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Complex Numbers Cheat Sheet

Real and Imaginary parts

Let $z = x + iy$ for $x, y \in \mathbb{R}$. $x$ is the real part of $z$ and denoted by $\Re(z)$. $y$ is the imaginary part of $z$ and denoted by $\Im(z)$.

The modulus is defined as $\abs{z} = \sqrt{x^2 + y^2}$.

Polar notation

$z = r e^{i \theta} = r (\cos \theta + i \sin \theta)$, for $r = \abs{z}$ and $\theta = \tan^{-1}(y/x)$. $\theta$ is called the argument of $z$ and denoted by $\mbox{arg}(z)$.

Conjugate

Let $z = a + ib$ for $a, b \in \mathbb{R}$. The conjugate, denoted by $\overline{z}$ is defined as $z = a - ib$. Use \overline{z} instead of \bar{z} in LaTeX.

Identities

Real (LaTex \Re) and imaginary (LaTeX \Im) parts:

\[\Re(z) = \frac{z + \overline{z}}{2}, \qquad \Im(z) = \frac{z - \overline{z}}{2i}\]

Modulus:

\[\abs{z}^2 = z \overline{z}\]

Euler’s identity:

\[e^{i\pi} + 1 = 0\]

Properties

Conjugate is invariant with the arithmetic operations of addition, subtraction, multiplication and division:

\[\begin{align} \overline{a + b} &= \overline{a} + \overline{b} \\ \overline{a - b} &= \overline{a} - \overline{b} \\ \overline{a * b} &= \overline{a} * \overline{b} \\ \overline{a / b} &= \overline{a} / \overline{b} \end{align}\]

Modulus

Properties

Triangle inequality.

\[\abs{a + b} \le \abs{a} + \abs{b}\]

Strict triangle inequality. If

\[\abs{a + b} = \abs{a} + \abs{b}\]

then $a = kb$ for some real $k$.

This is trivial for $a = 0$ or $b = 0$, so assume $a, b \ne 0$. Consider the square of the left side and use that $\abs{z}^2 = z \overline{z}$ $$ \abs{a + b}^2 = (a + b)(\overline{a + b}) = (a + b)(\overline{a} + \overline{b}) = \abs{a}^2 + \abs{b}^2 + a\overline{b} + \overline{a}b $$ We have that $a\overline{b} + \overline{a}b = a\overline{b} + \overline{a\overline{b}}$, so using $2\Re(z) = z + \overline{z}$, we have: $$ \abs{a + b}^2 = \abs{a}^2 + \abs{b}^2 + 2\Re(a\overline{b}) $$ expanding and using $\abs{z} = z \overline{z}$: $$ (a + b)(\overline{a + b}) = \abs{a}^2 + \abs{b}^2 + 2 $$ Now squaring the right side: $$(\abs{a} + \abs{b})^2 = \abs{a}^2 + \abs{b}^2 + 2\abs{a}\abs{b}$$ which implies that $$\abs{a}\abs{b} = \Re(a\overline{b})$$ we have that $\abs{a}\abs{b} = \abs{a}\abs{\overline{b}} = \abs{a\overline(b)}$ and thus: $$\abs{a\overline(b)} = \Re(a\overline{b})$$ which implies $a\overline(b)$ is real. Let $t = a\overline(b)$, so that: $$ a = \frac{t}{\overline(b)} $$ multiply the fraction by $b/b$ to obtain: $$ a = \frac{t}{\abs{b}^2} b $$ $k = t / \abs{b}^2$ is real.

Multiplication.

\[\abs{a \cdot b} = \abs{a} \cdot \abs{b}\]

Division.

If $b \ne 0$:

\[\abs{\frac{a}{b}} = \frac{\abs{a}}{\abs{b}}\]

Exponential. Let $x$ be a real value and $z$ a complex number. Then:

\[\abs{x^z} = x^{\Re(z)}\] Rebase $x^z$ as $e^{z \log x}$ and let $z = a + i b$ for $a, b \in \mathbb{R}$. Then: $$x^z = e^{a \log x} e^{i b \log x}$$ Apply modulus: $$\abs{x^z} = \abs{e^{a \log x}} \abs{e^{i b \log x}}$$ The second term is a complex number in polar notation with $r = 1$ and $\theta = b \log x$, so $\abs{e^{i b \log x}} = 1$ and hence: $$\abs{x^z} = \abs{e^{a \log x}}$$ Since all terms are reals, $$\abs{e^{a \log x}} = e^{a \log x}$$ Rebasing back: $$= x^{a} = x^{\Re(z)}$$