Cheat Sheet
General Information
Imaginary Unit
\[\begin{align*} i &= \sqrt{-1} \\ i^2 &= -1 \\ i^3 &= -i \\ i^4 &= 1 \\ i^5 &= i \\ i^{k+4n} &= i^k \\ \end{align*} \]Cartesian/Arithmetic/Algebraic Form
\[z=a+bi \]Where \(a,b \in \mathbb{R}\) are the real and imaginary parts of the complex number:
- The real part, \(a=Re(z)=\Re(z)\).
- And the imaginary part, \(b=Im(z)=\Im(z)\).
Set of Complex Numbers
\[\mathbb{C}=\{z|z=a+bi\text{ mit }a,b \in \mathbb{R}\} \]Equality of Complex Numbers
Complex numbers are equal if and only if their real and imaginary parts are equal.
\[a+bi=c+di \iff a=c \land b=d \]Complex/Gaussian Plane
A unique mapping and identification of complex numbers as points.
\[f: \mathbb{C} \to \mathbb{R}^2, a+bi \mapsto (a,b) \]Magnitude/Modulus of a Complex Number
The modules/magnitude of a complex number is the distance of the point from the origin, i.e. the length of the vector.
\[r=|z|=|a+bi|=\sqrt{z\overline{z}}=\sqrt{a^2+b^2} \]Argument of a Complex Number is the angle of the vector with respect to the x-axis.
\[\varphi = \arg(z) = \arg(a+bi) = \begin{cases} \arctan(\frac{b}{a}) &a > 0 \\ \pi + \arctan(\frac{b}{a}) &a < 0 \\ \frac{\pi}{2} &a=0,b > 0 \\ -\frac{\pi}{2} &a=0,b < 0 \\ \text{undefined} &a=0,b=0 \end{cases} \]Alternatively, but less common:
\[\varphi=arg(z)=\arg(a+bi) = \begin{cases} \arccos(\frac{a}{r}) &\text{für } b \geq 0 \\ -\arccos(\frac{a}{r}) &\text{für } b < 0 \end{cases} \]Polar Form
Using the absolute value and argument, a complex number can also be represented in the polar coordinate system.
Trigonometric/Goniometric Form
From the conversion from Cartesian to polar coordinates, we notice that we can also uniquely write and identify complex numbers in another way:
\[z=r(\cos(\varphi) + i \sin(\varphi))=r \text{ cis}(\varphi) \]where \(r=|z| \in \mathbb{R}_0^+\) and \(\varphi=\arg(z) \in [0,2\pi)\).
Exponential/Euler Form
From the power series of \(e^x,\sin(x)\) and \(\cos(x)\), we can form the Euler form:
\[e^{ix}=\cos(x)+i\sin(x)=\text{cis}(x) \]Operations
Negation
Negation corresponds to reflection across the origin in the complex plane. If \(z= a+bi\) then
\[-z=-a-bi \]Complex Conjugation
Complex conjugation is negation of only the imaginary part and corresponds to reflection across the x-axis. If \(z=a+bi\) then the komplexe Konjugation is
\[\overline{z}=a-bi \]Addition/Subtraction
Addition/subtraction is best done in Cartesian form, where the two vectors are added/subtracted.
\[s=z_1 + z_2=(a+bi)\pm (c+di)=(a \pm c) + i(c\pm d) \]Multiplication
Multiplication can be done in all forms, but it is easiest and fastest in the trigonometric or Euler form. In Cartesian form, multiplication becomes complicated with more than two factors.
- Cartesian form: \(z_1 \cdot z_2=(a+bi)(c+di)=(ac-bd) + i(ad+bc)\)
- Trigonometric form: \(z_1 \cdot z_2 = (r_1 \text{cis}(\varphi_1))(r_2 \text{cis}(\varphi_2))=r_1r_2\text{cis}(\varphi_1 + \varphi_2)\)
- Euler form: \(z_1 \cdot z_2 = r_1 e^{i\varphi_1} \cdot r_2 e^{i\varphi_2}=r_1 r_2 e^{i(\varphi_1 + \varphi_2)}\)
Exponentiation
Using the binomial formula, we can raise complex numbers to a power using Cartesian form, but because multiplication is so much easier in the other forms, exponentiation is also much easier in those forms.
- Cartesian form: \(z^n=(a +bi)^n = \sum_{k=0}^{n}{\binom{n}{k} a^{n-k}(bi)^k}\)
- Trigonometric form: Using De Moivre’s theorem \(z^n= (r \text{ cis}(\varphi))^n=r^n \text{ cis}(n\varphi)\)
- Euler form: \(z^n=(re^{i\varphi})^n=r^n e^{in\varphi}\)
Division
Division can also be done in all forms, but like multiplication, it is best not to use Cartesian form.
- Cartesian form: \(\frac{z_1}{z_2}=\frac{a+bi}{c+di}=\frac{ac+bd}{c^2+d^2} +i \frac{bc-ad}{c^2+d^2}\)
- Trigonometric form: \(\frac{z_1}{z_2}=\frac{r_1 \text{cis}(\varphi_1)}{r_2 \text{cis}(\varphi_2)}=\frac{r_1}{r_2} \text{cis}(\varphi_1 - \varphi_2)\)
- Euler form: \(\frac{z_1}{z_2}=\frac{r_1 e^{i\varphi_1}}{r_2 e^{i\varphi_2}}=\frac{r_1}{r_2} e^{i(\varphi_1 - \varphi_2)}\)
Roots of Complex Numbers
The process of taking roots is more complicated with complex numbers than with real numbers. Fortunately, taking roots is the inverse function of raising to a power, and we can convert roots to powers as follows: \(\sqrt[4]{x}=(x)^{\frac{1}{4}}\). Thus, we can proceed as we did with real numbers. However, when raising to a power, we did not have to take into account the periodicity of the cosine and sine functions, but when taking roots, we must do so because \(\cos(\frac{\varphi+2\pi k}{n})\) for \(k=0,1,...,n-1\) gives exactly \(n\) different values. Of course, we can also convert the formulas below into trigonometric form.
\[\sqrt[n]{re^{i\varphi}}=\sqrt[n]{r}(\cos(\frac{\varphi}{n})+i \sin(\frac{\varphi}{n}))=\sqrt[n]{r}e^{i\frac{\varphi}{n}} \]To calculate all solutions, we use the following formula
\[z_k=\sqrt[n]{r} e^{i(\frac{\varphi}{n}+\frac{2\pi k}{n})} \]where \(k=0,1,...n-1\) and we call the solution \(z_0\) the principal value of the root.
Logarithm
Thanks to the exponential/euler form, we can also take logarithms.
\[\ln(z)=\ln(re^{i\varphi})=\ln(r)+i\varphi \]If the natural logarithm is not used, we can make a base change at the end.
\[\log_2(2e^{i\frac{7\pi}{4}})=\frac{\ln(2e^{i\frac{7\pi}{4}})}{\ln(2)}=1+i\frac{\frac{7\pi}{4}}{ln(2)} \]Complex Numbers in the Exponent
We can also solve problems where the complex number is used as an exponent.
\[e^{1+\pi i}=e^1e^{\pi i}=e e^{(\pi i)} \] \[2^i=e^{\ln(2^i)}=e^{i \ln(2)} \]