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.

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