Komplekse tall på polar form kan skrives på to måter:
z = \textcolor{red}{r}e^{í \textcolor{blue}{\theta}} \\ \textnormal{ og } \\
z = \textcolor{red}{r}\Big(\cos(\textcolor{blue}{\theta}) + i \sin(\textcolor{blue}{\theta})\Big)- $i = \sqrt{-1}$ er imaginær enhet
- $\textcolor{red}{r} = |z|$ er avstanden fra origo og kalles modulus til $z$
- $\textcolor{blue}{\theta}$ er vinkelen i radianer og kalles argumentet til $z$
+ Eksempel: $z = 2e^{i \pi/3}$
Gitt et kompleks tall:
z = \textcolor{red}{2} e^{\textcolor{blue}{\frac{\pi}{3}}i}Modulus til $z$ (dvs. avstanden fra origo) er:
\textcolor{red}{r} = |z| = \textcolor{red}{2}Argumentet til $z$ (dvs. vinkelen i radianer) er:
\textcolor{blue}{\theta} = \arg(z) = \textcolor{blue}{\frac{\pi}{3}}+ De to måtene å skrive komplekse tall på polar form
Sammenhengen mellom de to måtene å skrive komplekse tall er gitt ved Eulers formel:
e^{\theta i} = \cos \theta + i \sin \thetaNoen ganger kalles $z = e^{\theta i}$ for eksponentiell form og $z = r(\cos \theta + i \sin \theta)$ for polar form. Men siden begge inneholder $r$ og $\theta$ er de begge på polar form.
En måte å vise sammenhengen er å bruke definisjonene av cosinus og sinus:
\cos \theta = \frac{e^{\theta i} + e^{-\theta i}}{2} \qquad \textnormal{ og } \qquad
\sin \theta = \frac{e^{\theta i} - e^{-\theta i}}{2i} \\
\begin{aligned} \\
\Rightarrow \quad \cos \theta + i \sin \theta & = \frac{e^{\theta i} + e^{-\theta i}}{2} + \cancel{i} \cdot \frac{e^{\theta i} - e^{-\theta i}}{2\cancel{i}} \\
\Rightarrow \quad \cos \theta + i \sin \theta & = \frac{e^{\theta i} + e^{-\theta i} + e^{\theta i} - e^{-\theta i}}{2} \\
\Rightarrow \quad \cos \theta + i \sin \theta & = \frac{e^{\theta i} + \cancel{e^{-\theta i}} + e^{\theta i} - \cancel{e^{-\theta i}}}{2} \\
\Rightarrow \quad \cos \theta + i \sin \theta & = \frac{2e^{\theta i}}{2} \\
\Rightarrow \quad \cos \theta + i \sin \theta & = e^{\theta i}
\end{aligned}Og, vips, har vi vist sammenhengen.
+ Hva er den kompleks konjugerte?
Hvis vi vil finne den kompleks konjugerte til $z = re^{i \theta}$, bytter vi bare fortegn foran $i$:
\overline{z} = \overline{\textcolor{red}{r} e^{\textcolor{blue}{\theta}i}} = \textcolor{red}{r} e^{\textcolor{blue}{-\theta}i}Streken over betyr bare at vi vil ha den kompleks konjugerte av det som står under.
