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)

JP
Answered by Jesse P. Further Mathematics tutor

5984 Views

See similar Further Mathematics A Level tutors

Related Further Mathematics A Level answers

All answers ▸

How to integrate ln(x)?


How do you find the determinant of a matrix?


The complex number -2sqrt(2) + 2sqrt(6)I can be expressed in the form r*exp(iTheta), where r>0 and -pi < theta <= pi. By using the form r*exp(iTheta) solve the equation z^5 = -2sqrt(2) + 2sqrt(6)i.


How do I solve x^2 + x - 6 > 0 ?


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 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences