Why does e^ix = cos(x) + isin(x)

If you look at the Taylor series expansion of ex: ex = 1 + x + x2/(2!) + x3/(3!) + x4/(4!)...

If you then make this eix, you get

1+ix - x2/(2!) - ix3/(3!) + x4/(4!)...

If we split this into real and imaginary, we see the real part is

1 - x2/(2!) + x4/(4!) - x6/(6!)...

The series expansion of cos(x) is

1 - x2/(2!) + x4/(4!) - x6/(6!)...

Therefore, the real part of eix is cos(x)

If we look at only the imaginary part of eix, we get

i(x - x3/(3!) + x5/(5!) - x/ (7!)

If we look at the series expansion of sin(x) we get

(x - x3/(3!) + x5/(5!) - x/ (7!)

Therefore the imaginary part of eix = sin(x)

Putting this together, we get eix = Re(eix) + Im(eix)

eix = cos(x) + isin(x)

Jesse P. A Level Maths tutor, A Level Further Mathematics  tutor, GCS...

1 year ago

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