How do you show #e^(it) = cos t + i sin t# ?

The exponential of a real number x, written as #e^x# , is defined by an sum of infinite series, as follows

#e^x = ∑_(k=0) ^ ∞ (x^k/(k!)) = 1 + x + (x^ 2/(2!)) + (x^3 /(3!)) + ...#

Also #costheta# and #sintheta# can be expressed as sum of infinite series as follows

#cos(θ) = 1 - (θ^2/(2!)) + (θ^4 /(4!)) + ...#

#sin(θ) = θ - (θ^3/(3!)) + (θ^5/(5!)) + ...#

The exponential of a complex number z is written #e^z# , and is defined in the same way as the exponential of a real number

#e^z = ∑_(k=0) ^ ∞ (z^k/(k!)) = 1 + z + (z^ 2/(2!)) + (z^3 /(3!)) + ...#

Let set #z=i*theta# in the previous relation to get

#e^(i*theta) = ∑_(k=0) ^ ∞ ((i*theta)^k/(k!)) = 1 + (i*theta) + ((i*theta)^ 2/(2!)) + ((i*theta)^3 /(3!)) + ...= [1 - (θ^2/(2!)) + (θ^4/(4!)) - ...] + i[θ - (θ^3/(3!)) + (θ^5/(5!)) + ...]= =cos(θ) + i sin(θ)#

Thus the proof concluded.

The identity #e^(itheta)=costheta+isintheta# is known as Euler's formula

Consider a point on the Complex plane at #cos t + i sin t#. This will lie on the unit circle for any Real value of #t#.

Next suppose the point moves anticlockwise around the unit circle at a rate of #1# radian per second. That is #t# increases at a rate of #1# per second.

The rate of change of the position (i.e. velocity) at time #t# will be a tangential vector, perpendicular to the radius and will have length #1#.

Looking at the #x# and #y# (or Real and imaginary) components we see that #d/(dt) cos t = - sin t# and #d/(dt) sin t = cos t#

We can then use Taylor's theorem to derive the Maclaurin series for #cos t# and #sin t#.

#f(t) = sum_(k=0)^oo f^((k))(0)/(k!) t^n#

For example, in the case of #f(t) = cos t# we have:

#f^((0))(0) = cos(0) = 1#

#f^((1))(0) = -sin(0) = 0#

#f^((2))(0) = -cos(0) = -1#

#f^((3))(0) = sin(0) = 0#

#f^((4))(0) = cos(0) = 1#

etc.

Hence:

#cos t = sum_(k = 0)^oo (-1)^k/((2k)!) t^(2k) = 1 - t^2/(2!) + t^4/(4!) -...#

Similarly:

#sin t = sum_(k = 0)^oo (-1)^k/((2k+1)!) t^(2k+1) = t - t^3/(3!) + t^5/(5!) -...#

If we then define:

#e^z = sum_(k=0)^oo 1/(k!) z^n = 1 + z/(1!) + z^2/(2!) + z^3/(3!) +...#

we find:

#e^(i t) = cos t + i sin t#

as Konstantinos showed.