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## Main Question or Discussion Point

Hi All and sorry if this is too easy a question but here goes....

Sines, Cosines and the rest of the trig functions are the ratio of two lengths and thus are dimensionless quantities.

That is if I plug in a value for t in sin(ωt) there are no units.

For example the solution of

x'' + ω^2 x = 0 with x(0) = x0 and x'(0) = v0 is given by

x(t) = x0 cos(ωt) + v0/ω sin(ωt)

The units come from the initial conditions not the sine or cosine.

So here is the question...

Same is true ( I believe ) when using the natural exponential function exp(t).

How does one simply explain this.

I tried to reason it out using eulers formula exp(iω) = cos(ω) + i sin(ω) figuring that again we get ratios of lengths,

however in the case where the real part is non-zero we get another exponential (which is not the ratio of lengths)

exp(a +ib) = exp(a)(cos(b) + i sin(b))

Is there a simple explaination as to why we "lose the units" when using the exponential function?

Sines, Cosines and the rest of the trig functions are the ratio of two lengths and thus are dimensionless quantities.

That is if I plug in a value for t in sin(ωt) there are no units.

For example the solution of

x'' + ω^2 x = 0 with x(0) = x0 and x'(0) = v0 is given by

x(t) = x0 cos(ωt) + v0/ω sin(ωt)

The units come from the initial conditions not the sine or cosine.

So here is the question...

Same is true ( I believe ) when using the natural exponential function exp(t).

How does one simply explain this.

I tried to reason it out using eulers formula exp(iω) = cos(ω) + i sin(ω) figuring that again we get ratios of lengths,

however in the case where the real part is non-zero we get another exponential (which is not the ratio of lengths)

exp(a +ib) = exp(a)(cos(b) + i sin(b))

Is there a simple explaination as to why we "lose the units" when using the exponential function?