Starting from Eulers formula derive the formula of de moivre

Starting from Euler\'s formula, derive the formula of de moivre: (cos theta + i sin theta)^n = (cos n theta + i sin n theta) and, n squareroot (cos theta + i sin theta) = (cos + i sin )

Solution

(3) De Movires formula deriver from Euler\'s formula

e^{ix} = cos x + i sinx
the exponential law for integer powers

[e^{ix}]n = e^{inx}
Then, by Euler\'s formula,

e^i(nx) = cos (nx) + i sin(nx)
Arrange the values

(cosx + i sinx)^2 = cos^2 x + 2isin xcos x-sin^2 x = (cos^2 x - sin^2 x) + i(2 sinx cosx)
= cos(2x) + i sin (2x)

final equality follows

cos^2 x - sin^2 x = cos(2x)
2sinx cosx = sin(2x) .
Proved

(4) a cos 2theta = cos^2 theta - sin^2 theta

put theta = ix

we get cos(2ix) = cos^2 (ix) - sin^2(ix)

cosh 2x = (cosh x)^2 - (i sinh x)^2

cosh 2x = cosh^2 x + sinh^2 x

(b) sin 2x = 2sinx cosx

we have to proof

so the same rule put theta = ix

we get sin(2ix) = 2 sin(ix) cos(ix)

or we can say that isinh 2x = 2 .sinh x .cosh x

hence sin 2x = 2sinx cosx

Proved