\omega^2 = \overline{\omega}$: The complex conjugate of $\omega$ is $\omega^2$.
• $1 + \omega + \omega^2 = 0$: As mentioned earlier, this is a fundamental identity and is often used in various proofs and factorizations.

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.