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)

Related Further Mathematics A Level answers

All answers ▸

Are we able to represent linear matrix transformations with complex numbers?


Prove that (AB)^-1 = B^-1 A^-1


What is the general solution to the equation d2y/dx2 + dy/dx - 2y = -3sinx + cosx (d2y/dx2 signals a second order derivative)


How do I prove that the differential of coshx is equal to sinhx?


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2024

Terms & Conditions|Privacy Policy