4) Suppose r1 and r2 are two complex numbers. Using Euler’s formula exp(α + iβ) =
exp(α)(cos(β) + isin(β)) where α, β are real, show that: (e^r1)(e^r2) = e^(r1+r2)
exp(α)(cos(β) + isin(β)) where α, β are real, show that: (e^r1)(e^r2) = e^(r1+r2)
-
Let r1 = x + yi and r2 = v + wi, where x, y, v, and w are real numbers.
The idea is to use Euler's formula, multiply out, and use the sum identities for sine and cosine.
(e^r1)(e^r2) = (e^(x+yi))(e^(v+wi))
= (e^x)(cos y + i sin y)(e^v)(cos w + i sin w)
= (e^x)(e^v)(cos y + i sin y)(cos w + i sin w)
= (e^(x+v))(cos y cos w + i cos y sin w + i sin y cos w + i^2 sin y sin w)
= (e^(x+v))(cos y cos w - sin y sin w + i(cos y sin w + sin y cos w))
= (e^(x+v))(cos(y+w) + i sin(y+w))
= e^[(x+v) + i(y+w)]
= e^(x+yi+v+wi)
= e^(r1+r2).
Lord bless you today!
The idea is to use Euler's formula, multiply out, and use the sum identities for sine and cosine.
(e^r1)(e^r2) = (e^(x+yi))(e^(v+wi))
= (e^x)(cos y + i sin y)(e^v)(cos w + i sin w)
= (e^x)(e^v)(cos y + i sin y)(cos w + i sin w)
= (e^(x+v))(cos y cos w + i cos y sin w + i sin y cos w + i^2 sin y sin w)
= (e^(x+v))(cos y cos w - sin y sin w + i(cos y sin w + sin y cos w))
= (e^(x+v))(cos(y+w) + i sin(y+w))
= e^[(x+v) + i(y+w)]
= e^(x+yi+v+wi)
= e^(r1+r2).
Lord bless you today!