Proof for law of complex exponents using only differential equation I just read that an elegant proof exists that the law of exponents also holds for complex numbers (a,b,z all complex): ea+b=eaeb, which only uses the definition that y=ezt is solution to dydt=zy, with initial condition y(0)=1, so in particular ez=y(1). I can only find a proofs which use the trig-representation of complex numbers.

Name: Proof for law of complex exponents using only differential equation I just read that an elegant proof exists that the law of exponents also holds for complex numbers (a,b,z all complex): ea+b=eaeb, which only uses the definition that y=ezt is solution to dydt=zy, with initial condition y(0)=1, so in particular ez=y(1). I can only find a proofs which use the trig-representation of complex numbers.
Uploaded: 2022-02-23T17:22:57+00:00
Description: Proof for law of complex exponents using only differential equation I just read that an elegant proof exists that the law of exponents also holds for complex numbers (a,b,z all complex): ea+b=eaeb, which only uses the definition that y=ezt is solution to dydt=zy, with initial condition y(0)=1, so in particular ez=y(1). I can only find a proofs which use the trig-representation of complex numbers.

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Proof for law of complex exponents using only differential equation  I just read that an elegant proof exists that the law of exponents also holds for complex numbers (a,b,z all complex):  ea+b=eaeb,  which only uses the definition that  y=ezt  is solution to dydt=zy,  with initial condition y(0)=1, so in particular ez=y(1).  I can only find a proofs which use the trig-representation of complex numbers.

eninsala06 · Accepted Answer

If you define ez as the unique solution to the ODE f′(z)=f(z) with initial condition f(0)=1, then you have by the product rule:(ezec−z)′=ezec−z+ez(−ec−z)=0Thus ezec−z is a constant. Using the initial condition e0=1 we find that ezec−z=ec. Now let z=a and c=a+b and c=a+b and the result follows.

Beverly Smith · Accepted Answer

Let g(z)=ea+zea. Then g′(z)=g(z) and g(0)=1. So g(z)=ez, and we have that ea+z=eaez.That assumes that ez is the only solution to f′(z)=f(z) with f(0)=1.

alenahelenash · Accepted Answer

Another way is to see that any f:C→C satisfying f′(z)=f(z) and f(0)=1 is analytic in C (entire) and admits a power series representation f(z)=∑n=0∞anznThe fact that f′(z)=f(z) and f(0)=1 easily give usf(z)=∑n=0∞znn!Now it is easy to verify that f indeed satisfies the above differential equation and initial conditions (and hence is the unique function) and that f(a+b)=f(a)f(b)

Proof for law of complex exponents using only differential equation I just read that an elegant

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