argument of product and quotientArgumentOfProductAndQuotient2008-01-23 13:34:502009-06-09 16:15:39Theoremcomplexmultiplicationdivision% this is the default PlanetMath preamble. as your knowledge
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Using the distributive law, we perform the multiplication
$$(\cos\varphi_1+i\sin\varphi_1)(\cos\varphi_2+i\sin\varphi_2)
= (\cos\varphi_1\cos\varphi_2-\sin\varphi_1\sin\varphi_2)+i(\sin\varphi_1\cos\varphi_2+\cos\varphi_1\sin\varphi_2).$$
Using the addition formulas of \PMlinkname{cosine}{GoniometricFormulae} and \PMlinkname{sine}{GoniometricFormulae} we still obtain the formula
\begin{align}
(\cos\varphi_1+i\sin\varphi_1)(\cos\varphi_2+i\sin\varphi_2)
= \cos(\varphi_1+\varphi_2)+i\sin(\varphi_1+\varphi_2).
\end{align}
The inverse number of\, $\cos\varphi_2+i\sin\varphi_2$\, is calculated as follows:
$$\frac{1}{\cos\varphi_2+i\sin\varphi_2} =
\frac{\cos\varphi_2-i\sin\varphi_2}{(\cos\varphi_2-i\sin\varphi_2)(\cos\varphi_2+i\sin\varphi_2)}
= \frac{\cos\varphi_2-i\sin\varphi_2}{\cos^2\varphi_2+\sin^2\varphi_2}$$
This equals\, $\cos\varphi_2-i\sin\varphi_2$,\, and since the cosine is an \PMlinkname{even}{EvenFunction} and the sine an odd function, we have
\begin{align}
\frac{1}{\cos\varphi_2+i\sin\varphi_2} = \cos(-\varphi_2)+i\sin(-\varphi_2).
\end{align}
The equations (1) and (2) imply
$$\frac{\cos\varphi_1+i\sin\varphi_1}{\cos\varphi_2+i\sin\varphi_2} =
(\cos\varphi_1+i\sin\varphi_1)(\cos(-\varphi_2)+i\sin(-\varphi_2))=
\cos(\varphi_1+(-\varphi_2))+i\sin(\varphi_1+(-\varphi_2)),$$
i.e.
\begin{align}
\frac{\cos\varphi_1+i\sin\varphi_1}{\cos\varphi_2+i\sin\varphi_2} =
\cos(\varphi_1-\varphi_2)+i\sin(\varphi_1-\varphi_2).
\end{align}
According to the formulae (1) and (3), for the complex numbers
$$z_1 = r_1(\cos\varphi_1+i\sin\varphi_1)\;\;\mbox{and}\;\;z_2 = r_2(\cos\varphi_2+i\sin\varphi_2)$$
we have
$$z_1z_2 = r_1r_2(\cos(\varphi_1+\varphi_2)+i\sin(\varphi_1+\varphi_2)),$$
$$\frac{z_1}{z_2} \;=\; \frac{r_1}{r_2}(\cos(\varphi_1-\varphi_2)+i\sin(\varphi_1-\varphi_2)).$$
Thus we have the
\textbf{Theorem.}\; The modulus of the product of two complex numbers equals the product of the moduli of the factors and the argument equals the sum of the arguments of the \PMlinkname{factors}{Product}.\; The modulus of the quotient of two complex numbers equals the quotient of the moduli of the dividend and the divisor and the argument equals the difference of the arguments of the dividend and the divisor.\\
\textbf{Remark.}\, The equation (1) may be by induction generalised for more than two factors of the left hand \PMlinkescapetext{side}; then the special case where all factors are equal gives de Moivre identity.\\
\textbf{Example.}\, Since
$$(2\!+\!i)(3\!+\!i) = 5\!+\!5i \;=\; 5e^{\frac{\pi}{4}},$$
one has
$$\arctan\frac{1}{2}+\arctan\frac{1}{3} \;=\; \frac{\pi}{4}.$$