# argument of product and quotient

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 cosine (http://planetmath.org/GoniometricFormulae) and sine (http://planetmath.org/GoniometricFormulae) we still obtain the formula

 $\displaystyle(\cos\varphi_{1}+i\sin\varphi_{1})(\cos\varphi_{2}+i\sin\varphi_{% 2})=\cos(\varphi_{1}+\varphi_{2})+i\sin(\varphi_{1}+\varphi_{2}).$ (1)

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 even (http://planetmath.org/EvenFunction) and the sine an odd function, we have

 $\displaystyle\frac{1}{\cos\varphi_{2}+i\sin\varphi_{2}}=\cos(-\varphi_{2})+i% \sin(-\varphi_{2}).$ (2)

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.

 $\displaystyle\frac{\cos\varphi_{1}+i\sin\varphi_{1}}{\cos\varphi_{2}+i\sin% \varphi_{2}}=\cos(\varphi_{1}-\varphi_{2})+i\sin(\varphi_{1}-\varphi_{2}).$ (3)
 $z_{1}=r_{1}(\cos\varphi_{1}+i\sin\varphi_{1})\;\;\mbox{and}\;\;z_{2}=r_{2}(% \cos\varphi_{2}+i\sin\varphi_{2})$

we have

 $z_{1}z_{2}=r_{1}r_{2}(\cos(\varphi_{1}+\varphi_{2})+i\sin(\varphi_{1}+\varphi_% {2})),$
 $\frac{z_{1}}{z_{2}}\;=\;\frac{r_{1}}{r_{2}}(\cos(\varphi_{1}-\varphi_{2})+i% \sin(\varphi_{1}-\varphi_{2})).$

Thus we have the

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 factors (http://planetmath.org/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.

Remark.  The equation (1) may be by induction generalised for more than two factors of the left hand ; then the special case where all factors are equal gives de Moivre identity  .

Example.  Since

 $(2\!+\!i)(3\!+\!i)=5\!+\!5i\;=\;5e^{\frac{\pi}{4}},$

one has

 $\arctan\frac{1}{2}+\arctan\frac{1}{3}\;=\;\frac{\pi}{4}.$
 Title argument of product and quotient Canonical name ArgumentOfProductAndQuotient Date of creation 2013-03-22 17:45:20 Last modified on 2013-03-22 17:45:20 Owner pahio (2872) Last modified by pahio (2872) Numerical id 8 Author pahio (2872) Entry type Theorem Classification msc 30-00 Classification msc 26A09 Synonym product and quotient of complex numbers Related topic Argument Related topic PolarCoordinates Related topic ModulusOfComplexNumber Related topic Complex Related topic EqualityOfComplexNumbers