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argument of product and quotient∗
pahio†
2013-03-22 0:16:14
Using the distributive law, we perform the multiplication
(cos ϕ1 +i sin ϕ1 )(cos ϕ2 +i sin ϕ2 ) = (cos ϕ1 cos ϕ2 −sin ϕ1 sin ϕ2 )+i(sin ϕ1 cos ϕ2 +cos ϕ1 sin ϕ2 ).
Using the addition formulas of cosine and sine we still obtain the formula
(cos ϕ1 + i sin ϕ1 )(cos ϕ2 + i sin ϕ2 ) = cos(ϕ1 + ϕ2 ) + i sin(ϕ1 + ϕ2 ).
(1)
The inverse number of cos ϕ2 + i sin ϕ2 is calculated as follows:
cos ϕ2 − i sin ϕ2
1
cos ϕ2 − i sin ϕ2
=
=
cos ϕ2 + i sin ϕ2
(cos ϕ2 − i sin ϕ2 )(cos ϕ2 + i sin ϕ2 )
cos2 ϕ2 + sin2 ϕ2
This equals cos ϕ2 − i sin ϕ2 , and since the cosine is an even and the sine an
odd function, we have
1
= cos(−ϕ2 ) + i sin(−ϕ2 ).
cos ϕ2 + i sin ϕ2
(2)
The equations (1) and (2) imply
cos ϕ1 + i sin ϕ1
= (cos ϕ1 +i sin ϕ1 )(cos(−ϕ2 )+i sin(−ϕ2 )) = cos(ϕ1 +(−ϕ2 ))+i sin(ϕ1 +(−ϕ2 )),
cos ϕ2 + i sin ϕ2
i.e.
cos ϕ1 + i sin ϕ1
= cos(ϕ1 − ϕ2 ) + i sin(ϕ1 − ϕ2 ).
cos ϕ2 + i sin ϕ2
(3)
According to the formulae (1) and (3), for the complex numbers
z1 = r1 (cos ϕ1 + i sin ϕ1 ) and z2 = r2 (cos ϕ2 + i sin ϕ2 )
we have
z1 z2 = r1 r2 (cos(ϕ1 + ϕ2 ) + i sin(ϕ1 + ϕ2 )),
∗ hArgumentOfProductAndQuotienti
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1
z1
r1
=
(cos(ϕ1 − ϕ2 ) + i sin(ϕ1 − ϕ2 )).
z2
r2
Thus we have the
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 factors. 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 side; then the special case where all factors are equal
gives de Moivre identity.
Example. Since
π
(2+i)(3+i) = 5+5i = 5e 4 ,
one has
arctan
1
π
1
+ arctan = .
2
3
4
2