Eksempler:
\begin{aligned}
\overline{\textcolor{red}{2} e^{\textcolor{blue}{\frac{\pi}{3}}i} } = \textcolor{red}{2} e^{\textcolor{blue}{-\frac{\pi}{3}}i} \\
\overline{\textcolor{red}{3} e^{\textcolor{blue}{-\frac{\pi}{4}}i} } = \textcolor{red}{3} e^{\textcolor{blue}{\frac{\pi}{4}}i} \\
\end{aligned}En kul greie:
z\overline{z} = \textcolor{red}{r}e^{\textcolor{blue}{\theta}i} \textcolor{red}{r}e^{\textcolor{blue}{-\theta}i} = \textcolor{red}{r^2} e^{(\textcolor{blue}{\theta - \theta})i} = \textcolor{red}{r^2} e^0 = \textcolor{red}{r^2}Eksempel hvis $z = \textcolor{red}{2}e^{\textcolor{blue}{\frac{\pi}{3}}i}$:
z \overline{z} = \textcolor{red}{2}e^{\textcolor{blue}{\frac{\pi}{3}}i} \cdot \textcolor{red}{2}e^{\textcolor{blue}{-\frac{\pi}{3}}i} = \textcolor{red}{2^2} e^{\left(\textcolor{blue}{\frac{\pi}{3} - \frac{\pi}{3}} \right)i} = \textcolor{red}{2^2} e^0 = \textcolor{red}{2^2} = 4 + Multiplikasjon av komplekse tall
Gitt to komplekse tall:
z_1 = \textcolor{red}{r_1} e^{\textcolor{blue}{\theta_1}i} \\
z_2 = \textcolor{red}{r_2} e^{\textcolor{blue}{\theta_2}i}Multiplikasjon ved hjelp av litt potensregler:
z_1 \cdot z_2 = \textcolor{red}{r_1} e^{\textcolor{blue}{\theta_1}i} \cdot \textcolor{red}{r_2} e^{\textcolor{blue}{\theta_2}i}
= \textcolor{red}{r_1r_2} e^{(\textcolor{blue}{\theta_1} + \textcolor{blue}{\theta_2})i}Eksempel:
\textcolor{red}{2}e^{\textcolor{blue}{\frac{\pi}{3}}i} \cdot \textcolor{red}{3}e^{\textcolor{blue}{\frac{\pi}{4}}i}
= \textcolor{red}{2} \cdot \textcolor{red}{3} e^{\left(\textcolor{blue}{\frac{\pi}{3}} + \textcolor{blue}{\frac{\pi}{4}} \right) i}
= 6e^{\frac{7\pi}{12}i}
Og, vips, er vi ferdige!
+ Divisjon av komplekse tall
Gitt to komplekse tall:
z_1 = \textcolor{red}{r_1} e^{i \textcolor{blue}{\theta_1}} \\
z_2 = \textcolor{red}{r_2} e^{i \textcolor{blue}{\theta_2}}Når vi vil dele to komplekse tall på hverandre, kan vi bruke potensregler:
\frac{z_1}{z_2} = \frac{\textcolor{red}{r_1}e^{\textcolor{blue}{\theta_1}i}}{\textcolor{red}{r_2} e^{ \textcolor{blue}{\theta_2}i}}
= \frac{\textcolor{red}{r_1} e^{\textcolor{blue}{\theta_1}i} e^{\textcolor{blue}{-\theta_2} i}}{\textcolor{red}{r_2}}
= \frac{\textcolor{red}{r_1} e^{(\textcolor{blue}{\theta_1 - \theta_2})i}}{\textcolor{red}{r_2}}
= \frac{\textcolor{red}{r_1}}{\textcolor{red}{r_2}} e^{(\textcolor{blue}{\theta_1 - \theta_2})i}Eksempel:
\frac{\textcolor{red}{2} e^{\textcolor{blue}{\frac{\pi}{3}}i} }{\textcolor{red}{3} e^{\textcolor{blue}{\frac{\pi}{4}}i} }
= \frac{\textcolor{red}{2} e^{\textcolor{blue}{\frac{\pi}{3}}i} \cdot e^{\textcolor{blue}{-\frac{\pi}{4}}i}}{\textcolor{red}{3}} = \frac{\textcolor{red}{2}}{\textcolor{red}{3}} e^{\left(\textcolor{blue}{\frac{\pi}{3} - \frac{\pi}{4}} \right) i}
= \frac{2}{3} e^{\frac{\pi}{12}i} Og, vips, er vi ferdige!
+ de Moivres formel
de Moivres formel sier at:
(\cos \theta + i \sin \theta)^{\textcolor{red}{n}} = \cos (\textcolor{red}{n} \theta) + i \sin (\textcolor{red}{n} \theta)fordi:
(\cos \theta + i \sin \theta)^{\textcolor{red}{n}} = (e^{\theta i})^{\textcolor{red}{n}} = e^{\textcolor{red}{n} \theta i} = \cos (\textcolor{red}{n} \theta) + i \sin (\textcolor{red}{n} \theta)Og, vips, har vi vist de Moivres formel.
+ Potenser av komplekse tall
Hvis $z=re^{\theta i}$ kan vi bruke litt potensregler:
z^{\textcolor{red}{n}} = \left( re^{\theta i} \right)^{\textcolor{red}{n}} = r^{\textcolor{red}{n}} e^{\textcolor{red}{n} \theta i}Eksempel: Hvis $z = 2e^{3i}$ blir
z^{\textcolor{red}{4}} = \left( 2e^{3 i} \right)^{\textcolor{red}{4}} = 2^{\textcolor{red}{4}} e^{\textcolor{red}{4} \cdot 3i} = 16e^{12i}Og, vips, er vi ferdige!