Die Multiplikation komplexer Zahlen in den Polarkoordinaten.

Es sei

\begin{array}{rcl} z_{1} & = r_{1} (cos ϕ_{1} + i sin ϕ_{1}) & = r_{1} e^{i ϕ_{1}}, \\ z_{2} & = r_{2} (cos ϕ_{2} + i sin ϕ_{2}) & = r_{2} e^{i ϕ_{2}} . \end{array}

Dann gilt nach Definition der Multiplikation und der Additionstheoreme für die Winkelfunktionen

\begin{array}{rcl} z_{1} z_{2} & = & r_{1} (cos ϕ_{1} + i sin ϕ_{1}) r_{2} (cos ϕ_{2} + i sin ϕ_{2}) \\ = & r_{1} r_{2} (\underset{cos (ϕ_{1} + ϕ_{2})}{\underset{︸}{(cos ϕ_{1} cos ϕ_{2} - sin ϕ_{1} sin ϕ_{2})}} + i \underset{sin (ϕ_{1} + ϕ_{2})}{\underset{︸}{(cos ϕ_{1} sin ϕ_{2} + cos ϕ_{2} sin ϕ_{1})}} \\ = & R (cos ψ + i sin ψ), R = r_{1} r_{2}, ψ = ϕ_{1} + ϕ_{2} . \end{array}

Das heisst

\begin{array}{rcl} | z_{1} z_{2} | & = & | z_{1} | \cdot | z_{2} |, & (1.41) \\ arg z_{1} z_{2} & = & arg z_{1} + arg z_{2} mod 2 π . & (1.42) \end{array}

Problem 1.11.4. Es seien $z_{1}, z_{2}$ wie oben gegeben, wobei $z_{2} \neq 0 .$ Zeigen Sie, da

z_{2}^{- 1} = r_{2}^{- 1} (cos (- ϕ_{2}) + i sin (- ϕ_{2})) = r_{2}^{- 1} e^{- i ϕ_{2}}

und folglich $\begin{array}{rcl} \frac{z_{1}}{z_{2}} & = & \frac{r_{1}}{r_{2}} (cos (ϕ_{1} - ϕ_{2}) + i sin (ϕ_{1} - ϕ_{2})) \\ = & \tilde{R} e^{i \tilde{ψ}}, R = r_{1} r_{2}^{- 1}, \tilde{ψ} = ϕ_{1} - ϕ_{2}, \end{array}$

oder mit anderen Worten $\begin{array}{rcl} | z_{1} ∕ z_{2} | & = & | z_{1} | ∕ | z_{2} |, \\ arg z_{1} z_{2}^{- 1} & = & arg z_{1} - arg z_{2} mod 2 π . \end{array}$

1.11.4 Die Multiplikation komplexer Zahlen in den Polarkoordinaten.