To show by means of *phasor analysis* that Eq.(A.2) always has a solution, we can express each component sinusoid as

re

Equation (A.2) therefore becomes

Thus, equality holds when we define

Since is just the polar representation of a complex number, there is always some value of and such that equals whatever complex number results on the right-hand side of Eq.(A.5).

As is often the case, we see that the use of Euler's identity and
complex analysis gives a simplified *algebraic* proof which
replaces a proof based on trigonometric identities.

[How to cite this work] [Order a printed hardcopy] [Comment on this page via email]

Copyright ©

Center for Computer Research in Music and Acoustics (CCRMA), Stanford University